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Section - 2
HESS’S LAW AND ITS APPLICATIONS
Let us consider a simple example before we study Hess’s law. Suppose we want to find the enthalpy change for the combustion of graphite (carbon) to carbon monoxide. i.e. we want to calculate ∆H for the following reaction:
2C (graphite)+O2 ( g ) → 2CO( g ).
...(1)
We may argue that we could let “one mole of this reaction” take place at constant pressure and measure the accompanying heat change. This heat change would be the required enthalpy change. Theoretically, our argument is correct but our argument is based on an assumption that all the graphite is converted to only carbon monoxide. However, this does not happen in practice. What happens in practice is that once carbon monoxide starts forming it reacts further with oxygen to yield carbon dioxide. In fact, it we perform the experiment in an excess of oxygen, we will obtain only CO2 and so the enthalpy change will be heat of complete combustion of graphite, i.e.
C (graphite) + O2 ( g ) → CO2 ( g )
...(2)
On the other hand if we perform the experiment in a limited quantity of oxygen, we obtain a mixture of CO and
CO2 , and the accompanying heat of reaction will be a value corresponding to a mixture of these products. So, either way we won’t be able to get the enthalpy change for the reaction (1). So, how can we obtain the enthalpy change for the preparation of pure carbon monoxide from graphite and oxygen ? This problem can be resolved by Hess’s law. Let us study it and see how it can solve such problems. We know that enthalpy is a state function. This means that enthalpy change for a chemical reaction is independent of the path by which the products are obtained. Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps. In other words, no matter how we go from reactants to products (whether one step or several), the enthalpy change for the overall chemical change is the same. Having stated the Hess’s law, let us see, if we can apply it to find the enthalpy change for reaction (1) above. Let us suppose that the combustion of graphite to carbon monoxide takes place in two separate steps :
2C (graphite) + 2O2 ( g ) → 2CO2 ( g ) first step 2CO2 → 2CO( g ) + O2 ( g )
second step
In the first step, we burn 2 moles of graphite in 2 moles of oxygen to produce 2 moles of CO2 . In the second step we decompose this CO2 to give 2 moles of CO and 1 mole of O2 . The net result is the combustion of 2 moles of graphite in 1 mole of oxygen to give 2 moles CO. We can obtain this result by adding the two steps, cancelling out 2 moles of CO2 and 1 mole of O2 on both sides of the Equation : 2C (graphite) + 2 O2 ( g ) → 2 CO2 ( g ) 2CO2 ( g )
→ 2CO( g ) + O2 ( g )
2C (graphite) + O2 ( g ) → 2CO( g ) Chemistry/Thermodynamics
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According to Hess’s law, the enthalpy change for the overall equation (which is the equation we want) equals the sum of the enthalpy changes for the two steps. So, if we can find the enthalpy changes for the two steps; their sum would give us the required enthalpy. We can determine the enthalpy change for the first step by simply burning graphite in an excess of oxygen. The result (found experimentally) is ∆H = −393.5 kJ per mole of CO2 formed. For 2 moles of CO2 , we multiply by 2. 2C ( graphite ) + 2O2 ( g ) → 2CO2 ( g ) ; ∆H = ( −393.5 kJ ) × 2. The second step, decomposition of CO2, is not an easy experiment. However, the reverse of this decomposition is simply the combustion of CO. We could determine the ∆H for that combustion by burning CO in an excess of oxygen. (In fact, the reaction is similar to the one for the combustion of graphite to CO2 . ) 2CO( g ) + O2 ( g ) → 2CO2 ( g ); ∆H = −566.0kJ From the properties of thermochemical equations we know that the enthalpy change for the reverse reaction is simply (–1) times the original reaction. 2CO2 ( g ) → 2CO( g ) + O2 ( g ); ∆H = (−566.0kJ ) × (−1) If we now add these two steps and all their enthalpy changes, we obtain the chemical equation and the enthalpy change for the combustion of CO, which is what we wanted
2C
(graphite) + 2O2 ( g ) → 2CO2 ( g ) → 2CO( g ) + O2 ( g )
2CO2 ( g ) 2C
(graphite) + O2 ( g ) → 2CO( g )
∆H1 = (−393.5kJ ) × (2) ∆H 2 = (−566.0kJ ) × (−1) ∆H 3 = −221.0kJ
So, we see that the combustion of 2 moles of graphite to give 2 moles of CO has an enthalpy change of –221.0 kJ. The figure below shows the “enthalpy diagram” showing the relationship among the enthalpy changes for this calculation: 2 mol C (graphite) + 2 mol O2 (g)
Enthalpy (k J)
2 mol CO(g) + 1 mol O2(g)
∆H2 = + 566.0 kJ
∆H1 = – 787.0 kJ
∆H3 = –221.0 kJ
2 mol CO2(g) fig.10: Enthalpy diagram illustrating Hess's law. The diagram shows the equality of the enthalpy change for the complete combustion of graphite to the sum of the enthalpy changes for the combustion of graphite to CO and the combustion of CO to CO2.
The above example illustrated how we can use Hess’s law to obtain the enthalpy change for a reaction that is difficult to determine by direct experiment. Chemistry/Thermodynamics
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Hess’s law is more generally useful, however, in that it allows us to calculate the enthalpy change for one reaction from the values for others, whatever their source. (We will illustrate this in the examples that follow.) Important points 1. Hess’s law is a consequence of the fact that enthalpy is a state function. 2. We can calculate ∆H for any process, as long as we find a route for which ∆H is known for each step. This important fact permits us to use a relatively small number of experimental measurements to calculate ∆H for a vast number of different reactions( we will illustrate this as we move ahead in this chapter.) 3. Will the final value of ∆H for a reaction depend on the way in which we break it down to use Hess’s law? Because H is a state function, we will always get the same value of ∆H for an overall reaction, regardless of how many steps we employ to get to the final products. 4. It should be noted that in applying Hess’s law, the individual steps need not be realisable in practice. They may be hypothetical reactions and the only requirement is that their chemical equations should balance. Example – 6 Suppose we are given the following data : S ( s ) + O2 ( g ) → SO2 ( g ); ∆H = −297 kJ
...(1)
2 SO3 ( g ) → 2SO2 ( g ) + O2 ( g ); ∆H = +198kJ
...(2)
How could we use these data to obtain the enthalpy change for the following equation ? 2 S ( s ) + 3O2 ( g ) → 2SO3 ( g )
...(3)
Critical thinking We need to multiply equations (1) and (2) by factors (perhaps reversing one or both equations) so that when we add them together we obtain equation (3). We can usually guess what we need to do to the first two equations to obtain the third one. Note that Equation (3) has a coefficient of 2 for S(s). This suggests that we should multiply equation (1) by 2 (and multiply the ∆H by 2). Note also that SO3 ( g ) in equation (3) is on the RHS. This suggests that we should reverse equation (2) (and multiply the ∆H by -1) Solution: Following the process described above, the whole problem boils down to this : + 2O2 ( g ) → 2 SO2 ( g )
∆H = ( −297 kJ ) × (2)
2SO2 ( g ) + O2 ( g ) → 2SO3 ( g )
∆H = (198kJ ) × ( −1)
2S (s)
2S (s)
+ 3O2 ( g ) → 2SO3 ( g )
∆H = −792kJ
The next example gives another illustration of how Hess’s law can be used to calculate the enthalpy change for a reaction from the enthalpy values for other reactions. In this case, the problem involves three equations from which we obtain a fourth. Although this problem is somewhat complicated than the one we just did, the basic procedure is the same : Compare the coefficients in the equations. See by what factors we need to multiply the equations whose ∆H ′ s we know to obtain the equation we want. Chemistry/Thermodynamics
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Example – 7 What is the enthalpy of reaction, ∆H , for the formation of tungsten carbide, WC, from the elements ? W ( s ) + C (graphite) → WC ( s).
(The enthalpy change for this reaction is difficult to measure directly, because the reaction occurs at 1400°C. However, the heats of combustion of the elements and of Tungsten Carbide can be measured easily and is given by :) (1)
2W ( s ) + 3O2 ( g ) → 2WO3 ( s ); ∆H = −1680.6kJ
(2)
C (graphite) + O2 ( g ) → CO2 ( g ); ∆H = −393.5kJ
(3)
2WC ( s ) + 5O2 ( g ) → 2WO3 ( s ) + 2CO2 ( g ); ∆H = −2391.6kJ
Critical thinking We need to multiply equation (1), (2) and (3) by factors so that when we add the three equations we obtain the desired equation for the formation of WC(s). To obtain these factors compare equations (1), (2) and (3) in turn with the desired equation. For instance, note that Equation (1) has 2W(s) on the left side, whereas the desired equation has W(s). Therefore, we multiply equation (1) (and its ∆H ) by 1 2 .
Solution: Multiplying Equation (1) by 1 2 , we obtain 3 W ( s ) + O2 ( g ) → WO3 ( s); ∆H = 1 × ( −1680.6kJ ) = – 840.3 kJ 2 2
Compare Equation (2) with the desired equation. Both have C(graphite) on the left side; therefore, we leave equation (2) as it is. Now, compare equation (3) with the desired equation. Equation (3) has WC(s) on the left side, whereas the desired equation has WC(s) on the right side. Hence we reverse equation (3) and multiply it (and its ∆H ) by 1 2 . 5 WO3 ( s ) + CO2 ( g ) → WC ( s ) + O2 ( g ); ∆H = − 1 × ( −2391.6kJ ) = 1195.8kJ 2 2
Chemistry/Thermodynamics
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Note that the ∆H is obtained by multiplying the value for equation (3) by − 1 2 . Now these three equations and the corresponding ∆H ′s are added together. W (s) +
3 O2 ( g ) → WO3 ( s ) ; 2
∆ H = − 840.3kJ
C (graphite) + O2 ( g ) → CO2 ( g ) ; WO3 ( s ) + CO2 ( g ) → WC ( s ) +
5 O2 ( s ) ; 2
W ( s ) + C (graphite) → WC ( s )
∆ H = − 393.5 kJ ∆ H = + 1195.8 kJ ∆ H = − 38.0 kJ
To give you more practice for mastering how to manipulate thermochemical equations in applying Hess’s law we include one more example with only its critical thinking. You are requested to provide the solution yourself. (The answer is given.)
Example – 8 Use thermochemical equations given below to determine ∆H (at 25°C) for the following reaction :
C (graphite) + 2 H 2 ( g ) → CH 4 ( g ) Given:
C (graphite) + O2 ( g ) → CO2 ( g ) ; ∆H = −395.3kJ
...(1)
1 H 2 ( g ) + O2 ( g ) → H 2O (l ) 2
...(2)
; ∆H = −285.8kJ
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2O(l ) ; ∆H = −890.3 kJ
Chemistry/Thermodynamics
...(3)
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Critical thinking 1.
We want one mole of C (graphite) as reactant, so we write down equation (1) as it is.
2.
We want two moles of H 2 ( g ) as reactants, so we multiply equation (2) by 2.
3.
We want one mole of CH 4 ( g ) as product, so we reverse equation (3).
4.
We do the same operations on each ∆H value.
5.
Then we add these equations term by term. The result is the desired thermochemical equation; all unwanted substances cancel. The sum of the ∆H values is the ∆H for the desired reaction.
Answer: C (graphite)+2 H 2 ( g ) → CH 4 ( g )
∆H = −74.8kJ
(You are requested to provide the solution yourself) STANDARD STATE Before we delve into standard enthalpies of formation, we will have to study and understand clearly, so as to what does the standard state of a substance means and why is it needed. We have seen earlier in this chapter that the magnitude of enthalpy change of any reaction depends upon the state (gas, liquid or solid) of reactants and products. However, this is not the only factor on which enthalpy change depends. It depends also upon the conditions of temperature and pressure of the reactants and products. So, we can make out from this that depending upon various varying conditions we will get different enthalpy changes for the same reaction. To understand so as to what problem this poses in our study of reactions, consider two arbitrary reactions. Suppose ∆H for the first reaction at 2 bar, 30°C is –150kJ and that for the second reaction at 5 bar , 200°C is –250 kJ. Then can we say that the second reaction is more exothermic than the first one ? The answer is NO. We can say that a certain reaction is more ( or less)exothermic (or endothermic) than another reaction only when we are comparing their ∆H's under same conditions of temperature and pressure. So, this means that, in order to compare the enthalpy change of different reactions we need to have a certain “standard” condition of temperature and pressure. The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data : 1 bar (1× 105 Pa ) and a specified temperature ( if the temperature is not specified we assume it to be 25°C.) So, this should mean that the physical state of a substance at a pressure of 1 bar and a specified temperature is the standard state of that substance at that particular temperature. But we still have an anomaly to deal with. Many substances exist in the same physical state (gas, liquid or solid) in two or more distinct forms. For example, oxygen in any of the physical states occurs both as dioxygen (commonly called simply oxygen.) with O2 molecules, and as ozone, with O3 molecules. So, at a specified temperature and pressure of 1 bar, which one of the two should be considered as the standard state of oxygen ? We resolve this anomaly as follows : Chemistry/Thermodynamics
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Oxygen with its two different forms under same conditions is said to exist in different allotropic forms. An allotrope is one of two or more distinct forms of an element in the same physical state. Further, we define reference form of an element as the stablest form (physical state and allotrope) of the element under standard thermodynamic conditions. With these two terms defined we have all that we need to define standard state without any ambiguity. If at a particular temperature and pressure of 1 bar, a substance exists in two or more allotropic forms, we choose the stablest (i.e. reference form) of them as the standard state of that substance at that particular temperature. So, now, we can say that the standard state of a substance is the physical state of the substance (at 1 bar and a specified temperature) in its reference form. Reference form of some substances (at 25°C) are mentioned here : Substance Hydrogen Oxygen Carbon Sulphur
Reference form H2, gas O2, gas C, graphite S, rhombic.
Important Points 1.
It is a general misconception to confuse standard thermodynamic conditions with STP for gases. They are not identical.
2.
Another widespread misconception is regarding the temperature in standard thermodynamic conditions. Please read the definition and note that although the standard pressure has been fixed at 1 bar, no fixed temperature is mentioned. So, substances can have various standard states at different temperatures. For example, standard state of H 2O at –5°C is solid; at 5°C it is liquid and at 105°C it is gas.
3.
However, if no temperature is mentioned, we assume it to be 25°C.
4.
The standard enthalpy change, ∆H °reaction (abbreviated generally as ∆H °r ), for a reaction, reactants → products refers to the ∆H when the specified no. of moles of reactants, all at standard states, are converted completely to the specified number of moles of products, all at standard states. We allow the reaction to take place, with changes in temperature or pressure if necessary; when the reaction is complete, we return the products to the same conditions of temperature and pressure that we started with, keeping track of energy or enthalpy changes as we do so. When we describe a process as taking place “at constant T and P”, we mean that the initial and final conditions are the same. Because we are dealing with state functions, the net change is the same as the change we would have obtained hypothetically with T and P held constant.
Chemistry/Thermodynamics
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STANDARD ENTHALPIES OF FORMATION If a compound is formed from its constituent elements, then the enthalpy change for the reaction is called enthalpy of formation, ∆H f . For example,
C (diamond) + O2 ( g ) → CO2 ( g );
∆H1 = −395.4 kJ
∆H1 is the enthalpy of formation of CO2 here as it is being formed from the elements that constitute it. But the following reaction gives enthalpy of formation of CO2 as well :
C (graphite) + O2 ( g ) → CO2 ( g );
∆H 2 = −393.5kJ
Under same thermodynamic conditions we can see that ∆H1 ≠ ∆H 2 . Besides for a “formation reaction” ( i.e. a reaction where a substance is formed from its constituent elements), the enthalpy of formation will vary with changing number moles of the reactants. To resolve these anomalies, we define standard enthalpy of formation : The standard enthalpy of formation (also called the standard heat of formation) of a substance is the enthalpy change for the formation of one mole of the substance in a specified state from its elements in their reference form and in their standard states. It is denoted by ∆H °f . To understand this definition, consider the standard enthalpy of formation of liquid water. Note that the stablest forms of hydrogen and oxygen at 1 atm and 25°C are H 2 ( g ) and O2 ( g ), respectively. These are therefore the reference forms of the elements. We can write the formation reaction for liquid water as follows :
2 H 2 ( g ) + O2 ( g ) → 2 H 2O(l ); ∆H = −571.6kJ The standard enthalpy change for this reaction is –571.6 kJ per two moles of H 2O formed. But according to the definition of ∆H °f we need the enthalpy change for only 1 mole of the substance being formed. So, we divide the thermochemical equation above by 2 to get : H 2 ( g ) + 1 O2 ( g ) → H 2O(l ); ∆H °f = −285.8kJ 2 So, the standard enthalpy of formation of liquid water at 25°C is –285.8 kJ. The standard enthalpies of formation of the elements in their reference state are zero, by definition. Therefore
∆H °f of pure graphite is zero. Note that the standard enthalpy of formation of an element will depend upon the form of the element. For example, the ∆H °f for diamond equals the enthalpy change from the stablest form of carbon (graphite) to diamond. The thermochemical equation is :
C (graphite) → C (diamond); ∆H °f = 1.9kJ Chemistry/Thermodynamics
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On the other hand, the ∆H °f for graphite equals zero. Here, we list ∆H °f of some compounds and elements. You are not expected to memorise this table. But it is expected that you study this table and at least memorise the reference states of elements (when there ∆H °f = 0). It will be useful later. (These are marked in bold italics.) Please also note that the ∆H °f of an electron and H + (aq ) are zero, by definition.
Standard Enthalpies of formation (at 25° C) Formula e–(g) H+ (aq) H(g) H2(g) Na(g) Na(s) NaCl (s) Ca(s) C(g) C(graphite) C(diamond) Si (s) Pb (s) N(g) N2(g) O(g) O2(g) O3(g) H2O(g) H2O (l) S(g) S2(g)
∆H °f ( kJ / mole) 0 0 218.0 0 107.8 0 -411.1 0 715.0 0 1.9 0 0 473 0 249.2 0 143 -241.8 -285.8 279 129
Formula S8 (rhombic) S8(monoclinic) F2(g) Cl2(g) Br2(g) Br2(l) I2(g) I2(s) B(β-rhombohedral) Sn (gray) Sn (white) P(g) P(red) P(white) P2(g) P4(g) Cu(s) Cu(g) Hg(g) Hg(l) Fe(s) Fe(l)
∆H °f ( kJ / mole) 0 2 0 0 30.91 0 62.44 0 0 3 0 333.9 0 69.8 146.2 128.9 0 341.1 61.30 0 0 13.13
Listing out standard enthalpies of formation, actually helps us calculate standard enthalpy of any reaction. Let us see how to use standard enthalpies of formation to find the standard enthalpy change for a reaction. We will first look at this problem from the point of view of Hess’ law. But when we are done we will note a pattern in the result, which will allow us to state a simple formula for solving this type of problem.
Chemistry/Thermodynamics
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Consider the equation
CH 4 ( g ) + 4Cl2 ( g ) → CCl4 (l ) + 4 HCl ( g );
∆H ° = ?
The enthalpies of formation of CH 4 ( g ), CCl4 (l ) and HCl ( g ) are :
C (graphite) + 2 H 2 ( g ) → CH 4 ( g );
∆H °f = −74.9 kJ
...(1)
C (graphite) + 2Cl2 ( g ) → CCl4 ( g );
∆H °f = −139 kJ
...(2)
1 H ( g ) + 1 Cl ( g ) → HCl ( g ); 2 2 2 2
∆H °f = −92.3 kJ
...(3)
We now apply Hess’s law. Since we want CH 4 to appear on the left and CCl4 and 4HCl on the right, we reverse equation (1) and add equation (2) and 4 × Equation (3). CH 4 ( g ) → C (graphite) + 2 H 2 ( g ) C (graphite) + 2Cl2 ( g ) → CCl4 (l ) 2Cl2 ( g ) + 2 H 2 ( g ) → 4 HCl ( g ) CH 4 ( g ) + 4Cl2 ( g )
→
CCl4 (l ) + 4 HCl ( g )
(−74.9 kJ ) × (−1) (−139 kJ ) × (1) (−92.3 kJ ) × (4) ∆H ° = −433 kJ
The set up of this calculation can be greatly simplified if we closely examine what we are doing. Note that the ∆H °f of each compound has been multiplied by its coefficient in the chemical equation whose ∆H ° we are calculating. Moreover, the ∆H °f for each reactant is multiplied by a negative sign. For example, the coefficient of HCl in the original equation is 4 and we see that we have multiplied the ∆H °f of HCl by 4 itself. We can symbolise the enthalpy of formation of a substance by writing the formula in parantheses following ∆H °f . Then our calculation of ∆H ° can be written as follows : ∆H ° = ∆H °f (CCl4 ) + 4.∆H °f ( HCl ) − ∆H °f (CH 4 ) + 4.∆H °f (Cl2 ) = [(−139)
+ 4.(−92.3) ] kJ − [(−74.9) + 4.(0)] kJ
= −433 kJ
Extrapolating on this observation, in general, we can calculate the ∆H ° for a reaction by the equation.
∆H ° = ∑ n ⋅ ∆H °f (products) − ∑ m ⋅ ∆H °f (reactants) Here “Σ” represents “sum of” and m and n are the coefficients of the substances in the chemical equation. Chemistry/Thermodynamics
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Important Points 1.
The superscript zero in ∆H °f signifies standard pressure, 1 bar. Negative values of ∆H °f describe exothermic formation reactions, whereas positive values for ∆H °f describe endothermic formation reactions.
2.
Remember that all our calculations are based on enthalpy changes, not on actual enthalpy values. So, defining ∆H °f of elements (in their reference forms) as zero does not mean that the enthalpy of the elements is zero. It only means that we are establishing a thermochemical reference point, from which all changes are measured.
VARIOUS TYPES OF HEAT OF REACTIONS Depending upon the type of reaction, heat of reactions are named after these types. We study the important ones here : (a)
Heat of combustion Enthalpy change accompanying the complete combustion of 1 mole of a substance in an excess of oxygen is known as the heat of combustion of that substance. Example :
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2O(l ); (b)
∆H = −887 kJ
Heat of formation We have already studied this under “standard enthalpies of reaction”.
(c)
Heat of solution The heat of solution of a substance is the heat evolved or absorbed when 1 mole of that substance is dissolved in a stated quantity of solvent (usually the solvent is water) Example: NaCl ( s) → Na + (aq) + Cl − (aq);
(d)
∆H = 5kJ
Heat of ionisation The heat of ionisation of a substance is the amount of heat absorbed when 1 mole of a compound completely dissociates into ions. Example : HCN (aq ) → H + (aq) + CN − (aq );
Chemistry/Thermodynamics
∆H = +46kJ
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Heat of neutralisation Heat of neutralisation is the heat change accompanying an acid neutralising a base forming 1 mole of
H 2O and salt. (another definition of heat of neutralisation is : it is the amount of heat liberated when 1 gmequivalent of an acid is completely neutralised by 1 gm-equivalent of base) Some important points about ∆H neutralisation are : (i) When a strong acid reacts with a strong base (i.e. they are completely ionised in water by themselves) then the ∆H neutralisation is same for any pair of strong acid and strong base. We can understand this by the following example: Neutralisation of a strong acid by a strong base should be written in ionic form as the actual reaction takes place in ionic form. Consider neutralisation of HClO4 ( a strong acid) by NaOH ( a strong base) written in ionic form :
H + (aq) + ClO4− (aq) + Na + (aq) + OH − → Na + (aq) + ClO4− (aq) + H 2O (l ) After cancelling we get the net ionic equation : H + (aq ) + OH − (aq ) → H 2O(l );
∆H ° = −57.27kJ
The ∆H ° = −57.27kJ per mole of liquid water formed. We can see that the reaction of any strong acid with any strong base boils down to the same net ionic equation as above. So, ∆H ° accompanying any strong acid and base will be same i.e. ∆H °neutralisation = −57.27kJ for reaction between a strong acid and strong base. (ii) However, when any of the acid or base is weak (i.e. they don’t completely ionise in the water), the accompanying heat released is less than 57.27kJ . This is because some of the heat is utilised in ionising the weak acid or base. Some examples of weak acids are HCN , CH 3COOH , etc. and examples of weak bases are NH 4OH , etc. (A detailed discussion on acids and bases will be taken up in the unit on ionic equilibrium.) (f)
Heat of vapourisation Vapouration is the change of a liquid to the vapour state. For example :
H 2O (l ) → H 2O ( g );
° ∆H vap = 44kJ
° the ∆H accompanying such a process is called heat of vapourisation. Here, ∆H vap (at 25°C) = 44kJ.
(g)
Heat of fusion Fusion (or melting) is the change of a solid to the liquid state. For example,
H 2O( s ) → H 2O(l );
∆H °fus
the ∆H accompanying such a process is called heat of fusion. Chemistry/Thermodynamics
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Heat of sublimation Sublimation is the change of solid directly to the vapour state. For example , CO2 ( s ) → CO2 ( g ); dry ice
° ∆H sublimation
the ∆H accompanying such a process is called heat of sublimation.
Please attempt the following basic problems before seeing their solution. •
The heat of fusion (also called heat of melting), ∆H fus , of ice is the enthalpy change for H 2O ( s ) → H 2O(l );
∆H fus .
Similarly, the heat of vaporisation, ∆H vap , of liquid water is the enthalpy change for H 2O (l ) → H 2O ( g );
∆H vap .
How is the heat of sublimation, ∆H sub , the enthalpy change for the reaction H 2O( s ) → H 2O( g );
∆H sub
related to ∆H fus and ∆H vap ? Solution: You can think of the sublimation of ice as taking place in two stages. First, the solid metals to liquid, then the liquid vaporises. The first process has an enthalpy ∆H fus . The second process has an enthalpy ∆H vap . Therefore, the total enthalpy, which is the enthalpy of sublimation, is the sum of these two enthalpies: ∆H sub = ∆H fus + ∆H vap
•
What is the enthalpy change for the preparation of one mole of liquid water from the elements, given the following equations? H 2 ( g ) + (1/ 2)O2 ( g ) → H 2O ( g ); H 2O (l ) → H 2O ( g );
Chemistry/Thermodynamics
∆H vap
∆H f
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Solution: You can imagine this process taking place in two steps : first, the preparation of water vapour from the elements, and second, the change of the vapour to liquid. Here are the equations : H 2 ( g ) + 1 O2 ( g ) → H 2O ( g ); 2
∆H f
H 2O ( g ) → H 2O (l );
−∆H vap
The last equation is the reverse of the vapourisation of water, so the enthalpy of the step is the negative of the enthalpy of vapourisation. The enthalpy change for the preparation of one mole of liquid water, ∆H is the sum of the enthalpy changes for these two steps : ∆H = ∆H f + ( −∆H vap ) = ∆H f − ∆H vap
BOND ENTHALPY AND BOND DISSOCIATION ENERGY Consider the breaking of an H2 molecule into two hydrogen atoms:
H 2 ( g ) → H ( g ) + H ( g ) ....(1) What has been essentially done in this process is that an H — H bond has been broken to produce two H atoms. This must have required energy. This energy has helped form atoms by dissociating a bond. We define this as: Energy required to break a particular type of covalent bond is called Bond dissociation energy. Let us consider the corresponding thermochemical equation of (1): H 2 ( g ) → H ( g ) + H ( g ); ∆H 0 = 435.8 kJ i.e. the correspending standard molar enthalpy change of (1) is + 435.8 kJ. This is Bond dissociation enthalpy of (1). The Bond dissociation enthalpy is defined as enthalpy change accompanying breaking of 1 mole of a bond of particular type in gaseous state. Now, consider the experimentally determined enthalpy changes for the dissociation of a C — H bond in CH4 and C2H6 (both in gas phase): H
H
H — C — H (g) H H
H H
H — C — C — H (g) H Chemistry/Thermodynamics
H — C (g) + H(g); ∆H = +435 kJ
H
H
H
H — C — C (g) + H(g); ∆H = +410 kJ H
H
LOCUS
37
We see that bond dissociation enthalpy for C — H bond in both the cases is different. This is owing to the different molecules from which the bonds have been broken. So, we cannot specify the exact bond dissociation enthalpy of a C — H bond. However, we see that the two values are pretty close to each other. This suggests that the enthalpy change for the dissociation of a C — H bond may be about the same in other molecules. Comparisons of this sort lead to the conclusion that we can obtain approximate values of energies of various bonds. We define the A — B bond energy as the average enthalpy change for the breaking of 1 mole of A — B bond in a molecule in the gas phase to form A and B in gas phase. It is also sometimes referred to as bond enthalpy. It is denoted by ∈ ( A — B ) . We make the distinction between bond dissociation energy and bond energy more clear in the following example: •
Given the following bond dissociation energies, calculate the average bond enthalpy for the Ti — Cl bond. ∆H ( kJ / mol )
TiCl4 ( g ) → TiCl3 ( g ) + Cl ( g )
335
TiCl3 ( g ) → TiCl2 ( g ) + Cl ( g )
423
TiCl2 ( g ) → TiCl ( g ) + Cl ( g )
444
TiCl ( g ) → Ti ( g ) + Cl ( g )
519
Solution: If the above dissociation process was written as one equation, it would be:
TiCl4 ( g ) → Ti ( g ) + 4Cl ( g ); ∆H = 1721 kJ Where the ∆H = 1721 kJ is sum of all the bond dissociation energies for breaking each of Ti − Cl bonds. The bond disssociation energies are different for subsequent bond breaking owing to different entities from which the bonds are broken. So, the enthalpy change accompanying this bond dissociation is actually a sum of various bond dissociation energies. But the Bond enthalpy is defined for 1 mole of bonds broken. But here 4 moles of Ti − Cl bonds have been broken. To obtain average bond enthalpy for 1 mole of Ti − Cl bonds we must divide 1721 by 4 to get the required value. ∴ , the average bond enthalpy =
1721 kJ = 430 ⋅ 25 kJ 4
So, we may conclude that the bond energy of a diatomic molcule like H2, O2, HCl, etc. is same as their bond dissociation energy but for bonds which are found only in polyatomic molecules, the bond dissociation energies will be different from the bond energy. (as illustrated in the example above). Chemistry/Thermodynamics
LOCUS
38
We can use Bond energies to estimate heats of reaction or enthalpy changes, ∆H , for gaseous reactions. To illustrate this, let us find the ∆H for the following reaction: CH 4 ( g ) + Cl2 ( g ) → CH 3Cl ( g ) + HCl ( g ) We can imagine that the reaction takes place in steps involving the breaking and forming of bonds. Starting with the reactants, we see that one C — H bond and the Cl — Cl bond break. i.e. H H — C — H + Cl — Cl H
H H — C + H + Cl + Cl H
The enthalpy change is ∈(C — H ) + ∈ (Cl — Cl ) . Now we reassemble the fragments to give the products: H H — C + H + Cl + Cl H
H H — C — Cl + H — Cl H
In this case, C — Cl and H — Cl bonds are formed, and the enthalpy change equals the negative of the bond energies i.e. − ∈ (C — Cl ) and − ∈ ( H — Cl ) . So, the net ∆H for the overall reaction may be written as: ∆H ! ∈ (C — H ) + ∈ (Cl — Cl )− ∈ (C — Cl )− ∈ ( H — Cl )
Because the bond energy concept is only approximate, the ∆H obtained above will also be approximate. (In fact the experimental value is –101 kJ and the value obtained by above method is –104 kJ for the above reaction. Approximate, but close! Isn’t it?) In general the enthalpy of a reaction is obtained (approximately) using: ∆H ° = ∑(Bond energy)reactants − ∑ (Bond energy)products
The above method of calculating ∆H reaction using bond enthalpies is used only when the thermochemical data are not known by another approach. Important points 1.
The definition of bond energies is limited to the bond breaking process only and doesn’t include any provision for changes of state. Thus it is valid only for substances in the gaseous state. Therefore the calculations of this section apply only when all substances in the reaction are gases. If liquids or solids were involved, then additional information such as heats of vapourisation and fusion would be needed to account for phase changes.
2.
Bond energies are perhaps of greatest value when we try to explain heats of reaction to understand the relative stabilities of compounds. In general a reaction is exothermic (gives off heat) if weak bonds are replaced by strong bonds. In the reaction we just discussed, two bonds were broken and replaced by two new, stronger bonds.
Chemistry/Thermodynamics
LOCUS
39
BORN-HABER CYCLE Lattice energy cannot be determined directly by experiment. It can, however, be calculated by envisioning the formation of an ionic compound as occurring in a series of well-defined steps. We can then use Hess’s law to put these steps together in a way that gives us the lattice energy for the compound. By so doing, we construct a Born-Haber cycle. Let us try to understand the Born-Haber cycle using the example of NaCl. In the Born-Haber cycle for NaCl we consider the formation of NaCl(s) from the element Na(s) and Cl2(g) by two different routes, as shown in the Figure: Na (g) + e + Cl(g) +
–
E(Cl) Na (g)+Cl (g) +
–
Na(g)+Cl(g) Na ( g ) +
° ∆H f [Na (g)]
Na ( s ) +
1 2 1 2
Cl2 ( g ) Cl2 ( g )
° ∆H f [NaCl (s)]
Lattie energy of NaCl
° ∆H f [Cl(g) ]
–Lattie energy of NaCl
Energy
I1(Na)
NaCl(s) A Born-Haber cycle shows the energetic relationships in the formation of ionic solids from the elements. The enthalpy of formation of NaCl(s) from elemental sodium and chlorine Equation (1) is equal to the sum of the energies of several individual steps (Equation (2) through (6)) by Hess's law. fig. 11
The enthalpy change for the direct route is the heat of formation of NaCl (s): 1 Na ( s ) + Cl2 ( g ) → NaCl ( s ) ∆H °f [ NaCl ( s )] = −410.9 kJ ...(1) 2 The indirect route consists of five steps. First we generate gaseous atoms of sodium by vapourising sodium metal. Then we form gaseous atoms of chlorine by breaking the bond in the Cl2 molecule. The enthalpy changes for these processes are available to us as enthalpies of formation : ,
Na( s) → Na( g )
∆H °f [ Na( g ) ] = 107.7 kJ
...(2)
1 Cl2 ( g ) → Cl ( g ) 2
∆H °f [Cl ( g ) ] = 121.7 kJ
...(3)
Chemistry/Thermodynamics
LOCUS
40
Notice that both of these processes are endothermic; energy is required to generate gaseous sodium and chlorine atoms. In the next two steps we remove the electron from Na(g) to form Na+(g) and then add the electron to Cl(g) to form Cl (g). The enthalpy for these processes equal the first ionisation energy of Na, I1(Na), and the electron affinity of Cl, denoted by E(Cl), respectively.: Na( g ) → Na + ( g ) + e −
∆H = I1 ( Na) = 496 kJ
...(4)
Cl ( g ) + e− → Cl − ( g )
∆H = E (Cl ) = −349 kJ
...(5)
Finally, we combine the gaseous sodium and chloride ions to form solid sodium chloride. Because this process is just the reverse of the lattice energy (breaking a solid into ions), the enthalpy change is the negative of the lattice energy, the quantity that we wish to determine: Na + ( g ) + Cl − ( g ) → NaCl ( s)
∆H = −∆H lattice = ?
...(6)
1 Cl2 ( g ). Thus, from Hess’s law 2 we know that the sum of the enthalpy changes for these five steps equals that for the direct path, indicated by the bold arrow, Equation (1) :
The sum of the five steps in the indirect path gives us NaCl(s) from Na(s) and
∆H °f [ NaCl ( s)] = ∆H °f [ Na( g )] + ∆H °f [Cl ( g ) ] + I1 ( Na) − E (Cl ) − ∆H . ⇒ − 411 kJ
= 108 kJ
+ 122 kJ
+ 496 kJ − 349 kJ − ∆H lattice
Solving for ∆H lattice : ∆H lattice = 108 kJ + 122 kJ + 496 kJ − 349 kJ + 411 kJ = 788 kJ Thus the lattice energy of NaCl is 788 kJ / mol.
Example – 9 The standard molar enthalpies of combustion of C2H2(g), C(graphite) and H2(g) are – 1299.63, – 393.51, – 285.85 kJ mol–1, respectively. Calculate the standard enthalpy of formation of C2H2(g). Critical thinking In order to proceed in such questions, it is suggested that you write down the “target” equation that will lead us to finding the required quantity. In this case, the “target” equation will be the equation representing the formation of C2H2(g).Thereafter, see how the given data can be used in conjunction with the “target” equation to obtain the required quantity.
Chemistry/Thermodynamics
LOCUS
41
Solution: Our “target” equation is,
2C (graphite) + H 2 ( g ) → C2 H 2 ( g ); ∆H
...(1)
and we have to determine its ∆H as its ∆H is actually the enthalpy of formation C2H2. Now, the given data can be written as: 5 C2 H 2 ( g ) + O2 ( g ) → 2CO2 ( g ) + H 2O (l ) 2
∆H ° = −1299.63 kJ / mol
...(2)
C (graphite) + O2 ( g ) → CO2 ( g )
∆H ° = −393.51 kJ / mol
...(3)
H 2 ( g ) + 1 O2 ( g ) → H 2O (l ) 2
∆H ° = −285.85 kJ / mol
...(4)
We can use Hess’s law to get Eq.(1) from Eq (2), (3) and (4) as follows:
(3) × 2 ⇒
2C (graphite) + 2O2 ( g ) → 2 CO 2 ( g )
(4) × 1 ⇒
∆H º = −285.85 kJ / mol H 2 ( g ) + 1 O 2 ( g ) → H 2O (l ) 2 5 2 CO 2 ( g ) + H 2O (l ) → O 2 ( g ) + C2 H 2 ( g ) ∆H º = 1299.63 kJ / mol 2 2C (graphite) + H 2 ( g ) → C2 H 2 ( g ) ∆H º = 226.76 kJ / mol
(2) × −1 ⇒
∆H º = 2 × −393.51 kJ / mol
Hence, ∆H °f (C2 H 2 , g ) = 226.76 kJ / mol.
Example – 10 Use the following information to determine ∆H °f for PbO (s, yellow) PbO (s, yellow) + CO (g) → Pb(s) + CO2 (s); ∆H r° = −65.69 kJ . Given:
∆H °f for CO2 (g) = –393.5 kJ/mol. ∆H °f for CO(g) = –110.5 kJ/mol. Critical thinking We can use Hess’s law in the form ∆Hr° = ∑ n ⋅∆H °f (products) − ∑ m ⋅∆H °f (reactants). Now we are given ° and the ∆H °f for all substances except PbO (s, yellow). We can solve for this unknown. ∆H reaction
Chemistry/Thermodynamics
LOCUS
42
Solution: ∆H r° = ∑ n ⋅ ∆H °f (products) − ∑ m ⋅ ∆H °f (reactants) ⇒
∆H r° = ∆H °f ( Pb, s ) + ∆H °f (CO2 , g ) − ∆H °f ( PbO ( s, yellow) + ∆H °f (CO, g )
⇒
−65.69 = 0 + (−393.5) − ∆H °f ( PbO( s, yellow) + ( −110.5)
Rearranging to solve for ∆H °f ( PbO( s, yellow)), we have,
∆H °f ( PbO( s, yellow) ) = 65.69 − 393.5 + 110.5 = –217.3 kJ/mol of PbO.
Example – 11 When Aluminium metal is exposed to atmospheric oxygen, it is oxidised to form Aluminium oxide. How much heat is released by the complete oxidation of 24.2 gram of Aluminium at 25°C and 1 atm ? The thermochemical equation is : 4Al (s) + 3O2 (s) → 2Al2O3 ; ∆H = –3352 kJ/mol Critical thinking The thermochemical equation tells us that 3352 kJ of heat is released for every “mole of reaction” i.e. for every 4 moles of Al that reacts. We convert 24.2 g of Al to moles, and then calculate the number of kilojoules corresponding to that number of moles of Al. Solution: For 24.2 grams of Al, heat released can be obtained by, 24.2 g ⋅ Al ×
1mol Al 27 g Al
×
1 mol reaction 4 mol Al
×
−3352 kJ 1 mol reaction
= –751 kJ This tells us that 751 kJ of heat is released to the surroundings during the oxidation of 24.2 grams of aluminium.
Example – 12 Given the following standard enthalpies of reactions. (i)
Enthalpy of formation of water = –285.8 kJ mol–1.
(ii)
Enthalpy of combustion of acetylene = –1299.6 kJ mol–1.
(iii)
Enthalpy of combustion of ethylene = –1410.8 kJ mol–1.
Calculate the heat of reaction for the hydrogenation of acetylene to ethylene at constant volume (25°C).
Chemistry/Thermodynamics
LOCUS
43
Critical thinking Hydrogenation of acetylene to ethylene can be written as:
C2 H 2 ( g ) + H 2 ( g ) → C2 H 4 ( g ) We have to find the heat of reaction for this reaction. But under what conditions? And under these conditions which state function change will you calculate to obtain the heat of the reaction? Solution: Our “target” equation is: (1) : C2H2(g) + H2(g) → C2H4(g) We can obtain the ∆H accompanying this reaction from the given data by applying Hess’s law. But that would give us heat of the reaction under constant pressure. The conditions under which we have to obtain the heat of reaction is constant volume. This can be obtained by calculating ∆U for the reaction. First, let us calculate the ∆H by applying Hess’s law as follows: Given data may be written as : (2) : H 2 ( g ) + 1 2 O2 ( g ) → H 2O (l )
∆H = −285.8 kJ / mol
(3) : C2 H 2 ( g ) + 5 2 O2 ( g ) → 2CO2 ( g ) + H 2O(l )
∆H = −1299.6 kJ / mol
(4) : C2 H 4 ( g ) + 3O2 ( g ) → 2CO2 ( s) + 2 H 2O(l )
∆H = −1410.8 kJ / mol
We can modify Equation (2), (3), (4) as follows to obtain Equation (1) (1) × 1 : H 2 ( g ) + 1 O 2 ( g ) → H 2O (l ) ∆H = − 285.8 kJ / mol 2 (2) × 1 : C2 H 2 ( g ) + 5 O 2 ( g ) → 2 CO 2 ( g ) + H 2O (l ) ∆H = −1299.6 kJ / mol 2 (3) × −1: 2 CO 2 ( s ) + 2 H 2O (l ) → C2 H 4 ( g ) + 3 O 2 ( g ) ∆H = +1410.8 kJ / mol H 2 ( g ) + C2 H 2 ( g )
→
C2 H 4 ( g )
∆H = − 174.6 kJ / mol
Now, to obtain ∆U we use the following equation: ∆H = ∆U + (∆n) RT. ⇒
∆U = ∆H – (∆n) RT.
...(5)
Here, change in number of moles gaseous substances ∆n is: 1–2 = –1. Substituting the appropriate values in the (5) above, we get,
Chemistry/Thermodynamics
LOCUS
44
∆U = [(–174.6 kJ/mol) – (–1) (8.314 × 10–3 kJ/mol.K) (298 K)] = (–174.6 + 2.48) kJ/mol = –172.12 kJ/mol (Note : In the calculation of ∆U above, we have taken the value of R as 8.314 × 10–3 kJ/mol.K in order to keep the unit of energy uniform throughout i.e. kJ).
Example – 13 An intimate mixture of ferric oxide, Fe2O3, and aluminium, Al, is used in solid fuel rockets. Calculate the fuel value per gram and fuel value per cm3 of the mixture. Given: ∆fH(Al2O3, s) = –1699 kJ mol–1; ∆fH (Fe2O3, s) = –833 kJ mol–1. Density of Fe2O3 = 5.2 g cm−3 ; Density of Al = 2.7 g cm −3 Critical thinking We are given that a mixture of Fe2O3 and Al are used as solid rocket fuel. What is its fuel value ? It is the heat released when Fe2O3 and Al react to form products according to the following thermochemical equation: Fe2O3 ( s ) + 2 Al ( s ) → Al2O3 ( s ) + 2 Fe( s ); ∆H = ? Solution: The target equation is: (1) Fe2O3 ( s ) + 2 Al ( s ) → Al2O3 ( s ) + 2 Fe( s ); ∆H reaction And we are given, ∆H f ( Al2O3 , s) = −1669 kJ / mol ∆H f ( Fe2O3 , s ) = −833 kJ / mol
Using this we can calculate the ∆Hreaction of (1) as follows: ∆H reaction = ∆ f ( Al2O3 , s) + 2 × ∆H f ( Fe, s) − ∆H f ( Fe2O3 , s) − ∆H f ( Al , s)
= (–1669 + 0 –(–833) – 0) kJ/mol = – 836 kJ/mol. Now, in order to calculate fuel value per gram of the fuel (i.e. the reactants) we must divide this ∆H by the molar mass of reacting species (each multiplied by appropriate stoichiometric coefficients.) i.e. Total mass of the reactants = molar mass (Fe2O3) + 2 × molar mass (Al) Chemistry/Thermodynamics
LOCUS
45
= 160 g + 2 × 27 g = 214 g ∴ fuel value per gram =
836 kJ / mol 214 g
= 3.906 kJ/mol. In order to calculate the fuel value per cm3 of the mixture, we must determine the volume of the mixture, molar volume of Fe2O3 ( s ) = molar mass of Fe2O3 density of Fe2O3 =
160 g 5.2 g ⋅ cm −3
= 30.77 cm3 molar value of
Al ( s ) =
=
molar mass of Al density of Al 27 g 2.7 g cm −3
= 10 cm3. ∴ the volume of the mixture = molar volume (Fe2O3) + 2 × molar volume (Al)
= (30.77 + 2 × 10) cm3. = 50.77 cm3. Therefore fuel value per cm3 of fuel =
836 kJ 50.77 cm3
= 16.47 kJ/cm3.
Example – 14 From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpy of formation of CO2(g), H2O(l) and propene (g) are –393.5, –285.8 and 20.42 kJ mol–1, respectively. The enthalpy of isomerisation of cyclopropane to propene is –33.0 kJ mol–1.
Chemistry/Thermodynamics
LOCUS
46
Critical thinking Our “target” Equation is CH2
CH2
H2C
9 ( g ) + O2 ( g ) → 3CO2 ( g ) + 3H 2O (l ); ∆H = ? 2
...(1)
The ∆H accompanying this reaction is the required enthalpy of combustion. We have to get to this equation from the given equations using Hess’s law. Solution: Given : ∆H f (CO2 , g ) = −393.5 kJ / mol ∆H f ( H 2O, l ) = −285.8 kJ / mol ∆H f (CH 3CH = CH 2 ) = 20.42 kJ / mol CH2 isomerisation
CH3 CH — CH2 (g);
∆rH = –33.0 kJ/mol
...(2)
CH2
H2C
It is not possible to get the ∆H for equation (1) from this data directly. However, if we analyse closely, we see that ∆Hcombustion for propene is obtainable and the combustion reaction for propene could be combined with equation (2) to obtain the equation (1) and hence its ∆Hcombustion. So, combustion reaction for propene is : 9 CH 3CH = CH 2 ( g ) + O2 ( g ) → 3CO2 ( g ) + 3H 2O (l ) 2
∆H for this reaction can be obtained as : ∆H c ( propene) = 3 ⋅ ∆H f (CO2 , g ) + 3∆H f ( H 2O, l ) − ∆H f (CH 3CH = CH 2 ) = [3 ⋅ ( −393.5) + 3( −285.8) − (20.42)]kJ / mol = −2058.32 kJ / mol
Chemistry/Thermodynamics
...(3)
LOCUS
47
Now, if we add reactions (2) and (3) we get equation (1): CH2
isomerisation
CH3CH = CH 2(g)
∆H = – 33 kJ/mol
CH2
H2C
9 CH 3 CH = CH 2 ( g ) + O2 ( g ) → 3CO2 ( g ) + 3H 2O (l ) 2
∆H = −2058.3kJ / mol
____________________________________________________________________________________
9 + O2 ( g ) → 3CO2 ( g ) + 3H 2O (l ); ∆H = −2091.32kJ / mol 2
Hence, enthalpy of combustion of cyclopropane is – 2091.32 kJ/mol
Example – 15 Use the given bond energies to estimate the heat of reaction at 298 K for the following reaction. (All bonds are single bonds). Br2 ( g ) + 3F2 ( g ) → 2 BrF3 ( g ) Given:
∈Br − Br = 192 kJ / mol ∈F − F = 159 kJ / mol
∈Br − F = 197 kJ / mol
Critical thinking Each BrF3 molecule contains three Br – F bonds. So, two moles of BrF3 contain six moles of Br – F bonds. Three moles of F2 contain a total of three moles of F – F bonds, and one mole of Br2 contains one mole of Br – Br bonds. We only have to use bond energy form of Hess’s law i.e. ∆H ° = ∑ B.E.(reactants) − ∑ B.E.(products) (note: this can be applied in gas phase reactions only.) Chemistry/Thermodynamics
LOCUS
Solution:
48
∆H °reaction = [∈Br − Br +3. ∈F − F ] − [6. ∈ Br − F ] = 192 + 3 (159) – 6(197) = – 513 kJ/mol
Example – 16 The enthaplies of the following reactions are shown alongwith. 1 1 H 2 ( g ) + O2 ( g ) → OH ( g ) 2 2
∆H = 42.09 kJ mol −1
H 2 ( g ) → 2H ( g )
∆H = 435.89 kJ mol −1
O2 ( g ) → 2O( g )
∆H = 495.05 kJ mol −1
Calculate the O — H bond energy for the hydroxyl radical.
Critical thinking As discussed in the theory, we have to calculate the enthalpy change for the following reaction (which will be same as bond energy of the hydroxyl radical): OH ( g ) → O ( g ) + H ( g ); ∆H
As pointed out in example - 9 above, this is our “target” equation. How will you proceed to obtain the required ∆H now? Solution: Our “target” equation is: (1) OH ( g ) → O ( g ) + H ( g ); ∆H = ? We are given (2)
1 1 H 2 ( g ) + O2 ( g ) → OH ( g ) 2 2
∆H = 42.09 kJ / mol
(3) H 2 ( g ) → 2 H ( g )
∆H = 435.89 kJ / mol
(4) O2 ( g ) → 2O( g )
∆H = 495.05 kJ / mol
Chemistry/Thermodynamics
LOCUS
49
We can obtain equation (1) from (2), (3), (4) by applying Hess’s law as follows: 1 1 H 2 ( g ) + O2 ( g ) 2 2
(1) × −1:
OH ( g ) →
1 : 2 1 (3) × : 2
1 H2 (g ) → H (g) 2 1 O2 ( g ) → O ( g ) 2 OH ( g ) → H ( g ) + O( g )
(2) ×
∆H = −42.09 kJ / mol 1 × 435.89 kJ / mol 2 1 ∆H = × 495.05 kJ / mol 2 ∆H = 423.38 kJ / mol ∆H =
Example – 17 Use the given bond energies to estimate the heat of reaction at 298 K for the following reaction: C 3H 8( s ) H
H
+ Cl 2 ( g )
H
H — C — C — C — H + Cl — Cl H
Given:
H
H
C 3 H 7 Cl ( g ) H
H
+ HCl (g)
H
H — C — C — C — Cl + H — Cl H
H
H
∈C − H = 414 kJ / mol ∈Cl −Cl = 243 kJ / mol ∈C −Cl = 330 kJ / mol ∈H −Cl = 431 kJ / mol
Critical thinking Two moles of C — C bonds and seven moles of C — H bonds are the same before and after the reaction, so we do not need to include them in the bond energy calculation. The only reactant bonds that are broken are one mole of C — H bonds and one mole of Cl — Cl bonds. On the product side, the only new bonds formed are one mole of C — Cl bonds and one mole of H — Cl bonds. We need to take into account only the bonds that are different on the two sides of the equation. As in the last example, we add and subtract the appropriate bond energies. Solution: ∆H r0 = [∈C − H + ∈Cl −Cl ] − [∈C −Cl + ∈H −Cl ] = [414 + 243] − [330 + 431] = −104 kJ / mol
Example – 18 When 2.1 g of iron combines with sulphur, 3.77 kJ are evolved. Calculate the heat of formation of iron sulphide. Chemistry/Thermodynamics
LOCUS
50
Critical thinking The reaction involved can be written as:
Fe( s) + S ( s) → FeS ( s);
° ∆H reaction =?
Recall that this is a thermochemical equation and has a molar interpretation. Now,. ∆H ° for the reaction will be: ° ∆H reaction = ∆H °f ( FeS , s ) – ∆H °f ( Fe, s ) − ∆H °f ( S , s )
∵,
∴,
the last two terms are ∆H °f of elements, they are zero. ° ∆H °f ( Fe S , s ) is same as ∆H reaction .
The thermochemical equation above can be interpreted as : 1 mol of Fe gives 1 mol of FeS and is ° accompanied by ∆H reaction (which is same as ∆H °f ( FeS , s )). How will you use this information with the ° given data in the problem to obtain ∆H reaction ?
° Solution: 1 mol of Fe reacts to release ∆H reaction . in this reaction ° i.e. 56 grams of Fe reacts to release ∆H reaction .
But we are given that, 2.1 grams of Fe reacts to release
3.77 kJ
⇒
1 gram of Fe reacts to release
3.77 kJ 2.1
⇒
56 grams of Fe reacts to release
3.77 × 56 kJ = 100.5 kJ 2.1
° Hence, ∆H reaction = ∆H °f ( FeS , s ) = 100.5 kJ / mol.
Chemistry/Thermodynamics
LOCUS
51
Here is a reference list of ∆H °f values that you will need to solve the various problems of Try Yourself - II and Exercises at end.
Substance
∆H °f (at 298 K) in kJ/mol
C2 H 2 ( g )
226.8
C2 H 4 ( g )
52.3
C2 H 6 ( g )
–89.7
CO2 ( g )
–393.5
CO ( g )
–110.5
H 2O ( g )
–241.8
H 2O(l )
–285.8
CuO (s)
–162.0
Chemistry/Thermodynamics
LOCUS
52
TRY YOURSELF - II Q. 1
Does the value of ∆H ° for a reaction depend on the presence of catalysts in a system ? Substantiate your answer.
Q. 2
Upon the complete combustion of ethylene, C2H4 (with the formation of liquid water), 6226 kJ were evolved. Find the volume of the oxygen that entered into the reaction in standard conditions.
Q. 3
In the reduction of 12.7 g of copper (II) oxide with coal (with the formation of CO), 8.24 kJ are absorbed. ° Determine ∆H 298 of formation of CuO.
Q. 4
Water gas is a mixture of equal volumes of hydrogen and carbon monoxide. Find the amount of heat evolved in the combustion of 112 litres of water gas taken in standard conditions.
Q. 5
° Find ∆H 298 of formation of ethylene using the following data:
C2 H 4 ( g ) + 3O2 ( g ) → 2CO2 ( g ) + 2 H 2O( g ); ∆H ° = −1323 kJ
Q. 6
C (graphite) + O2 ( g ) → CO2 ( g );
∆H ° = −393.5 kJ
1 H 2 ( g ) + O2 ( g ) → H 2O ( g ); 2
∆H ° = −241.8 kJ
Ethylamine undergoes an endothermic gas phase dissociation to produce ethylene (or ethene) and ammonia. H H H | | H—C—C—N H | | H H
∆
H
H
C— —C H
H
+
N | H H H
0
∆Hreaction = + 54.68 kJ/mol
The following average bond energies per mole of bonds are given : C — H = 414 kJ; C— C = 347 kJ C — C = 611 kJ; N — H = 389 kJ. Calculate the C — N bond energy in ethylamine. Q. 7
The standard enthalpies of formation at 298 K for CCl4 ( g ), H 2O( g ), CO2 ( g ) and HCl ( g ) ° are –106.7, –241.8, –393.7 and –92.5 kJ mol–1, respectively. Calculate ∆H 298 for the reaction,
CCl4 ( g ) + 2H 2O( g ) → CO2 ( g ) + 4HCl ( g )
Chemistry/Thermodynamics
LOCUS
Q. 8
53
Diborane is a potential rocket fuel which undergoes combustion according to the reaction B2 H 6 ( g ) + 3O2 ( g ) → B2O3 ( s ) + 3H 2O( g ) From the following data, calculate the enthalpy change for the combustion of diborane.
Q. 9
2 B( s) + (3 / 2)O2 ( g ) → B2O3 ( s)
∆H = −1273 kJ mol −1.
H 2 ( g ) + (1/ 2)O2 ( g ) → H 2O (l )
∆H = −286 kJ mol −1
H 2O (l ) → H 2O( g )
∆H = 44 kJ mol −1
2 B( s) + 3H 2 ( g ) → B2 H 6 ( g )
∆H = 36 kJ mol −1
Bond energies of F2 and Cl2 are respectively 36.6 and 58.0 kcal per mole. If the heat liberated in the reaction F2 + Cl2 → 2 FCl is 26.6 kcal, calculate the bond energy of F — Cl bond.
Q. 10 From the following data, calculate the heat evolved in the formation of 2.5 litres of carbon monoxide from its elements:
C + O2 → CO2 ∆H = −94,500 cal 2CO + O2 → 2CO2 ∆H = −136, 000 cal. Q. 11 Calculate ∆H °f of C6 H12O6 ( s ) from the following data: º of C6 H12O6 ( s ) = −2816 kJ / mol ∆H combustion
Chemistry/Thermodynamics
