6 .4
Hess's law
Learni ng outcomes , On completion of this section, you should be able to: ■ explain and use the term ‘enthalpy change of formation' ■ state Hess's law ■ use Hess's law to calculate enthalpy changes of formation, reaction and combustion.
Enthalpy change of formation Standard enthalpy change of formation: The enthalpy change when 1 mole of a compound is formed from its elements under standard conditions. Symbol AH f. Example: C(graphite) + 2H,(g) -» CH„(g) AHf [CH 4(gj] = -74.8 kj mob1
C(graphite) is used here because graphite is the most stable form of carbon. The value of AHf of an element in its normal state at room temperature is zero.
Hess's law
Did you know?
Hess's law states that the total enthalpy change for a chemic al reaction is independent of the route by w hich the reaction proceeds.
Hess's law is an example of the First law of Thermody namics as applied
Figure 6.4.1 illustrates Hess's law. It shows that the enthalpy change is the same, whether you go by the direct route or the indirect route.
to chemical reactions. The First law states that Energy cannot be crea ted
or destroyed.
We can use Hess's law to calculate enthalpy changes which cannot be found directly by experiment, e.g. the enthalpy change of formation of propane.
Deducing enthalpy change of reaction using AH f values An enthalpy cycle (Hess cycle) is constructed as shown in Figure 6.4.2. a h ;
reactants
* shows the indirect route via intermediates X and Y
Figure 6.4 .7 The enthalpy change is the same whichever route is followed
a h ;
products xa h ;
reactants'
a h ;
NHj + HCl a h ; ( n h 3)x
products
n h 4c i
/ a h ; ( n h 4c i )
+a h ; ( h c i )
elements in standard states
j N2+
2H2+ jC l2
Figure 6.4.2
Using H ess's law we can see tha t AHfreactants + A Hf = AHfproducts So AHf = AHf products - AHfrcactants
Worked example 1 Calculate AHf for the reaction: S 0 2(g) + 2H,S(g) —» 3S(s) + 2H 20(1) AHf values: SO,(g) = - 29 6.8 kj m o l1, H,S(g) = ^ O A k Jm o l1, H20(1) = -285.8kJmol-‘ The stages are: ■ Write the balanced equation. Draw the enthalpy cycle with the elements at the bottom as shown. ■ Apply Hess's law. S 0 2(g) + 2H2S(g) -___________ > 3S(s) + 2HzO(l) A H ;[ S0 2] \ X +2x a h ; [ h 2s ] \ '
/
2
X
n
3S(s) + 0 2(g) + 2H2(g)
Figure 6.4.3
AHf [H20]
Chapter 6 Energetics ■
According to Hess's law: AH ? [S02] + 2AH ? [H2Sj + AH ? = 2AH f [H2Oj -29 6.8 + 2 x (-20.6) + AH ? = 2 x (-285.8) * so AH ? = 2 x (-285.8) + 296.8 + (2 x 20.6) so AH ? = -2 33.6 kjmol 1
Deducing enthalpy change of reaction using AH f values The enthalpy cycle is constructed as shown below and a worked example shown.
: igure 6.4.4
Worked example 2 Calculate AH f for the reaction: 3C(graphite) + 4H,(g) —>C 3H8(g) AH? values: C(graphite) = -3 9 3. 5 kj m ok 1, H,(g) = -2 85 .8 kjm ok1, C,H8(g) = -2219.2kJmoh1
3C(graphite) + 4H2(g)
a h ;
3 x AHc*[C (g r)r
C3H8(g)
/AWdC 3H8(g)]
+ 4 X A H T[ H M 3 C 0 2(g) + 4H20 (l) :
gure 6. 4. 5
According to Hess's law: AH ? + AH ? [C3H8] = 3AH ? [C(graphite)] + 4AH ? [H2(g)( AH ? + (-2 219.2) = 3 x (-393.5) + 4 x (-285.8) so AH ? = 3 x (- 393.5) + 4 x (-285.8) + 221 9.2 AH? = - 1 0 4 . 5 k j m o k 1 Note: In this reaction, AH ? is also the enthalpy change of formation of propane.
Key points ■
Entha lpy change of formatio n refers to the formation of 1 mole
^ Exam tips ; rmemberthat:
■ AH? of an element is 0. ■ In enthalpy cycles, make sure that you follow the arrows round in the correct direction. ■ Look Carefully at the state symbols when looking up enthalpy changes, e.g. A H ? [H20(gJ] = -241.8 kj rook 1 but A H f [H20(lj] = -285.8 kj rook 1
of a compound from its elements under standard conditions. ■ Hess's law states that the total
enthalpy change i n a reacti on is i ndependent of the r oute. ■ Hess's law can be used to calculate enthalpy changes which cannot be determined directly by experiment, e.g. AH , from AH, or A Hc of reactants and products.
67
6.5
More enthalpy changes
Learni ng outcomes On completion of this section, you should be able to: ■ construct Hess cycles using bond
An enthalpy cycle using bond energies We can use the enthalpy cycle in Figure 6.5.1 to calculate an enthalpy change using bond energies. It is often easier, however, to use the 'balance sheet' method (see Section 6.2). .
energies ■ explain the term enthalpy change of hy dration, A£/hyd ■ explain the term lattice energy,
AHlat, ■ use Hess cycles to determine AHsol or A Hhyd.
Figure 6.5.1
Worked example 1 Calculate AH f for the complete combustion of ethene to carbon dioxide and water. £(C=C) +610, £(C-H) +410, £(0=0) +496, E(C=0) +740, £(H-0) +460 H. C=C H7
H * +3 0 = 0 - = ^ — 2 0 = C = 0 + 2 H — O— H XH
£(c=c) + 4 x F(C-H) + 3 x £(0=0)
4 x £(C=0) + 4 x f(H-O) 2C(g) + 60(g) + 4H(g)
Figure 6.5.2
According to Hess's law: AH f = £(reactants) + £(products) Note: The values of £(products) are negative because bond formation is exothermic.
£(C=C) + 4 x £(C-H) + 3 x £( 0 = 0 ) + 4 x £(C= 0) + 4 x £(H-0) + 610 + 4 x 410 + 3 x 496 + 4 x (-740) + 4 x (-460) A H f = -1062kJmoH
Didyou know?
Enthalpy change of hydration and lattice energy
We can calculate lattice energy in
Standard enthalpy change of hydration: Th e enthalpy change when 1 mole of a specified gaseous ion dissolves in enough water to form an infinitely dilute solution under standard conditions. Symbol A H fyd.
theory from the size and charge of the ions and the way they are packed. This som etimes gives slightly ■ different v alues from the lattice energy obtained by experiment. 'Lattice energy’ is a commonly used term and is often used to mean
Examples: Mg2+(g) + aq -> Mg2+(aq) AHf yd = -19 2 0kjm ol-1 • Ch(g) + aq —>Ck(aq) AHfyd = -3 64 kj mol-1 Lattice energy: The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions. Symbol AHf..
the same as lattice enthalpy. The
Example: Ca2+(g) + 2C l“(g) —>C aC l2(s) AH^tt = - 2 2 5 8 kj m ol -1
difference between the two is so
Lattice energies are always exo therm ic because energy is always released when new 'bonds' are made.
small that we often use the two terms to mean the same thing.
Enthalpy change of solution from A Hfatt and AH fyd •Ve can calculate the enthal py change of solution or the enthalpy change of hydration using an enthalpy cycle.
Figure 6.5.3
Here is a worked example: Calculate the enthalpy change of solution of magnesium chloride using the following data: AH^tt[MgCl2] = - 2 5 9 2 kj mol-1, AH fyi [Mg2+] = -1 92 0 kjm ol-1( AHfyd [Ch] = - 36 4 kj mol"1
Figure 6.5.4
According to Hess's law: AH£tt[MgCl2] + AHfol[MgCl2] = AH?yd[Mg2+] + 2 x AHfyd[Ck] So AHfol[MgCl2] = AHfyd [Mg2+] + 2 x AH(fyd [Cl ] - AH£tt[MgC l2] -19 20 + 2 x (-364) (-2592) So AH^ijMgCl,] = - 5 6k Jm ol _1 The value of the enthalpy change of hydration for a particular ion can also be found from this type of enthalpy cycle if we know: the lattice energy the enthalpy change of solution the enthalpy change of hydration of one of the ions.
Key points ■ AH , can be found by using a Hess cycle involving bond energies of reactants and products. ■ Enthalpy change of hydration, AHhyd is the enthalpy change when 1 mole of gaseous ions dissolves in water to form a very dilute solution. ■ Lattice energy, A/-/latt is the en thalp y change when 1 mole of an ionic compound is formed from its gaseous ions. ■ A Hess cycle can be drawn relating the ent halpy ch anges AH sol, AH hyd and
AHlat,..
6 .6
Lattice energy: the Born-Haber cycle
Learni ng outcomes On completion of this section, you should be able to: ■ explain the ter ms ‘enthalpy change of atomisation' and ‘electron affinity' ■ calculate lattice energy from an appropriate Born-Haber cycle.
Some more enthalpy changes In order to calculate lattice energy, we need to use ionisation energies, enthalpy change of formation and two other enthalpy changes. These two are: Standard enthalpy change of atomisation: Th e enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state under standard conditions. Symbol AHft.
Examples: Na(s) -n>Na(g) }C l2(g) -4 Cl(g)
AHaf = + 10 7. 3k jm oh ‘ AHat = + 122 kj m ol '1
Values of AH £ are always positive (endothermic) because energy has to be absorbed to break bonds holding the atoms together. It is important to remember that the enthalpy change of atomisation is per atom formed. Electron affinity: The first electron affinity is the enthalpy change when 1 mole of electrons are added to a mole of gaseous atoms to form 1 mole of gaseous ions X-. Symbol AH%x.
Examples: Cl(g) + e- -» Ch(g) AH^, = —34 8 kj mol-1 S(g) + e- -» S-(g) AHfal = -200kJmol-> As with ionisation energies, successive electron affinities are used to form anions with multiple charges: O(g) + e~ -> 0~(g) O'(g) + e- -» 0 2'(g)
0
AHZi = -1 41 kj mol'1 AHfal = + 798 kj m ol1
Exam tips
It is important that you put the correct signs (- or +) when you carry out enthalpy change calculations to find lattice energy. Make sure you know these. A H%i is generally negative but A H fa2 is always positive, because the incoming electron is being added to a negatively charged ion. The repulsive forces have to be overcome. A Hf* values and AH^ values are always positive. A H ft, are always negative. A H f values may be positive or negative.
Chapter 6 Energetics
Born-Haber cycles We can construct an enthalpy cycle and use Hess's law to determine the , lattice energy of an ionic solid.
Figure 6.6 .7
AHf consists of several steps:. atomise metal -+ atomis e non -metal —>ionise metal a toms —> ionise non-metal atoms So for NaBr: AHf = AHf [Na] + AHf [Br] + AH f [Na] + AHfal[Br] By Hess's law: AHf + AHftl = AHf So AHftt = AHf - AHf Ve can show alj these changes on an energy level diagram. This is called a Born-Haber cycle. The Born-Haber cycle can be used to determine the lattice energy of sodium bromide as shown in Figure 6. 6.2 . N a f g) + Br(g) + e~ AHfal[Br] -325 Nafg ) + Br(g) AHjt[Na) +494
Aw;[iBrJ‘ +112
AH * [Na] +107
Na(g) + Br(g) Na(g) +?Br?(g) Na(s) +?Brz(l)
0
AWfti [Na Br]
It is easier to determine lattice
AHf *[Na Br] -361
Exam tips
energy by calculating the sum of
Na Br(s)
the atomisation energies, ionisation
figure 6.6 .2 A Born-Haber cycle to determine the lattice energy of sodium bromide. All Sgures are in k j mol~1
From the enthalpy change values shown on the Born-Haber cycle we can -ee that: AH f = AH f[Na] + AHaf[B r] + AH f[Na j + AHfal[Br]' \ AHf = + 107 + 112 + 494 + (-325) kj mob* = +388 kj mob1 AHflt = AH f - AHf = 361 - ( + 388) = -74 9k Jm ob 9Vi-
Key points
*
Enthalpy change of atomisation, AHat, is the enthalpy change when 1 mole of gaseous atoms is formed from the element. Electron affinity, AHea is the enthalpy change wh en 1 mole of electrons is added to 1 mole of gaseous atoms. A Born-Haber cycle is an enthalpy cycle which includes the enthalpy changes AHlatt, AHf, A Hav AH, and AHea.
energies and electron affinities first (AHf). Then apply AHftt = A H f - A H f .-You are less likely to make mistakes with the signs or be confused by too m any enthalpy changes this way.
6.7
More about lattice energy
Learning outcomes On completion of this section, you should be able to: ■ understand howto construct Born-Haber cycles involving ions with multiple charges ■ explain the effect of ionic charge and radius on the magnitude of lattice energy.
The Born-Haber cycle for magnesium oxide Magnesium forms an ion with a 2+ charge. The oxide ion has a charge :: 2 . So when constr ucting a Born -Ha ber cycle for magnesium oxide we have to take into account: ■
the first and the second ionisa tion energies of magn esium and
a
the first and second electron affinities of oxygen.
The enthalpy cycle and Born-Ha ber cycle to calculate the lattice e n e r g y or magnesium oxide are shown below. The relevant enthalpy changes (all in kjmol-1) are: AH?[MgO] = - 6 0 2 , AH?[Mg] = +7 36 , AH?a2[O] = + 798
AHa?[Mg] = +1 48 , AH,5[Mg] = +1 45 0,
AHa? [0 ] = +2 49 , AH?,[0] = - 1 4 1 ,
Mg2+(g) + O2 (g) -------- Mg O(s)
AH * several stepsN Mg(s) + j0 2(g) Figure 6. 7.7 Enthalpy cycle to determine the la ttice ene rgy of magnesium oxide
Mg2+(g) + O2 (g)
0
Exam tips
■ Take care with the signs for the first and second electron affinities. When added together they may give a positive value. ■
Remem ber that an arrow, 1, pointing downwards means that energy is given out and an arrow, T, pointing upwards means that energy is absorbed. Take care to draw these correctly if there is a second electron affinity involved.
72
Figure 6.7.2 Born-Haber cycle to determine the lattice energy of magnesium oxide
From the enthalpy change values given above we can see that: AH? = AHa?[Mg] + AHa? [0 ] + AH?[Mg] + AH,? [Mg] + AH?al[0 ] + AH?a2[0 ] AH? =( +1 48 ) + (+ 249) + (+ 736) + (+ 1450) + (-1 41 ) + ( + 798) = +3240kjmoi~1 AH?tt = AH? - AH? = -602 - (+ 3240) = -38 42k Jm ol-‘
Born-Haber cyc les for AlCl3, Na20 and Al20 3 Alum inium chloride, A1C13/h as an Al3+ ion and 3 C l ions. So in the calculation: * AH®[C1] and AH^JCl] must be multiplied by 3. The energy required to form Al3+(g) from Al(g) is AH® + A
+ AHf
Sodium(i) oxide, Na20 , has 2 Na+ ions and 1 O2 ion. So in the calculation: AH2’[Na| and AH,®[Na] must be multiplied by 2. Th e energy required to form 0 2'(g) fr om O(g) is AHfal + AH®2 Alum inium oxide, Al20 3, has 2 Al3+ ions and 3 O 2- ions. So in the calculation: AH®[Alj must be multiplied by 2 and AH®[02] must be multiplied by 3. The energy required to form Al3+(g) from Al(g) is AH® + AHf + AH® Th e energy required to form O2 (g) from O(g) is AH®j + AH®2
What affects the value of lattice energies? Lattice energy arises from the electrostatic force of attraction between oppositely charged ions. The value depends on: % The size of the ions. Small ions are attracted more strongly to ions of the opposite charge than large ions. The charge density is higher on smaller ions. So, for any given anion e.g. Cl", the lattice energy gets less exothermic as the size of the cation increases: L iC l = - 8 4 8 k j m o b 1 N aC l = - 7 8 0 k J m o l - ‘
KC1 = -7 1 1 k J m o l - ‘
The same effect is seen if the cation size is kept constant: LiF = —1031 kj mo l"1 LiCl = - 8 4 8 k J m o l ‘
LiBr = —803 kj mol-1
The charge on the ions. Ions with a high charge are attracted more strongly to ions of the opposite charge than ions with a low charge. If we compare the lattice energies of LiF and MgO, which have similar sized ions, we see that the lattice energy of MgO which has doubly charged ions is very much higher than LiF which has singly charged ions:
AH£«[LiF] = - 1 0 49 kj mol"1 AH£tt[MgO] = -3 9 2 3 kj m ol'1 The way the ions are arranged in the ionic lattice. The arrangement of the ions also affects the value of the lattice energy. It has less effect, however, than ionic charge and size.
Exam-style questions - Module 1 Answers to a ll exam -style questions can be found on the accompanying CD 10 25.0 cm 3of aqueous iron(n) sulphate required
M u l t ip l e - c h o i c e q u e s t io n s
20.5 cm3of 0.02 moldrrr3potassium manganate(vn) 1
Wh at is the relative atom ic mass (Ar) of bromine, given that the relative abundance of Br 79 is 50.5%
for complete reaction. The ionic equation for the reaction is:
and Br81 is 49.50% ?
2
A
(79 x 0.4 95 )-(8 1 x 0.505)
B
(79 x 0.4 95) + (81 x 0.505 )
C
(79 x 0.5 05 )-(81 x 0.495)
D
(79 x 0.505) + (81 x 0.495)
5Fe 2+(aq) + M n 0 4 (aq) + 8H +(aq) —» 5Fe3+(aq) + Mn2+(aq) + 4H 20 (l ) Wha t is the concentration (g m o l1) of the solution of iron(n) sulphate? (Ar values: Fe = 56; S = 32; O = 16)
Which of the following particles will be deflected to the greatest degree by an electric fietd? A
Alpha particles
B 3
25
20.5
C Electrons
Protons
D
B 0.02 x —
Neutr ons
25.0
X is a radioisotope. Which of the following
undergoes beta decay? 52 Y
A
4
27
B
'
*
D
52 Y
28
C
'
48 Y 24
d
'
U
56 Y
for the copper(i) ion? [Ar] 3d 10 4 s 1
C 5
B [Ar] 3d 9 4 s 2
[Ar] 3d 9 4 s 1
D
0.02x^5^x5x152 25.0 25 0 D 0.0 2 x ——— x 5 x 152 20.5
S t r u c t u re d q u e s t i o n s 11 a
[Ar] 3d 10
Which of the following equations represents the second ionisation energy of sodium?
6
x l x 152 5
28 '
Which of the following is the electronic configuration A
5
C
represents the resulting nuclide, Y, when ^ X
M
A 0.02 x ^^5 x 1 x 152
Describe what happens in terms of loss or gain of electrons AND change in oxidation number to an oxidising agent.
b
[2]
A standard solution of potassium manganate(vn) can be used to find the concentration of
A
Na(g ) -> Na*(g)
+ e~C
Na+(g)Na 2+(g)+ e
B
Na(g ) —>Na+(g)
+ 2e D
Na +(g) ->Na 2+(g) + 2e~
hydrogen peroxide. The two half equations are M n0 4~+ 8H ++ 5e
An elem ent has the following five successive
H20 2
ionisation energies measured in kj mol-1:
i
Mn2++ 4H 20
0 2 + 2H ++ 2e-
Using the half equation s above, write a
1st
736
4th
10 500
balanced equation to show the reaction
2nd
1450
5th
13 600
3rd
7740
between p otassium mangangate(vn) and hydrog en peroxide. ii
Wh at group of the Periodic Table is the element likely to belong to? A
V
B IV
C III
[2]
By making reference to the changes in oxidation numbers of the elements, show that potassium manganatefvn) is the oxidising
D II
agent and hydrogen peroxide is the reducing 7
Whic h of the following halides of potassium woutd have the largest lattice energy? A
8
KF
B KCl
C KBr
agent. c
D Kl
When aqueous bromine was added to an aqueous solution of the potassium halide, KX , a brown
Which of the following describes the shape of the
solution was observed.
car bo nate ion, C 0 32~?
i
Write the formula of the halide ion, X.
ii
Using you r answer in part i above, write a
A
octahedral
C
B
trigonal planar
pyramidal
D tetrahe dral
Wha t is the molar mass of 0.0 7g of a gas which occupies a volume of (1
A
74
120
cm 3 at r.t.p.?
mole of gas occupies a volume of 24dm 3at r.t.p.) 14
B 16
C 28
D 32
[1]
balanced equation forth? reaction between aqueous bromine and KX.
9
[4]
[ 2 ]
iii Wh ich of the eleme nts, Br2or X2, has the greater oxidising ability? Explain your answer using an appropriate half equation.
[4'
