CAPE 2015: SUGGESTED ANSWERS MODULE 1: FUNDAMENTALS IN CHEMISTRY 1. (a)(i) Define the term ‘bond energy’
[1
mark] The bond energy represents: The quantity of energy evolved when the bond is formed. Also, the bond energy is the amount of energy absorbed when the bond is broken. (ii) The bond lengths and bond energies of carbon-carbon single, double and triple bonds are listed below. C-C
C=C
C=C (triple)
Bond lengths
1.54 Ȃ
1.34 Ȃ
1.20 Ȃ
Bond energy
348 KJ mol-1
614 KJ mol-1
839 KJ mol-1
State the relationship between the strength of a covalent bond and its length [1 mark] The shorter the bond, the stronger it is. That means that more energy will be required to break it. This is why the carbon-carbon triple bond is the shortest, and thus has the highest bond energy (839 KJ mol-1) (b) Bond energies can be used to estimate the enthalpies of reactions in which bonds are broken and new bonds are formed. Consider the gas-phase reaction between methane (CH4) and chlorine to produce methylchloride (CH 3Cl) and hydrogen chloride (i) Write a balanced equation to represent the reaction above [2 marks] CH4 (g) + Cl2 (g)
CH 3Cl (g) + HCl (g)
(ii) Using bond energy values from Table 1, calculate the enthalpy change of reaction ∆Hrxn for the equation required in (b) (i) [3 marks]
What really happened: CH4 (g) + Cl2 (g) By looking at the equation:
CH 3Cl (g) + HCl (g)
What was broken: (*Energy is taken in to break a bond) One of the C-H bonds from CH4 = (+410 KJ) The Cl-Cl bond = (+244 KJ) What was formed: (*Energy is released when a bond is formed) A new C-Cl bond was formed = (-340 KJ) A H-Cl bond was formed = (-431 KJ) We now construct a cycle based on what happened: CH4 (g) + Cl2 (g)
∆H2
CH 3Cl (g) + HCl (g)
∆H1
∆H3
∆H2
CH3 + H + Cl + Cl
∆H3
Using ∆H1 = ∆H2 + ∆H3 = (654) + (-771) = -117 KJ (iii) Is the reaction in (b)(i) exothermic or endothermic? [1 mark] It would be exothermic, since ∆H is negative (iv) On the axes provided below, draw the energy profile diagram for the reaction in
Energy (KJ mol1 )
(b)(i) [2 marks] This of course depends on your ∆H value…
Reactants CH4 (g) + Cl2 (g
EA
Products ∆H
CH3Cl (g) + HCl (g) of reaction (c) A student is asked Progress to determine the enthalpy of the neutralization reaction between 75cm3 of 1.00M hydrochloric acid and 75 cm 3 of 1.00 M potassium
hydroxide solution. Outline the experimental steps (including calculations) required to obtain and accurate value. [Assume that the densities of the solutions of acid and base and their heat capacities are equal to the density (1 g cm -3) and heat capacity (4.18 J g-1oC-1) of water] Steps:
Measure 75cm3 of 1.00M hydrochloric acid into a polystyrene cup Stir gently for three minutes, taking the temperature reading every minute for 3 minutes Record the constant temperature as the initial temperature Add 75cm3 of 1.00M potassium hydroxide and stir for the next seven minutes, recording the temperature of the mixture at 1 minute intervals. Record the
highest temperature as the final temperature The temperature change (∆T) = initial temp=final temp Calculations: Using EH = mc∆T where m= mass of solution= (75 +75 = 150g) and c = 4.18 J g-1oC-1 The number of moles of HCl (or NaOH) used: 1000 cm3 = 1 mole 1cm3= 1/1000 75 cm3 = 1/1000 x 75 = 0.075 moles We say 0.075 moles = EH (from above) So 1 mole = EH/0.075 (this would be the molar enthalpy of neutralization) MODULE 2: KINETICS AND EQUILIBRIA 2. (a) Define EACH of the following terms: (i) Standard electrode potential of a half-cell
[2
marks] The potential of that half cell relative to a standard hydrogen half-cell under standard conditions. (ii)
Standard cell potential of an electrochemical cell
[2
marks] The sum of the potential of two half cells connected together in such a way that reduction occurs in one half cell and oxidation in the other OR This is the maximum potential difference obtained when two half cells are connected together under standard conditions OR
This is a cell potential against which the cell potential of other elements can be measured
(b) Consider the following (unbalanced) equation which describes the process that is taking place in an electrochemical cell under standard conditions: Al (s) + Sn2+ (aq) → Al3+ (aq) + Sn (s) As you can see, Al is being oxidized (change from 0 to +3). Oxidation occurs at the anode (An Ox) Also, Sn is being reduced (change from +2 to 0). Reduction occurs at the cathode (Red Cat) (i)
Write the ionic half-equation for the reaction taking place at EACH of the electrodes [2 marks]
ANODE: Al (s) (aq) → Al3+ (aq) + 3e CATHODE: Sn2+ (aq) + 2e → Sn (s) (ii)
Write the cell diagram
[1
mark] The cell notation would be: Al|Al3+ || Sn2+|Sn (iii)
Draw a well labeled diagram of the electrochemical cell. Include the direction of electron flow [6 marks]
Electron flow V Tin rod (cathode)
Aluminium rod (anode) Salt bridge
(iv)
Solution B: 1 moldm-3 Sn2+ ions
Solution A: 1 moldm3 Al3+ ions
For EACH electrode shown in Table 2, select the E Θ value to determine EΘ cell
[2 marks] Electrode Reaction Ag+ + e
Ag
EΘ at 298 K (250C) (volts) +0.80
Al3+ + 3e
Al
-1.66
Ba2+ + 2e
Ba
-2.90
Sn2+ +2e
Sn
-0.14
Sn2+
+0.15
Sn4+ +2e
Maybe I am not understanding correctly,
but.. is the question phrased
correctly? What were they asking for? The overall emf of the cell…? Al Al3+ + 3e EΘ = +1.66 2+ Sn +2e Sn EΘ = -0.14 2Al + 3Sn2+ 2 Al3+ + 3Sn
(Multiply this eq’n by 2) (Multiply this eq’n by 3) EΘ = (+1.66) + (-0.14) = +1.52 V
MODULE 3: CHEMISTRY OF THE ELEMENTS 3. The Group IV elements exhibit oxidation states of +2 and +4 in their compounds. (a) Complete Table 3 for the oxides of carbon and lead. [4 marks] TABLE 3: OXIDES OF GROUP IV ELEMENTS CO
CO2
PbO
PbO2
Acid/Base nature
neutral
Acidic
Amphoteric
Amphoteric
Thermal stability
Unstable: Readily oxidized to dioxide
Stable
Stable
Unstable: decomposes to form PbO
Oxidation state of Group IV element
+2
+4
+2
+4
How to get there? In CO2, the oxidation number of carbon would be:
x + (2 x -2) = 0 x -4 = 0 x = +4 (b) (i) Explain the relative stabilities of the +2 oxidation states of the oxides of carbon and lead [4 marks] Carbon is most stable in the +4 state (as CO 2) and least stable in the +2 state (as CO) Lead is most stable in the +2 state (as PbO) and least stable in the +4 state (as PbO2)
Stability
This is illustrated on the graph below: (Optional!)
+2 state
+4 state
C
Si
G e
Sn
Group IV Elements
Pb
Lead is a metal, and so prefers to lose two electrons to be stable. This gives Pb an oxidation number of +2. Carbon being a non- metal prefers to share electrons. It is less energetically feasible for C to lose two electrons and exist in the +2 state (ii) Use the electrode potential value in the electrode reaction for lead ions to explain the relative stabilities of the +2 and +4 oxidation states of lead: [2 marks] Pb4+ + 2e
Pb2+ EΘ =+1.80 V
The EΘ value for Pb going from the +4 to the +2 state is large and positive. This means it is very energetically feasible for Pb to be converted from the +4 to the +2 state. This suggests that for Pb, the +2 state is much more feasible than the +4.
(c) Describe what should be observed when (i)
Concentrated sodium hydroxide is added to solid lead (IV) oxide [1 mark]
**Since PbO2 is amphoteric, it will react with both acids and bases to form a salt and water A white precipitate (which is the salt) will be formed and may dissolve in the (colorless) water which is also formed (ii)
Concentrated hydrochloric acid is added to solid lead (IV) oxide [2 marks]
Both lead (IV) chloride and water will be formed. Since lead chloride is insoluble, there will be a white precipitate in a colorless liquid (water) (d) Describe a test to identify Pb2+ ions in solution [2 marks]
Potassium iodide (KI) can be added to the ions in solution. If Pb 2+ is present, there will be a bright yellow precipitate. Pb2+ (aq) + 2 I- (aq) → PbI2 (s)
MODULE 1: FUNDAMENTALS IN CHEMISTRY
4. (a) The atoms of certain elements contain nuclei in which the ratio of neutrons to protons is greater than 1. These nuclei tend to emit radiation in order to bring the ratio closer to 1. (i)
Describe the THREE types of radiation that an unstable atom may emit. Include in your answer the symbols and penetrating power of EACH type of radiation [6 marks]
EMISSION
SYMBOL
RELATIVE PENETRATING POWER
α Particles
Helium nuclei
1
4 2 He 2 β Particles
Electrons
100
0e 1 γ Rays (ii)
Electromagnetic waves
10,000
Americium-241 (Am-241) decays via alpha particle emission. Write the nuclear equations to show the new element that forms when an atom
of
241
95
Am decays via the emission of 2 alpha particles [2 marks] +
241
4
+
4
233
2+ orbitals of principal 2+ (b) (i) Draw quantum number 2. Am the diagrams of the atomic He2+ He2+ Pa
2 Include 95 x, y and z axes in your drawing
2
91
[2 marks] Principal quantum number is the same as the shell number. In the second shell, there are two orbitals: a 2s and a 2p.
(ii) Comment on the similarity and difference in the electronic configuration of K, Sc and Zn2+ given below
K: 1s2 2s2 2p6 3s2 3p6 4s1 Sc: 1s2 2s2 2p6 3s2 3p6 3d1 4s2 Zn2+: 1s2 2s2 2p6 3s2 3p6 3d10
Similarity: All of the elements have more than 18 electrons, and they have the same arrangement of electrons up until the 3p orbital (1s 2 2s2 2p6 3s2 3p6). This is sometimes represented as [Ar] followed by the rest of the electronic configuration Difference: Sc and Zn both have electrons in the d orbitals, but K does not. This is because Sc and Zn are transition metals and K is a Group 1 metal OR K and Sc have electrons in the 4s orbital, but Zn 2+ does not. This is because the configuration of Zn is 1s2 2s2 2p6 3s2 3p6 3d10 4s2. For Zn2+, 2 electrons were lost from the 4s orbital first, making it 1s 2 2s2 2p6 3s2 3p6 3d10
MODULE 2: KINETICS AND EQUILIBRIA 5. A buffer consisting of H2PO4- and HPO42- helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. (a) Using the buffer system mentioned above, describe how the solution maintains and almost constant pH even when small amounts of acid or alkali are added to the solution [6 marks] This buffer consists of a solution of a weak acid in the presence of one of its salts (that is, H2PO4- and Na2HPO4). Two reactions will be happening: Equation 1:
H2PO4-
H+ + HPO42-
Equation 2:
Na2HPO4 2 Na+ + HPO42-
If acid (H+) is added, Equation 1 is affected. This is because this equation also contains H+ ions. According to Le Chatelier, in Equation 1, the equilibrium will shift to the left. This gets rid of the added H + ions and pH returns to normal.
If alkali is added, (OH-), Equation 1 is still affected. Two things happen: 1. The added OH- ions combine with H+ to form H2O, which is neutral. The OH- has been removed. However, in the process we have also lost some H+ ions. 2. In order to correct this, according to Le Chatelier, in Equation 1, the equilibrium will shift to the right. This allows the H + ions to be replaced. The pH returns to normal. Optionally: Equation 2 has not yet been used. This equation produces a large amount of HPO42- ions. In the event that a large amount of acid (H +) is added to the solution, then the HPO42- from Equation 2 will be used to trap the added acid (H+).
(b) Calculate the pH of a soft drink in which the major buffer ingredients are 6.5 g of NaH2PO4 and 8.0 g of Na2HPO4 per 355 cm3 of solution [Ka (H2PO4-) = 6.4 x 10-8 moldm-3] [Relative atomic mass: H=1, Na = 23, O = 16, P = 31]
Using the formula: [H+] = Ka [acid] [salt] And we know that pH = -lg [H+] Since we are told that Ka (H2PO4-) = 6.4 x 10-8 moldm-3, if it has a Ka (acid dissociation constant) value then this is the acid! (NaH 2PO4) Which makes the other species the base (Na 2HPO4)
So the concentration of the acid would be: 1 mole (NaH2PO4) = (23 x 1) + (1 x 2) + (31 x 1) + (16 x 4) = 120 g 120 g = 1 mole 1g = 1/120 6.5g = 1/120 x 6.5 = 0.05417 moles
355 cm3 = 0.05417 moles 1 cm3 =
0.05417
1000cm3 =
/355
0.05417
/355 x 1000 = 0.1526 moles…. Concentration: 0.1526 moldm-3
So the concentration of the salt would be: 1 mole (Na2HPO4) = (23 x 2) + (1 x 1) + (31 x 1) + (16 x 4) = 142 g 142 g = 1 mole 1g = 1/142 8g = 1/142 x 8 = 0.0563 moles
355 cm3 = 0.0563 moles 1 cm3 =
0.0563
1000cm3 =
/355
0.0563
/355 x 1000 = 0.1587 moles…. Concentration: 0.1587 moldm-3
Plug values into formula: [H+] =
(6.4 x 10-8 moldm-3) x (0.1526 moldm-3) = 6.1 x 10-8 moldm-3 (0.1587moldm-3)
Work out pH: pH = -lg [[H+] = -lg [6.1 x 10-8 moldm-3] = 7.21
(c) Many chemical reactions occur in living systems such as the human body. Discuss the importance of biological buffers to the maintenance of a healthy body. (Include an example of a chemical reaction of a blood buffer)
[4 marks]
In living things, the pH of the blood must be kept constant, or else they will die. The normal pH of the blood is 7.4 There are two reactions of interest in the buffering system of blood: Equation 1:
CO2 (aq) + H2O (l)
H2CO3 (aq) (carbonic
acid) Equation 2:
H2CO3 (aq)
→
H + + HCO3 -(aq)
(bicarbonate) If a person has been running rapidly for a long time, he usually starts gasping for breath. This means that he is struggling to inhale oxygen and exhale carbon dioxide. If carbon dioxide is allowed to build up in the blood, it combines with water to form carbonic acid. This carbonic acid rapidly breaks down into H + and CO32- ions. The H+ ions can cause the blood to become acidic. This can be lethal to cells. By exhaling carbon dioxide rapidly, there will be less reactant in Equation 1. According to Le Chatelier, the equilibrium will shift to the left. This causes the carbonic acid to break up into CO 2 and H2O, instead of harmful H+ ions. The CO2 produced can then be exhaled.
MODULE 3: CHEMISTRY OF THE ELEMENTS 6. (a) The atomic and ionic radii of the Group II elements gradually increase down the group. Outline the reasons for this trend [3 marks] The atomic radius can be visualized as the distance between the nucleus and the outermost shell. As the Group is descended the atoms have an increase in the number of electrons. Be only has 2 electrons, while Ba has 56 (optional). Since Ba has
more electrons, there will also be more shells, and so the distance between the nucleus and the outermost shell increases. The Group II elements lose 2 electrons to become stable, thus forming cations with a charge of +2. Since they all lose one shell, there is still an increase in the number of shells as the Group is descended. Hence the radius of the ion formed also increases.
(b) Account for the variation in the melting points of the Group II elements from magnesium to barium [4 marks] There is actually a decrease in melting points as the Group is descended. All of the Group 2 elements are metals and so are held together by metallic bonds. This is the attraction between the cation and the sea of delocalized (lost) electrons. It would make sense to reason out that, as the Group is descended, there are more electrons and so there would be more metallic bonds, hence an increase in melting points. However, this is not so for Group 2. What happens is that since the ionic radius is becoming larger, the attraction for the delocalized electrons becomes weaker. Thus there is a decrease in melting points. **I am not sure if you were supposed to know this… or if CAPE wanted you to just reason it out and go with what is expected…
(c)Table 4 shows the observations when 0.1 moldm -3 solutions of the metal ions are treated with 1 moldm-3 Na2SO4 TABLE 4: OBSERVATIONS OF REACTIONS Solution of Group II cations
Observations with 1 moldm-3 Na2SO4
0.1 moldm-3 Mg2+
No precipitate
Ca2+
Thin, white precipitate
Sr2+
White precipitate
Ba2+
Thick, white precipitate
Account for the trend indicated in Table 4 marks]
[5
When the ions are reacted with sodium sulphate, the Group 2 sulphates are being formed. It is observed that as the group is descended, there is a general decrease in solubility of the sulphates. This is seen from the observations, as BaSO 4 is a thick white precipitate (less soluble) than MgSO 4 which is no precipitate (very soluble) We can explain this by considering the sulphates being dissolved in water as two steps: Step 1: The ions in the solid must be separated: MgSO4 (s) Mg2+ (g) + SO42- (g) (Remember: An ionic lattice is being broken up. The energy needed for this is the reverse of the lattice energy -ΔHlatt). Step 2: The separated ions must be hydrated (completely surrounded) by water Mg2+ (g) + SO42- (g) + (aq) Mg2+ (aq) + SO42- (aq) The smaller the ionic radius, the less hydration energy will be required. Hence BeSO4 will require the least hydration energy, while BaSO 4 will require the most. It makes sense that the greater the total energy for these processes, the less soluble the compound will be. As the group is descended, solubility of the sulphates decreases.
(d) Explain the variation in the thermal decomposition of the nitrates of the Group II elements [3 marks] The oxide ion is smaller than the nitrate ion. It will follow that the oxide ion is more stable than the nitrate ion. This explains why the nitrates of the Group II elements decompose under heat to form Group II oxides. 2 X(NO3)2 (s) 2 XO (s) + 4 NO2 (g) + O2 (g) Where X is any Group 2 element (Be, Mg, Ca, Sr or Ba) As the Group is descended, it requires more heat to decompose the nitrates. This is because the stability of the nitrates increases as Group II is descended. The more reactive an element is, when it forms a compound, that compound will be very stable. Since reactivity increases as the Group is descended, the most reactive Group 2 element would be Ba. Therefore Ba(NO 3)2 would be most stable
No matter how many mistakes you have made, this paper was ONLY 40%. You still have a fighting chance to get that distinction. CAPE accepts many answers, and I have only given a few of them here. And marks CAN be given for different things than I have suggested… Keep believing in yourself… Yes, you can still do it!
