Practice Paper 2 topic 16 and 17

Some Practice Short Answer Questions – Kinetics and Equilibrium
1. Methanol may be produced by the exothermic reaction of carbon monoxide gas and hydrogen gas.
CO(g) + 2H2(g)
(a)
CH3OH(g)
∆HO = –103 kJ
State the equilibrium constant expression, Kc, for the production of methanol.
(1)
(b)
State and explain the effect of changing the following conditions on the amount of methanol present at
equilibrium:
(i)
increasing the temperature of the reaction at constant pressure.
(2)
(ii)
increasing the pressure of the reaction at constant temperature.
(2)
(c)
The conditions used in industry during the production of methanol are a temperature of 450 °C and
pressure of up to 220 atm. Explain why these conditions are used rather than those that could give an
even greater amount of methanol.
(2)
(d)
A catalyst of copper mixed with zinc oxide and alumina is used in industry for this production of
methanol. Explain the function of the catalyst.
(1)
(Total 8 marks)
2. The equation for the main reaction in the Haber process is:
N2(g) + 3H2(g)
(i)
2NH3(g)
∆H is negative
Determine the equilibrium constant expression for this reaction.
(1)
(ii)
State and explain the effect on the equilibrium yield of ammonia with increasing the pressure and the
temperature.
(4)
(iii)
In practice, typical conditions used in the Haber process involve a temperature of 500 °C and a
pressure of 200 atm. Explain why these conditions are used rather than those that give the highest
yield.
(2)
(iv)
At a certain temperature and pressure, 1.1 dm3 of N2(g) reacts with 3.3 dm3 of H2(g). Calculate the
volume of NH3(g), that will be produced.
(1)
(v)
Suggest why this reaction is important for humanity.
(1)
(vi)
A chemist claims to have developed a new catalyst for the Haber process, which increases the yield of
ammonia. State the catalyst normally used for the Haber process, and comment on the claim made by
this chemist.
(2)
(Total 11 marks)
3. Consider the following graph of ln k against
1
(temperature in Kelvin) for the second order decomposition
T
of N2O into N2 and O.
N2O → N2 + O
1
/ 10  3 K–1
T
(a)
State how the rate constant, k varies with temperature, T.
(1)
(b)
Determine the activation energy, Ea, for this reaction.
(3)
(c)
The rate expression for this reaction is rate = k [N2O]2 and the rate constant is
0.244 dm3 mol–1 s–1 at 750 °C.
A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose. Calculate the rate when
10 % of the N2O has reacted.
(2)
(Total 6 marks)
4. Alex and Hannah were asked to investigate the kinetics involved in the iodination of propanone.
They were given the following equation by their teacher.

H ( aq)

 CH2ICOCH3(aq) + HI(aq)
CH3COCH3(aq) + I2(aq) 
Alex’s hypothesis was that the rate will be affected by changing the concentrations of the propanone and the
iodine, as the reaction can happen without a catalyst. Hannah’s hypothesis was that as the catalyst is involved
in the reaction, the concentrations of the propanone, iodine and the hydrogen ions will all affect the rate.
They carried out several experiments varying the concentration of one of the reactants or the catalyst while
keeping other concentrations and conditions the same, and obtained the results below.
Composition by volume of mixture / cm3
Experiment
(a)
1.00 mol dm–3
CH3COCH3(aq)
Water
1
10.0
2
Initial rate
/ mol dm–3 s–
1.00 mol dm–3
H+(aq)
5.00 × 10–3
mol dm–3 I2 in KI
60.0
10.0
20.0
4.96 × 10–6
10.0
50.0
10.0
30.0
5.04 × 10–6
3
5.0
65.0
10.0
20.0
2.47 × 10–6
4
10.0
65.0
5.0
20.0
2.51 × 10–6
1
Explain why they added water to the mixtures.
(1)
(b)
(i)
Deduce the order of reaction for each substance and the rate expression from the results.
(2)
(ii)
Comment on whether Alex’s or Hannah’s hypothesis is correct.
(1)
(c)
Using the data from Experiment 1, determine the concentration of the substances used and the rate
constant for the reaction including its units.
(3)
(d)
(i)
This reaction uses a catalyst. Sketch and annotate the Maxwell-Boltzmann energy distribution
curve for a reaction with and without a catalyst on labelled axes below.
(3)
(ii)
Describe how a catalyst works.
(1)
(Total 11 marks)
KEY
1. (a) (Kc =)
(b)
[CH 3 OH]
;
[CO][H 2 ] 2
Do not award mark if incorrect brackets are used or brackets are missing.
(i)
(ii)
(c)
(d)
1
amount (of methanol)/product decreases / less methanol;
(forward reaction) exothermic / reverse reaction endothermic / OWTTE;
2
amount (of methanol)/product increases / more methanol;
3 gas molecules/mol → 1 / decrease in volume / fewer gas
molecules on right hand side/products / more gas molecules
on left hand side/reactants;
2
high pressure expensive / greater cost of operating at high pressure;
lower temperature – lower (reaction) rate;
increases rate of forward and reverse reactions (equally) / lowers
activation energy/Ea (of both the forward and reverse reaction
equally) / provides alternative path with lower activation energy/Ea;
Accept reactants adsorb onto the catalyst surface and bonds
weaken resulting in a decrease in the activation energy.
2
1
[8]
2. (i)
(ii)
(iii)
(Kc =)
[NH 3 ] 2
[N 2 ][ H 2 ]3
(ignore units);
1
Increasing the pressure:
Yield increases / equilibrium moves to the right / more ammonia;
4 gas molecules → 2 / decrease in volume / fewer gas molecules
on right hand side;
Increasing the temperature:
Yield decreases / equilibrium moves to the left / less ammonia;
Exothermic reaction / OWTTE;
4
Higher temperature increases rate;
Lower pressure is less expensive / lower cost of operating at low
pressure / reinforced pipes not needed;
2
Do not award a mark just for the word “compromise”.
(iv)
2.2 (dm3);
1
Penalize incorrect units.
(v)
(vi)
Fertilizers / increasing crop yields;
Production of explosives for mining;
1 max
Fe/iron;
Allow magnetite/iron oxide.
Claim is not valid since catalysts do not alter the yield/position
of equilibrium / only increase the rate of reaction;
2
3. (a) k increases with increase in T / k decreases with decrease in T;
Do not allow answers giving just the Arrhenius equation or involving
ln k relationships.
(b)
gradient = –Ea/R;
–30000 (K) = –Ea/R;
Allow value in range –28800–31300 (K).
Ea =(30000 × 8.31=) 2.49 × 105 J mol–1 /249 kJ mol–1;
Allow value in range 240–260 kJ mol–1.
Allow [3] for correct final answer.
(c)
4. (a) to maintain a constant volume / OWTTE;
(i)
(ii)
(c)
2
1
[H+] order 1, [CH3COCH3] order 1, [I2] order 0;
(rate =) k[H+] [CH3COCH3];
Award [2] for correct rate expression.
Allow expressions including [I2]0.
2
neither were correct / Alex was right about propanone and wrong
about iodine / Hannah was right about propanone and hydrogen ions
but wrong about iodine / OWTTE;
1
[CH3COCH3] = 0.100 mol dm–3 and [H+] = 0.100 mol dm–3;
k=
4.96  10 6
= 4.96 × 10–4;
(0.100  0.100 )
mol–1 dm3 s–1;
Ignore calculation of [I2].
No ECF here for incorrect units.
(d)
3
0.9 × 0.200 = 0.180 (mol dm–3);
rate = (0.244 × (0.180)2 =) 7.91 × 10–3 mol dm–3 s–1;
Award [2] for correct final answer.
Award [1 max] for either 9.76 × 10–3 mol dm–3 s–1 or
9.76 × 10–5 mol dm–3 s–1.
(b)
1
(i)
axes correctly labelled x = energy/velocity/speed, y = number/% of
molecules/particles/probability;
graph showing correct curve for Maxwell-Boltzmann distribution;
If two curves are drawn, first and second marks can still be scored,
but not third.
Curve(s) must begin at origin and not go up at high energy.
3
(ii)
two activation energies shown with Ecat shown lower;
Award the mark for the final point if shown on an enthalpy
level diagram.
3
catalyst provides an alternative pathway of lower energy / OWTTE;
Accept catalyst lowers activation energy (of reaction).
1

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