BThe addition of a catalyst lowers the activation energy of the forward reaction but not that of the reverse reaction.[1]
A catalyst speeds up the rate of a reaction by providing an alternative pathway with a lower activation energy (A is correct). A catalysed and an uncatalysed reaction have the same enthalpy change (C is correct).
The activation energies for both the forward and backward reactions are lowered by the same amount (B is incorrect). Hence, a catalyst does not change the position of equilibrium (D is correct). Instead, it shortens the time to reach equilibrium.
B8.27 × 10−5[1]
At equilibrium, [NH3] = [HNO3] = 0.01130.83 × 66.8100 = 0.00909 mol dm−3
Kc = [NH3][HNO3] = (0.00909)(0.00909) = 8.27 × 10−5 mol2 dm−6
C2 and 3 only[1]
B2P ⇌ 2Q + R[1]
Let a and b be the stoichiometric coefficients of P and Q: aP ⇌ bQ + R. Equilibrium amounts: P (2 − ax), Q bx, R x.
At equilibrium, total amount = 2 − ax + bx + x = 2 + x, so 2 − ax + bx = 2. Therefore a = b, i.e. P and Q have the same stoichiometric coefficients.
Kc = [CO2][H2][CO][H2O][1]
Note[H2O] must be included, as this is a homogeneous system: all the species are gases.
| CO(g) | H2O(g) | CO2(g) | H2(g) | |
|---|---|---|---|---|
| initial / mol | 0.200 | 0.200 | 0 | 0 |
| change / mol | −0.2 × 0.775 | −0.2 × 0.775 | +0.2 × 0.775 | +0.2 × 0.775 |
| equilibrium / mol | 0.045 | 0.045 | 0.155 | 0.155 |
Kc = (0.155)2(0.045)2[1]
= 11.86 ≈ 11.9[1]
−18.4 × 103 = −(8.31)(600) ln Kc
ln Kc = 3.690, so Kc = 40.04 ≈ 40.0[1]
NoteConvert ΔG to J mol−1 first.
The dimerisation will be favoured by high pressure. At high pressure, the position of equilibrium will shift to decrease pressure by decreasing the total number of gaseous molecules present. Since there are two molecules of gas on the left-hand side of the equation and one on the right, the forward reaction is favoured, and the position of equilibrium shifts to the right to form more N2O4.[1]
The dimerisation will be favoured by low temperature. At low temperature, the position of equilibrium will shift to increase temperature. Since the forward reaction is exothermic, the forward reaction is favoured, and the position of equilibrium shifts to the right to form more N2O4.[1]
Tube in iced water: pale yellow. Tube in hot water: brown.[1]
In iced water, the low temperature favours the exothermic forward reaction, so the position of equilibrium shifts to the right to form more colourless N2O4. In hot water, the high temperature favours the endothermic backward reaction, so the position of equilibrium shifts to the left to form more brown NO2.[1]
Equilibrium amounts / mol: H2O 4 − x, CO 4 − x, CO2 x, H2 x. Let the volume of the vessel be V dm3.
Kc = (x/V)2((4 − x)/V)2 = 9[1]
Solving, x = 3 mol. Hence, the equilibrium amount of CO = 1 mol[1]
(i) At a lower temperature, the rate at which equilibrium is attained is slower. By Le Chatelier's Principle, since ΔH < 0, the position of equilibrium shifts to the right and produces heat so as to increase the temperature. Hence, the amount of CO2 is higher at equilibrium. Kc is also increased.[1]
(ii) A decrease in volume, which leads to an increase in concentration, favours neither the backward nor the forward reaction, since the number of molecules on both sides of the reaction is the same. The amount of CO2 produced is the same. The rate at which equilibrium is attained increases.[1]
(iii) In the presence of a suitable catalyst, the rate at which equilibrium is attained is faster. However, there is no effect on the amount of CO2 formed, and Kc remains the same.[1]
Advantage: by Le Chatelier's principle, a lower temperature favours the forward exothermic reaction, and hence the yield of ammonia increases.[1]
Disadvantage: if the temperature is too low, the reaction is too slow.[1]
ΔG = −2.303RT log Kc, so log Kc = (−71.14 × 1000) ÷ (−2.303 × 8.31 × 298) = 12.47[1]
Kc = 1012.47 = 2.95 × 1012 mol−½ dm3/2[1]
NoteR is in J K−1 mol−1, so ΔG must be in J mol−1, not kJ mol−1.