IB DP Chemistry Reactivity 2.3 How far? The extent of chemical change HL Paper 2- Exam Style Questions - New Syllabus
Question
Hydrogen is manufactured from methane by a process called steam reforming:
\(\displaystyle \mathrm{CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)} \qquad \Delta H = +206~\text{kJ mol}^{-1}\)
(a) Deduce the equilibrium constant, \(K_c\), expression for the reaction. [1]
(b) Predict, with a reason, the effect of increasing the temperature on the position of equilibrium. [1]
(c) Explain why the reaction rate increases with temperature, adding annotations to the following Maxwell–Boltzmann graph to assist your explanation. [3]
(d) Annotate this Maxwell–Boltzmann distribution graph in (c) to show the effect of a catalyst. [1]
(e) The reactants, \(\mathrm{CH_4}\) and \(\mathrm{H_2O}\), are passed over nickel metal. The following mechanism is proposed:
\(\mathrm{CH_4 + Ni \rightarrow Ni{\text -}CH_2 + H_2}\) (slow)
\(\mathrm{H_2O + Ni{\text -}CH_2 \rightarrow Ni{\text -}CHOH + H_2}\) (fast)
\(\mathrm{Ni{\text -}CHOH \rightarrow Ni{\text -}CO + H_2}\) (fast)
\(\mathrm{Ni{\text -}CO \rightarrow Ni + CO}\) (fast)
\(\mathrm{H_2O + Ni{\text -}CH_2 \rightarrow Ni{\text -}CHOH + H_2}\) (fast)
\(\mathrm{Ni{\text -}CHOH \rightarrow Ni{\text -}CO + H_2}\) (fast)
\(\mathrm{Ni{\text -}CO \rightarrow Ni + CO}\) (fast)
(i) Deduce, giving a reason, the role of nickel in the reaction. [1]
(ii) Deduce the rate equation corresponding to the mechanism. [1]
(iii) Suggest why experimental confirmation of the rate equation would not prove that the mechanism is correct. [1]
(iv) Sketch a graph of the relationship between the rate of this reaction and the concentration of methane. [1]
(ii) Deduce the rate equation corresponding to the mechanism. [1]
(iii) Suggest why experimental confirmation of the rate equation would not prove that the mechanism is correct. [1]
(iv) Sketch a graph of the relationship between the rate of this reaction and the concentration of methane. [1]
![Rate vs. [CH4] Graph](https://sp-ao.shortpixel.ai/client/to_webp,q_glossy,ret_img,w_303,h_200/https://www.iitianacademy.com/wp-content/uploads/2024/10/Screenshot-2024-10-07-154017.png)
(f) Calculate the standard entropy change, \(\Delta S^\circ\), of the forward reaction, in \(\text{J K}^{-1}\).
\(\mathrm{CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)} \qquad \Delta H = +206~\text{kJ mol}^{-1}\)
Use data from the Chemistry Data Booklet . Actual standard molar entropy values used: \(S^\circ(\mathrm{CH_4(g)}) = 186~\text{J K}^{-1}\text{ mol}^{-1}\), \(S^\circ(\mathrm{H_2O(g)}) = 188.8~\text{J K}^{-1}\text{ mol}^{-1}\), \(S^\circ(\mathrm{CO(g)}) = 197.7~\text{J K}^{-1}\text{ mol}^{-1}\), \(S^\circ(\mathrm{H_2(g)}) = 130.7~\text{J K}^{-1}\text{ mol}^{-1}\). [1]
(g) Calculate the Gibbs free energy, \(\Delta G\), and the equilibrium constant \(K_c\), for the forward reaction, at \(1500~\text{K}\). Use data booklet equations/constants : \(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ\), \(\Delta G^\circ = -RT\ln K\), \(R = 8.31~\text{J K}^{-1}\text{ mol}^{-1}\).
(If you were unable to obtain an answer for part (f) use \(227~\text{J K}^{-1}\), but this is not the correct value.) [2]
(h) Determine the temperature at which the forward reaction becomes spontaneous. Use the relationship \(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ\) and set \(\Delta G^\circ = 0\). [2]
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
Balanced reaction: \(\mathrm{CH_4 + H_2O \rightleftharpoons CO + 3H_2}\). For homogeneous gases, \[ K_c=\frac{[\mathrm{CO}]\,[\mathrm{H_2}]^3}{[\mathrm{CH_4}]\,[\mathrm{H_2O}]}\,. \] A1
[1 mark]
Balanced reaction: \(\mathrm{CH_4 + H_2O \rightleftharpoons CO + 3H_2}\). For homogeneous gases, \[ K_c=\frac{[\mathrm{CO}]\,[\mathrm{H_2}]^3}{[\mathrm{CH_4}]\,[\mathrm{H_2O}]}\,. \] A1
[1 mark]
(b)
The forward reaction is endothermic (\(\Delta H=+206~\mathrm{kJ\,mol^{-1}}\)), so increasing temperature shifts equilibrium to the right (more products) by Le Châtelier’s principle. A1
[1 mark]
The forward reaction is endothermic (\(\Delta H=+206~\mathrm{kJ\,mol^{-1}}\)), so increasing temperature shifts equilibrium to the right (more products) by Le Châtelier’s principle. A1
[1 mark]
(c)
At higher \(T\), the Maxwell–Boltzmann curve flattens and shifts right: the area under the curve beyond \(E_a\) increases, so a greater fraction of collisions have \(E\ge E_a\); additionally, particles move faster so collision frequency increases. (Annotate: new curve to the right; shade increased area \(E\ge E_a\); label \(E_a\).) (M1) A1 A1
[3 marks]
At higher \(T\), the Maxwell–Boltzmann curve flattens and shifts right: the area under the curve beyond \(E_a\) increases, so a greater fraction of collisions have \(E\ge E_a\); additionally, particles move faster so collision frequency increases. (Annotate: new curve to the right; shade increased area \(E\ge E_a\); label \(E_a\).) (M1) A1 A1
[3 marks]
(d)
A catalyst provides an alternative pathway with lower \(E_a\). On the MB plot, draw the same distribution but show a lower \(E_a\) threshold; the shaded “above-\(E_a\)” area increases at the same \(T\). A1
[1 mark]
A catalyst provides an alternative pathway with lower \(E_a\). On the MB plot, draw the same distribution but show a lower \(E_a\) threshold; the shaded “above-\(E_a\)” area increases at the same \(T\). A1
[1 mark]
(e)
Given mechanism: \[ \begin{aligned} &\mathrm{CH_4 + Ni \rightarrow Ni{\text -}CH_2 + H_2} \quad (\text{slow})\\ &\mathrm{H_2O + Ni{\text -}CH_2 \rightarrow Ni{\text -}CHOH + H_2} \quad (\text{fast})\\ &\mathrm{Ni{\text -}CHOH \rightarrow Ni{\text -}CO + H_2} \quad (\text{fast})\\ &\mathrm{Ni{\text -}CO \rightarrow Ni + CO} \quad (\text{fast}) \end{aligned} \]
(i) Role of Ni: Heterogeneous catalyst—provides a surface/alternative pathway; appears in mechanism but is regenerated and not consumed. A1
(ii) Rate equation: Rate is determined by the slow step \(\mathrm{CH_4 + Ni \to \dots}\). Treating the catalyst surface concentration as constant, \(\text{Rate}=k[\mathrm{CH_4}]\) (first order in \(\mathrm{CH_4}\), overall first order). A1
(iii) Different mechanisms can lead to the same rate law; matching the rate equation is necessary but not sufficient to prove the mechanism. A1
(iv) Sketch: Rate vs \([\mathrm{CH_4}]\) is a straight line through the origin (first-order dependence). A1
(ii) Rate equation: Rate is determined by the slow step \(\mathrm{CH_4 + Ni \to \dots}\). Treating the catalyst surface concentration as constant, \(\text{Rate}=k[\mathrm{CH_4}]\) (first order in \(\mathrm{CH_4}\), overall first order). A1
(iii) Different mechanisms can lead to the same rate law; matching the rate equation is necessary but not sufficient to prove the mechanism. A1
(iv) Sketch: Rate vs \([\mathrm{CH_4}]\) is a straight line through the origin (first-order dependence). A1
[1 + 1 + 1 + 1 = 4 marks]
![Rate vs. [CH4] Graph Answer](https://sp-ao.shortpixel.ai/client/to_webp,q_glossy,ret_img,w_205,h_156/https://www.iitianacademy.com/wp-content/uploads/2024/10/Screenshot-2024-10-07-154718.png)
(f)
Data (Section 12, standard molar entropies): \(S^\circ(\mathrm{CH_4})=186\), \(S^\circ(\mathrm{H_2O(g)})=188.8\), \(S^\circ(\mathrm{CO})=197.7\), \(S^\circ(\mathrm{H_2})=130.7\) \(\text{J K}^{-1}\text{ mol}^{-1}\).
\[ \Delta S^\circ = \big(S^\circ_{\mathrm{CO}} + 3S^\circ_{\mathrm{H_2}}\big) – \big(S^\circ_{\mathrm{CH_4}} + S^\circ_{\mathrm{H_2O}}\big) = (197.7 + 3\times130.7) – (186 + 188.8) = 215~\text{J K}^{-1}\text{ mol}^{-1}. \] A1
[1 mark]
(g)
Use \(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ\), \(\Delta G^\circ = -RT\ln K\), \(R=8.31~\text{J K}^{-1}\text{ mol}^{-1}\).
\[ \Delta G^\circ = 206000 – (1500)(215) = 206000 – 322500 = -116500~\text{J mol}^{-1} = -116.5~\text{kJ mol}^{-1}. \] \[ K = e^{-\Delta G^\circ/(RT)} = \exp\!\left(\frac{116500}{(8.31)(1500)}\right) \approx \exp(9.35) \approx 1.2\times10^{4}. \] M1 A1
[2 marks]
Use \(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ\), \(\Delta G^\circ = -RT\ln K\), \(R=8.31~\text{J K}^{-1}\text{ mol}^{-1}\).
\[ \Delta G^\circ = 206000 – (1500)(215) = 206000 – 322500 = -116500~\text{J mol}^{-1} = -116.5~\text{kJ mol}^{-1}. \] \[ K = e^{-\Delta G^\circ/(RT)} = \exp\!\left(\frac{116500}{(8.31)(1500)}\right) \approx \exp(9.35) \approx 1.2\times10^{4}. \] M1 A1
[2 marks]
(h)
Spontaneous when \(\Delta G^\circ < 0\). Threshold at \(\Delta G^\circ = 0\): \[ 0=\Delta H^\circ – T\Delta S^\circ \;\Rightarrow\; T=\frac{\Delta H^\circ}{\Delta S^\circ} =\frac{206000~\text{J mol}^{-1}}{215~\text{J K}^{-1}\text{ mol}^{-1}} \approx 958~\text{K}. \] So the forward reaction becomes spontaneous for \(T \gtrsim 958~\text{K}\). M1 A1
[2 marks]
Spontaneous when \(\Delta G^\circ < 0\). Threshold at \(\Delta G^\circ = 0\): \[ 0=\Delta H^\circ – T\Delta S^\circ \;\Rightarrow\; T=\frac{\Delta H^\circ}{\Delta S^\circ} =\frac{206000~\text{J mol}^{-1}}{215~\text{J K}^{-1}\text{ mol}^{-1}} \approx 958~\text{K}. \] So the forward reaction becomes spontaneous for \(T \gtrsim 958~\text{K}\). M1 A1
[2 marks]
Total Marks: 14