Hydrogen gas may be prepared industrially by heating hydrocarbons, such as a methane, with steam: CH — Physical Chemistry — Thermodynamics Chemistry Question
The industrial preparation of hydrogen
Hydrogen gas may be prepared industrially by heating hydrocarbons, such as a methane, with steam:
CH4(g) + H2O(g) 3 H2(g) + CO(g) (A)
Given the following thermodynamic data, calculate the ∆rG° for reaction A at 298 K and hence a value for the equilibrium constant, Kp.
Thermodynamic data at 298 K:
CH4(g): ∆fH° = –74.4 kJ mol–1, S° = 186.3 J K–1 mol–1
H2O(g): ∆fH° = –241.8 kJ mol–1, S° = 188.8 J K–1 mol–1
H2(g): S° = 130.7 J K–1 mol–1
CO(g): ∆fH° = –110.5 kJ mol–1, S° = 197.7 J K–1 mol–1
Model Answer
∆rH° = −110.5 − (−74.4) − (−241.8) = 205.7 kJ mol−1
∆rS° = 197.7 + 3 × 130.7 − 186.3 − 188.8 = 214.7 J mol−1 K−1
∆rG° = ∆rH° − T∆rS° = 205700 − 298 × 214.7 = 141700 J mol−1 = 141.7 kJ mol−1
∆rG° = − RT ln Kp
Kp = exp(−∆rG° / RT) = exp(−141700 / (8.314 × 298)) = 1.44 × 10−25
How will the equilibrium constant vary with temperature?
Model Answer
As the reaction is endothermic increasing the temperature will result in shifting the equilibrium towards the products, i.e. increasing the equilibrium constant.
The industrial preparation can be carried out at atmospheric pressure and high temperature, without a catalyst. Typically, 0.2 vol % of methane gas remains in the mixture at equilibrium.
Assuming the reaction started with equal volumes of methane and steam, calculate the value of Kp for the industrial process which gives 0.2 vol. % methane at equilibrium.
Model Answer
For ideal gases, vol % is the same as the mole fraction.
If 0.2 vol % CH4 remains then there must be 0.2 vol % H2O as well.
Remaining 99.6% corresponds to the products H2 and CO in ratio 3 : 1. Therefore there is 24.9 % CO and 74.7 % H2.
Kp = (0.747^3 × 0.249 × 101325^2) / (0.002 × 0.002) = 26640 × 100000 = 2.664 × 10^9
Use your answer from 5.3 together with the integrated form of the van’t Hoff isochore to estimate the temperature used in industry for the preparation of hydrogen from methane.
Model Answer
ln(K2/K1) = -(∆rH° / R) × (1/T2 - 1/T1)
1/T2 = 1/T1 - (R / ∆rH°) × ln(K2/K1)
T2 = 1580 K