Question

Hydrogen is produced commercially by the reaction of methane with steam:

${\mathbf{\text{CH}}}_{\mathbf{\text{4}}}{\mathbf{\text{(g) +H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{CO(g) + 3H}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

a. Calculate$\mathbf{∆}{\mathit{H}}^{\mathbf{0}}$ and$\mathbf{∆}{\mathit{S}}^{\mathbf{0}}$ for this reaction (use the data in Appendix ).

b. What temperatures will favour product formation assuming standard conditions and assuming that$\mathbf{∆}{\mathit{H}}^{\mathbf{0}}$ and$\mathbf{∆}{\mathit{S}}^{\mathbf{0}}$ do not depend on temperature?

Short answer

(a) For the reaction ${\text{CH}}_{\text{4}}{\text{(g) +H}}_{\text{2}}\text{O(g)}\rightleftharpoons {\text{CO(g) + 3H}}_{\text{2}}\text{(g)}$ the value for $∆H^0=206.5KJ$ and $∆S^0=-0.216kJ/K$.

(b) Assuming the standard conditions and also that $∆H^0$and$∆S^0$ do not depend on temperature, the temperatures greater than 956 K will favour product formation.

Step-by-step solution

Enthalpy and Entropy Change

Theenthalpy changefor a given reaction can be calculated by subtracting the enthalpies of the formation of the reactants from the enthalpies of the formation of the products.

$∆{\text{H}}_{\text{reaction}}\text{=}\sum \text{∆H(products)-}\sum {\text{∆H}}_{\text{f}}^{}\text{(reactants)}$

Remember that elements in their standard states are not included in the AH reaction calculations.

Theentropy changefor a given chemical reaction can be calculated by subtracting the standard entropy values of the products and those of the reactants.

$∆S_{\text{reaction}}\text{=}\sum \text{∆S(products)-}\sum {\text{∆S}}_{\text{f}}^{}\text{(reactants)}$

For a process occurring at constant temperature and pressure –

$\text{∆G =}\text{H∆- T∆S}$

Calculation table for enthalpy and entropy change

The values required to calculate the standard enthalpy and standard entropy

are shown in the table below.

The chemical reaction for which we calculate standard enthalpy and standard

entropy is shown in the figure below.

${\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)$

Standard Enthalpy

The standard enthalpy can be represented as –

$$\begin{gathered} {\text{∆H}}_{\text{reaction}}\text{=}\sum {\text{∆H}}_{\text{f}}^{}\text{(products) -}\sum {{\text{H}}_{\text{f}}}^{}\text{(reactants)} \\ ∆{\text{H}}_{\text{reaction}}\text{=}{\text{H∆}}_{\text{f}}^{}{\text{(CO) + 3∆H}}_{\text{f}}^{}\left({\text{H}}_{\text{2}}\right){\text{-∆H}}_{\text{f}}^{}\left({\text{CH}}_{\text{4}}\right){\text{-∆H}}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right) \\ ∆{\text{H}}_{\text{reaction}}\text{= 1mol×}\left(-110.5kJ/mol\right)\text{+3mol×0kJ/mol-1mol×}\left(-75kJ/mol\right)\text{-1mol×}\left(-242kJ/mol\right) \\ ∆{\text{H}}_{\text{reaction=206.5kJ}} \end{gathered}$$

Standard Entropy

The standard entropy can be represented as –

$$\begin{gathered} {\text{∆S}}_{\text{reaction}}\text{=}\sum {\text{∆S}}_{\text{f}}^{}\text{(products) -}\sum {\text{S}}^{}\text{(reactants)} \\ ∆S{\text{=H∆}}_{\text{f}}^{}{\text{(CO) + 3∆S}}_{\text{f}}^{}\left({\text{H}}_{\text{2}}\right){\text{-∆S}}_{\text{f}}^{}\left({\text{CH}}_{\text{4}}\right){\text{-∆S}}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right) \\ ∆S_{\text{reaction}}\text{= 1mol×}(-198J{K}^{-1}mo{l}^{-1})\text{+3mol×131}J{K}^{-1}mo{l}^{-1}\text{-1mol×}186J{K}^{-1}mo{l}^{-1})\text{-1mol×}189J{K}^{-1}mo{l}^{-1} \\ ∆S_{\text{reaction=216JK=0.216kJ/K}} \end{gathered}$$

Temperature that favour product formation

For a process occurring at constant temperature and pressure –

$\text{∆G =}\text{H∆- T∆S}$

The spontaneous processes are –

$$\begin{gathered} ∆G<0 \\ ∆H-T∆S<0 \\ ∆H<T∆S \\ T>\frac{∆H}{∆S} \\ \frac{∆H}{∆S}=\frac{-206.5kJ}{-0.216kJ/K} \\ T>956K \end{gathered}$$

Therefore, temperatures above 956 K will favour product formation.