Question

Hydrogen gas is being considered as a fuel for automobiles. There are many chemical means for producing hydrogen gas from water. One of these reactions is

${\text{C(s) +H}}_{\text{2}}\text{O(g)}\to {\text{CO(g) +H}}_{\text{2}}\text{(g)}$

In this case the form of carbon used is graphite.

a. Calculate ${\text{ΔH}}^{°}$and${\text{ΔS}}^{°}$for this reaction using data from Appendix 4 .

b. At what temperature is${\text{ΔG}}^{°}\text{=0}$ for this reaction? Assumeanddo not depend on temperature.

Short answer

1. The required solution is$\text{Δ}{{\text{H}}_{\text{reaction}}}^{°}\text{=131.5kJ}$ and$\text{Δ}{{\text{S}}_{\text{reaction}}}^{°}\text{=0.134kJ/K}$
2. The value of temperature is 981 k.

Step-by-step solution

Concept Introduction

The standard reaction enthalpy for forming a compound from its elements (atoms or molecules) in their most stable reference states at the given temperature (298.15K) and pressure (1bar) is the enthalpy of formation.

Write the formula of enthalpies of formation and standard entropy

a)

By subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products, the enthalpy change for a given reaction can be computed.

$\text{Δ}{{\text{H}}_{\text{reaction}}}^{°}=\sum \Delta H_f^{°}\left(\text{products}\right)\text{-}\sum \Delta H_f^{°}\left(\text{reactants}\right)$

Note that the calculations for the $\text{Δ}{{\text{H}}_{\text{reaction}}}^{°}$do not include elements in their normal states.

By subtracting the standard entropy values of the products and reactants, the entropy change for a specific chemical reaction may be computed.

$\text{Δ}{{\text{S}}_{\text{reaction}}}^{°}=\sum S_f^{°}\left(\text{products}\right)\text{-}\sum S_f^{°}\left(\text{reactants}\right)$

Now, writing the reaction tocalculate the standard enthalpy and standard entropy

The table below shows the values needed to calculate the standard enthalpy and standard entropy.

The chemical reaction for which the standard enthalpy and standard entropy are calculated is depicted below.

${\text{C(s) +H}}_{\text{2}}\text{O(g)}\to {\text{CO(g) +H}}_{\text{2}}\text{(g)}$

Then, calculate standard enthalpy

$$\begin{gathered} \text{Δ}{{\text{H}}_{\text{reaction}}}^{°}=\sum \Delta H_f^{°}\left(\text{products}\right)\text{-}\sum \Delta H_f^{°}\left(\text{reactants}\right) \\ \text{Δ}{{\text{H}}_{\text{reaction}}}^{°}\text{=Δ}{{\text{H}}_f}^{°}\left(\text{CO}\right){\text{+ΔH}}_f^{°}\left(H_2\right){\text{-ΔH}}_f^{°}\left(H_{2}O\right)\text{-Δ}{{\text{H}}_f}^{°}\left(C\right) \\ \text{Δ}{{\text{H}}_{\text{reaction}}}^{°}\text{=1mol×}\left(\text{-110.5kJ/mol}\right)\text{+1mol×}\left(\text{0kJ/mol}\right)\text{-1mol×}\left(\text{-242}\right)\text{kJ/mol-1mol×0kJ/mo} \\ \text{Δ}{{\text{H}}_{\text{reaction}}}^{°}\text{=131.5kJ} \end{gathered}$$

Calculate standard entropy:

$$\begin{gathered} \text{Standardentropy} \\ \text{Δ}{{\text{S}}_{\text{reaction}}}^{°}=\sum S_f^{°}\left(\text{products}\right)\text{-}{\sum S_f}^{°}\left(\text{reactants}\right) \\ \text{Δ}{{\text{S}}_{\text{reaction}}}^{°}\text{=}{{\text{S}}_f}^{°}\left(\text{CO}\right){\text{+S}}_f^{°}\left(H_2\right)\text{-}{{\text{S}}_f}^{°}\left(H_{2}O\right)\text{-}{{\text{S}}_f}^{°}\left(C\right) \\ \text{Δ}{{\text{S}}_{\text{reaction}}}^{°}\text{=1mol×}\left({\text{198JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{+1mol×}\left({\text{131JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right){\text{-1mol×189JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{-1mol}}^{\text{-1}}{\text{×6JK}}^{\text{-1}}\text{m} \\ \text{Δ}{{\text{S}}_{\text{reaction}}}^{°}\text{=134J/K=0.134kJ/K} \end{gathered}$$

Therefore the required answer of standard enthalpy and standard entropy is 131.5 kj and 0.134 KJ /K.

Calculate the value of temperature

b)

For a process that takes place at a fixed temperature and pressure.

$\Delta G=\Delta H-T\Delta S$$\text{ΔG=ΔH-TΔS}$

$\text{ForΔG=0,theΔH-TΔS=0}$

As a result, we can calculate the temperature at which the free Gibbs energy equals zero.

$\text{ΔH=TΔS}$

$$\begin{gathered} \text{T=}\frac{\text{ΔH}}{\text{ΔS}} \\ \text{T=}\frac{\text{131.5kJ}}{\text{0.134kJ/K}} \\ \text{T=981K} \end{gathered}$$

Therefore, the value of temperature is 981 k.