Question

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

${\mathbf{\text{3H}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) +N}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{2NH}}}_{\mathbf{\text{3}}}\mathbf{\text{(g)}}$

a. Using data from Appendix 4 , calculate$\mathbf{∆}{\mathit{H}}^{\mathbf{0}}$ ,$\mathbf{∆}{\mathit{S}}^{\mathbf{0}}$ , and $\mathbf{∆}{\mathit{G}}^{\mathbf{0}}$for the Haber process reaction.

b. Is this reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume that$\mathbf{∆}{\mathit{H}}^{\mathbf{0}}$ and$\mathbf{∆}{\mathit{S}}^{\mathbf{0}}$ do not depend on temperature.

Short answer

(a) For the reaction${\text{3H}}_2{\text{(g) +N}}_{\text{2}}\text{(g)}\to {\text{2NH}}_3\text{(g)}$the value for${\text{∆H}}^0\text{= - 92kJ}$, ${\text{∆S}}^0\text{= - 0.199kJ/K}$and $∆G^0=-34kJ$.

(b) Yes, the Haber process followed for the reaction makes the reaction spontaneous at standard conditions.

(c) Assuming the standard conditions and also that $∆H^0$and$∆S^0$ do not depend on temperature, at temperatures greater than 462K the reaction is spontaneous.

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.

${{\text{∆H}}^0}_{\text{reaction}}\text{=}\sum \text{∆}{{\text{H}}_{\text{f}}}^0\text{(products) -}\sum {\text{∆H}}_{\text{f}}^0\text{(reactants)}$

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.

${{\text{∆S}}^0}_{\text{reaction}}\text{=}\sum \text{∆}{{\text{S}}_f}^0\text{(products) -}\sum {\text{∆S}}_{\text{f}}^0\text{(reactants)}$

For a process occurring at constant temperature and pressure –

$\text{∆G =}\text{H∆- T∆S}$

Calculation table for enthalpy change, entropy change and free energy

The values required to calculate the standard enthalpy, standard entropy,

and standard free energy are shown in the table below

The chemical reaction for which we calculate the standard enthalpy, standard

entropy, and standard free energy is shown in the figure below.

${\text{3H}}_2{\text{(g) +N}}_{\text{2}}\text{(g)}\to {\text{2NH}}_3\text{(g)}$

Standard Enthalpy

The standard enthalpy can be represented as –

$$\begin{gathered} {{\text{∆H}}^0}_{\text{reaction}}\text{=}\sum {\text{∆H}}_{\text{f}}^0\text{(products) -}\sum {{\text{∆H}}_{\text{f}}}^0\text{(reactants)} \\ ∆{{\text{H}}^0}_{\text{reaction}}{\text{= 2∆H}}_{\text{f}}^0{\text{(NH}}_{\text{3}}{\text{) - 3∆H}}_{\text{f}}^0\left({\text{H}}_{\text{2}}\right){\text{-∆H}}_{\text{f}}^0\left({\text{N}}_{\text{2}}\right) \\ {{\text{∆H}}^0}_{\text{reaction}}\text{=2mol}\left(-46KJ/mol\right)\text{-3mol×0KJ/mol-1mol×0KJ/mol} \\ ∆{\text{H}}_{\text{reaction}}=-92KJ \end{gathered}$$

Standard Entropy

The standard entropy can be represented as –

$$\begin{gathered} {{\text{∆S}}^0}_{\text{reaction}}\text{=}\sum {\text{∆S}}_{\text{f}}^0\text{(products) -}\sum {{\text{∆S}}_{\text{f}}}^0\text{(reactants)} \\ ∆{S^0}_{\text{reaction}}{\text{= 2∆S}}_{\text{f}}^0{\text{(NH}}_{\text{3}}{\text{) - 3∆S}}_{\text{f}}^0\left({\text{H}}_{\text{2}}\right){\text{-∆S}}_{\text{f}}^0\left({\text{N}}_{\text{2}}\right) \\ {{\text{∆S}}^0}_{\text{reaction}}\text{=12mol}(193J{K}^{-1}mo{l}^{-1})\text{-3mol×131}J{K}^{-1}mo{l}^{-1}\text{-1mol×192}J{K}^{-1}mo{l}^{-1}\text{l} \\ ∆S_{\text{reaction}}=-0.199kJ/K \end{gathered}$$

Standard Free Energy

The standard free energy can be represented as –

$$\begin{gathered} {{\text{∆G}}^0}_{\text{reaction}}\text{=}\sum {\text{∆G}}_{\text{f}}^0\text{(products) -}\sum {{\text{∆G}}_{\text{f}}}^0\text{(reactants)} \\ ∆G_{\text{reaction}}{\text{= 2∆G}}_{\text{f}}^0{\text{(NH}}_{\text{3}}{\text{) - 3∆G}}_{\text{f}}^0\left({\text{H}}_{\text{2}}\right){\text{-∆G}}_{\text{f}}^0\left({\text{N}}_{\text{2}}\right) \\ {{\text{∆G}}^0}_{\text{reaction}}\text{=2mol}(-17KJ/mol)\text{-3mol×0}K/Jmol\text{-1mol×0}KJ/mol \\ ∆G_{\text{reaction}}=-34kJ \end{gathered}$$

Therefore, $∆H^0=-92KJ$,$∆S^0=-0.199kJ/K$ and $∆G^0=-34KJ$is obtained.

Spontaneous Reaction

The calculated ${{\text{∆G}}^0}_{\text{reaction}}$tells us whether the reaction is spontaneous under standard conditions or not.

The general rule is, the more negative the free energy of a reaction, the more spontaneous the process.

${{\text{∆G}}^0}_{\text{reaction}}=-34kJ$

Since,${{\text{∆G}}^0}_{\text{reaction}}$ is negative, we can conclude that the Haber process is a spontaneous process.

Therefore, the Haber process is a spontaneous process.

Temperature that favour product formation

For a process occurring at constant temperature and pressure –

$\text{∆G =∆H - T∆S}$

The spontaneous processes are –

$$\begin{gathered} ∆G<0 \\ ∆H-T∆S<0 \\ ∆H<T∆S \\ T>\frac{∆H}{∆S} \\ \frac{∆H}{∆S}=\frac{-92kJ}{-0.199kJ/K} \\ T>462K \end{gathered}$$

Therefore, temperatures above 462 will favour product formation and the reaction is spontaneous above the mentioned temperature.