Chemical Thermodynamics: Gibbs Free Energy of Formation 1

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Name	Status	Author	Last Modified
Chemical Thermodynamics: Gibbs Free Energy of Formation 1	Ready to use	Frances Docherty	30/03/2025 18:18
Chemical Thermodynamics: Gibbs Free Energy of Formation 2	Ready to use	Frances Docherty	30/03/2025 18:19
Chemical Thermodynamics: Gibbs Free Energy of Formation 3	Ready to use	Frances Docherty	30/03/2025 18:20
Chemical Thermodynamics: Non-Standard Chemical Potential 1	Ready to use	Frances Docherty	30/03/2025 18:43
Chemical Thermodynamics: Non-Standard Chemical Potential 2	Ready to use	Frances Docherty	30/03/2025 18:44

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			Name	Type	Generated Value

n2	number	3.5
h2	number	4
nh3	number	4.3
q	number	0.2873058351
standard_kj	number	-16.4
gibbs	number	-19490.0476269736
gibbskj	number	-19.490047627
standard	number	-16400

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3.5

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The standard Gibbs free energy of formation of ammonia is found to be $\mathrm{\var{standard_kj} \space kJ \space mol^{-1}}$ at $\mathrm{298\space K}$ .

What is the Gibbs free energy of formation of ammonia when the partial pressures of the $\mathrm{N_2}$ , $\mathrm{H_2}$ and $\mathrm{NH_3}$ (treated as perfect gases) are $\mathrm{\var{dpformat(n2,1)}\space bar}$ , $\mathrm{\var{dpformat(h2,1)}\space bar}$ , and $\mathrm{\var{dpformat(nh3,1)}\space bar}$ respectively?

Gap 0. $\mathrm{kJ\space mol^{-1}}$

Will the quantity of ammonia increase or decrease in this case?

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The Gibbs free energy of a reacting mixture, $\mathrm{\Delta G}$ , is given by the following equation:

$\mathrm{\Delta G=\Delta G^\circ + RTln(Q)}$ ,

where $\mathrm{\Delta G^\circ}$ is the standard Gibbs free energy, in $\mathrm{J \space mol^{-1}}$ ; $\mathrm{R=8.314\space J \space K^{-1}\space mol^{-1}}$ is the universal gas constant; $\mathrm{T}$ is the temperature of the system, in $\mathrm{K}$ ; and $\mathrm{Q}$ is the reaction quotient, given by:

$\mathrm{Q=\frac{[C]^c[D]^d}{[A]^a[B]^b}}$ for the reaction $\mathrm{aA\space +\space bB\space\rightleftharpoons\space cC\space +\space dD}$ .

We are asked to calculate $\mathrm{\Delta G}$ for the standard formation of ammonia. By considering the stoichiometries of hydrogen and nitrogen within ammonia, we can fairly easily deduce the chemical equation for this formation:

$\mathrm{N_2\space +\space 3H_2\space\rightleftharpoons\space 2NH_3}$ .

We must, however, pay attention to the fact that we were given $\mathrm{\Delta G^\circ}$ , for the standard formation of ammonia, i.e. the formation of 1 mole of ammonia under standard conditions. Thus, we half the stoichiometries of our equation:

$\mathrm{0.5N_2\space +\space 1.5H_2\space\rightleftharpoons\space NH_3}$ .

This gives us a reaction quotient of:

$\mathrm{Q=\frac{[NH_3]}{[N_2]^{0.5}[H_2]^{1.5}}}$ .

Since all three substances are gaseous at $\mathrm{298\space K}$ , we swap out their 'concentrations' for their partial pressures:

$\mathrm{Q=\frac{p_{_{NH_3}}}{p_{_{N_2}}^{0.5}p_{_{H_2}}^{1.5}}}$ .

Substituting this expression for $\mathrm{Q}$ into our original equation, we now have:

$\mathrm{\Delta G=\Delta G^\circ + RTln(\frac{p_{_{NH_3}}}{p_{_{N_2}}^{0.5}p_{_{H_2}}^{1.5}})}$ .

We are given the standard Gibbs free energy of formation, $\mathrm{\Delta G^\circ =\var{standard_kj} \space kJ \space mol^{-1}}$ , which we convert by multiplying by one thousand, to $\mathrm{\Delta G^\circ =\var{standard} \space J \space mol^{-1}}$ .

We are also given the temperature of the system, $\mathrm{T=298\space K}$ ; and the partial pressures of all three gases, $\mathrm{p_{_{NH_3}}=\var{dpformat(nh3,1)}\space bar}$ , $\mathrm{p_{_{N_2}}=\var{dpformat(n2,1)}\space bar}$ , and $\mathrm{p_{_{H_2}}=\var{dpformat(h2,1)}\space bar}$ .

We can thus calculate $\mathrm{\Delta G}$ for our reaction:

$\mathrm{\Delta G=\var{standard} + (8.314\times 298\times ln[\frac{\var{dpformat(nh3,1)}}{{\var{dpformat(n2,1)}}^{0.5}\times{\var{dpformat(h2,1)}}^{1.5}}])=\var{sigformat(gibbs,3)}\space J\space mol^{-1}}$ .

To quote our answer in the units asked for, we divide by one thousand to obtain:

$\mathrm{\Delta G=\var{sigformat(gibbskj,3)}\space kJ\space mol^{-1}}$ .

The Gibbs free energy we have calculated is negative, which tells us that the reaction will proceed spontaneously in the forward direction, i.e. ammonia will be formed. Thus, the quantity of ammonia will increase.

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Chemical Thermodynamics: Gibbs Free Energy of Formation 1

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Chemical Thermodynamics: Gibbs Free Energy of Formation 1

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