Chemical Thermodynamics: Gibbs Free Energy of Formation 3

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Created this as a copy of Gibbs Free Energy of Formation - Alt 3 (O).

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	2.8
h2	number	2.9
nh3	number	2
q	number	0.2420215194
standard_kj	number	-16.5
gibbs	number	-20015.0023380056
gibbskj	number	-20.015002338
enthalpy	number	-46.16
entropy	number	-99.5302013423
standard	number	-16500
enth	number	-46160

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The standard enthalpy and entropy of formation of ammonia are $\mathrm{\var{enthalpy} \space kJ \space mol^{-1}}$ and $\mathrm{\var{sigformat(entropy,4)} \space J \space mol^{-1}}$ , respectively, 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 will eventually be using the $\mathrm{\Delta G^\circ}$ , 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 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 are also given the standard enthalpy and entropy for the formation of ammonia: $\mathrm{\Delta H^\circ=\var{enthalpy} \space kJ \space mol^{-1}}$ , which we convert by multiplying by one thousand, to $\mathrm{\Delta H^\circ =\var{enth} \space J \space mol^{-1}}$ ; and $\mathrm{\Delta S^\circ=\var{sigformat(entropy,4)} \space J \space mol^{-1}}$ . From these values, we can calculate the standard Gibbs free energy of formation, using the formula:

$\mathrm{\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ}$

$\mathrm{\Rightarrow\space\space\space\Delta G^\circ =\var{enth}-(298\times \var{entropy})=\var{standard} \space J \space mol^{-1}}$

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

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

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 3

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Chemical Thermodynamics: Gibbs Free Energy of Formation 3

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