Question

A chemist fills a reaction vessel with 7.92 atm nitrogen (N2) gas, 2.02 atm hydrogen (H2) gas, and 2.11 atm ammonia (NH3) gas at a temperature of 25.0°C
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction

N2(8) +3H2 2NH3 (g)

Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Answer 1

Answer: The Gibbs free energy of the given reaction is -40 kJ

Explanation:

The given chemical equation follows:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NH_3(g))})]-[(1\times \Delta G^o_f_{(N_2(g))})+(3\times \Delta G^o_f_{(H_2(g))})]$

We are given:

$\Delta G^o_f_{(NH_3(l))}=-16.45kJ/mol\\\Delta G^o_f_{(H_2(g))}=0kJ/mol\\\Delta G^o_f_{(N_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -32.9 kJ/mol = -32900 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $25^oC=[273+25]K=298K$

$Q_{eq}$ = Ratio of concentration of products and reactants at any time = $\frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}$

$p_{NH_3}=2.11atm$

$p_{N_2}=7.92atm$

$p_{H_2}=2.02atm$

Putting values in above equation, we get:

$\Delta G=-32900J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(2.11)^2}{(2.02)^3\times 7.92}))\\\\\Delta G=-39553.04J/mol=--39.55kJ=-40kJ$

Hence, the Gibbs free energy of the given reaction is -40 kJ