Question

2nh₃(g) + 3o₂(g) + ch₄(g) → 2hcn(g) + 6h₂o(g)
calculate the gibbs free energy change for
the reaction from the thermodynamic data.

substance	δgᵣ° (kj/mol)
ch₄(g)	-51
hcn(g)	125
h₂o(g)	-229

δgᵣₓₙ° = ? kj
enter either a + or - sign.

Explanation:

Step1: Recall the formula for Gibbs Free Energy change of reaction

The formula for the standard Gibbs free energy change of a reaction (\(\Delta G^\circ_{rxn}\)) is the sum of the Gibbs free energy of formation of the products minus the sum of the Gibbs free energy of formation of the reactants, each multiplied by their stoichiometric coefficients. Mathematically, it is:
\(\Delta G^\circ_{rxn} = \sum (n \cdot \Delta G_f^\circ)_{\text{products}} - \sum (n \cdot \Delta G_f^\circ)_{\text{reactants}}\)

Step2: Identify reactants, products, and their stoichiometric coefficients

• Reactants: \(2\text{NH}_3(g)\), \(3\text{O}_2(g)\), \(1\text{CH}_4(g)\)
• For \(\text{O}_2(g)\), the standard Gibbs free energy of formation (\(\Delta G_f^\circ\)) is \(0\) kJ/mol (since it is an element in its standard state).
• Products: \(2\text{HCN}(g)\), \(6\text{H}_2\text{O}(g)\)

Step3: Calculate the sum of \(\Delta G_f^\circ\) for products

For products:

• \(2\text{HCN}(g)\): \(n = 2\), \(\Delta G_f^\circ = 125\) kJ/mol

Contribution: \(2 \times 125 = 250\) kJ

• \(6\text{H}_2\text{O}(g)\): \(n = 6\), \(\Delta G_f^\circ = -229\) kJ/mol

Contribution: \(6 \times (-229) = -1374\) kJ

Sum of products: \(250 + (-1374) = -1124\) kJ

Step4: Calculate the sum of \(\Delta G_f^\circ\) for reactants

For reactants:

• \(2\text{NH}_3(g)\): \(n = 2\), \(\Delta G_f^\circ = 17\) kJ/mol

Contribution: \(2 \times 17 = 34\) kJ

• \(3\text{O}_2(g)\): \(n = 3\), \(\Delta G_f^\circ = 0\) kJ/mol

Contribution: \(3 \times 0 = 0\) kJ

• \(1\text{CH}_4(g)\): \(n = 1\), \(\Delta G_f^\circ = -51\) kJ/mol

Contribution: \(1 \times (-51) = -51\) kJ

Sum of reactants: \(34 + 0 + (-51) = -17\) kJ

Step5: Calculate \(\Delta G^\circ_{rxn}\)

Using the formula:
\(\Delta G^\circ_{rxn} = (\text{Sum of products}) - (\text{Sum of reactants})\)
\(\Delta G^\circ_{rxn} = (-1124) - (-17)\)
\(\Delta G^\circ_{rxn} = -1124 + 17\)
\(\Delta G^\circ_{rxn} = -1107\) kJ

Answer:

\(-1107\)