QuestionMay 5, 2025

Consider the reaction N_(2)(g)+3H_(2)(g)arrow 2NH_(3)(g) Using the standard thermodynamic data in the tables linked above, calculate Delta G_(rxn) for this reaction at 298.15K if the pressure of each gas is 45.27 mm Hg. ANSWER: square kJ/mol

Consider the reaction N_(2)(g)+3H_(2)(g)arrow 2NH_(3)(g) Using the standard thermodynamic data in the tables linked above, calculate Delta G_(rxn) for this reaction at 298.15K if the pressure of each gas is 45.27 mm Hg. ANSWER: square kJ/mol
Consider the reaction N_(2)(g)+3H_(2)(g)arrow 2NH_(3)(g) Using the standard thermodynamic data in the tables linked above, calculate Delta G_(rxn) for this reaction at 298.15K if the pressure of each gas is 45.27 mm Hg. ANSWER: square kJ/mol

Solution
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Answer

-32.8 \text{ kJ/mol} Explanation 1. Convert pressure to atm 45.27 \text{ mm Hg} = 0.0596 \text{ atm} using the conversion 1 \text{ atm} = 760 \text{ mm Hg}. 2. Calculate reaction quotient Q Q = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(0.0596)^2}{(0.0596)(0.0596)^3} = 1. 3. Use \Delta G = \Delta G^\circ + RT \ln Q \Delta G^\circ is calculated from standard free energies of formation: \Delta G^\circ = [2(-16.4)] - [0 + 3(0)] = -32.8 \text{ kJ/mol}. R = 8.314 \times 10^{-3} \text{ kJ/mol K}, T = 298.15 \text{ K}. \Delta G = -32.8 + (8.314 \times 10^{-3} \times 298.15 \times \ln 1) = -32.8 \text{ kJ/mol}.

Explanation

1. Convert pressure to atm<br /> $45.27 \text{ mm Hg} = 0.0596 \text{ atm}$ using the conversion $1 \text{ atm} = 760 \text{ mm Hg}$.<br />2. Calculate reaction quotient $Q$<br /> $Q = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(0.0596)^2}{(0.0596)(0.0596)^3} = 1$.<br />3. Use $\Delta G = \Delta G^\circ + RT \ln Q$<br /> $\Delta G^\circ$ is calculated from standard free energies of formation: $\Delta G^\circ = [2(-16.4)] - [0 + 3(0)] = -32.8 \text{ kJ/mol}$.<br /> $R = 8.314 \times 10^{-3} \text{ kJ/mol K}$, $T = 298.15 \text{ K}$.<br /> $\Delta G = -32.8 + (8.314 \times 10^{-3} \times 298.15 \times \ln 1) = -32.8 \text{ kJ/mol}$.
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