Consider the reaction given below:
H₂ (g) + F₂ (g) →2 HF (g) for which D G ° _rxn = - 541.4 kJ/mol What is the Gibbs Free Energy Change, D G _rxn, under non-standard conditions for this reaction at 25 ° C if the partial pressure of each substance is listed below?
P_H2 = 8.2 atm P_F2 = 4.6 atm P_HF = 0.22 atm

Question

Quizplus · Accepted Answer

The Gibbs Free Energy Change under non-standard conditions ($ \Delta G $) can be calculated using the equation $ \Delta G = \Delta G^\circ + RT \ln Q $, where $ \Delta G^\circ $ is the standard Gibbs Free Energy Change, $ R $ is the gas constant (8.314 J/(mol·K)), $ T $ is the temperature in Kelvin, and $ Q $ is the reaction quotient. Given $ \Delta G^\circ = -541.4 $ kJ/mol, $ T = 25 $ °C (or 298 K), and the partial pressures, $ Q = \frac{(P_{HF})^2}{P_{H_2} \cdot P_{F_2}} = \frac{(0.22)^2}{8.2 \cdot 4.6} $. Plugging these values into the equation and converting units of $ \Delta G^\circ $ to J/mol will give the correct $ \Delta G $ under non-standard conditions.

Consider the Reaction Given Below