For the reaction N₂O₃(g)→ NO(g)+ NO₂(g)Δ_rG° = -4.78 kJ/mol.Calculate K_eq at 25 °C. A)1.023 B)1.002 C)1.005 D)0.9484 E)0.9981

Question

Quizplus · Accepted Answer

We can use the expression for the equilibrium constant, which is:

K_eq = e^(-ΔG°/RT)

where R is the gas constant and T is the temperature in Kelvin. Plugging in the values given,

ΔG° = -4.78 kJ/mol = -4780 J/mol
R = 8.314 J/(mol·K)
T = 25°C = 298 K

K_eq = e^(-(-4780)/(8.314·298)) = 1.002

Therefore, the answer is B (1.002).

For the Reaction N2O3(g)→ NO(g)+ NO2(g)ΔrG° = -4