In the gas phase, formic acid forms a dimmer, 2HCOOH(g) $\rarr$ (HCOOH)₂(g). For this reaction, $\Delta$H° = -60.1 kJ/mol and $\Delta$G° = -13.9 kJ/mol at 25°C. Find the equilibrium constant (K_p)for this reaction at 75 °C.

Question

Quizplus · Accepted Answer

We can use the van't Hoff equation to calculate the equilibrium constant at a different temperature:

ln(K2/K1) = -ΔH°/R (1/T2 - 1/T1)

where K1 is the equilibrium constant at the initial temperature (25°C), ΔH° is the enthalpy change, R is the gas constant, T1 is the initial temperature in Kelvin, and T2 is the final temperature in Kelvin.

Plugging in the values given:

ln(K2/1) = -(60.1 kJ/mol)/(8.314 J/mol·K) (1/298 K - 1/348 K)

ln(K2) = -12.82

K2 = e^(-12.82)

K2 = 8.33

Therefore, the equilibrium constant (K) at 75°C is 8.33. The answer is D.

In the Gas Phase, Formic Acid Forms a Dimmer, 2HCOOH(g) →\rarr→

In the Gas Phase, Formic Acid Forms a Dimmer, 2HCOOH(g) $\rarr$