In the gas phase, formic acid forms a dimer, 2HCOOH(g) (HCOOH)₂(g). For this reaction, ΔH° = -60.1 kJ/mol and ΔG° = -13.9 kJ/mol at 25°C. Find the equilibrium constant (K_P) for this reaction at 75 °C.

Question

Quizplus · Accepted Answer

First, we need to find the equilibrium constant (K) of the reaction at 25°C using the formula:

ΔG° = -RTln(K)

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and ln is the natural logarithm. So we have:

-13.9 kJ/mol = -8.314 J/mol·K × (298 K) × ln(K)

Solving for K, we get:

K = e^(-13.9 kJ/mol / (8.314 J/mol·K × 298 K)) = 8.33

Now we can use the van 't Hoff equation to find the equilibrium constant at 75°C:

ln(K2/K1) = ΔH°/R × (1/T1 - 1/T2)

where K1 is the equilibrium constant at T1, K2 is the equilibrium constant at T2, ΔH° is the enthalpy change of the reaction, and R is the gas constant. Converting the temperatures to Kelvin and substituting the values, we have:

ln(K2/8.33) = (-60.1 kJ/mol) / (8.314 J/mol·K) × (1/298 K - 1/348 K)

Solving for K2, we get:

K2 = 273

Therefore, the answer is D: 8.33.

In the Gas Phase, Formic Acid Forms a Dimer, 2HCOOH(g)