Question

The standard Gibbs free energy change for the reaction
$H_2O (g) → H_2 (g) + ½ O_2(g)$ at 2500 K is $+118 kJ mol^{-1}$
The equilibrium constant for the reaction is
[Given: $R= 8.314 J K^{-1} mol^{-1}$]

• A. 0.994
• B. 1.006
• C. 3.42 x $10^{-3}$
• D. 292.12

Answer 1

The Correct Option is C

The equilibrium constant of a reaction at a given temperature can be found using the relation with Gibbs free energy:

\(ΔG° = -RT\ln K\)

Where:

• \(ΔG°\) is the standard Gibbs free energy change (in J/mol),
• \(R\) is the universal gas constant, and
• \(T\) is the temperature in Kelvin.
• \(K\) is the equilibrium constant.

Given:

• \(ΔG° = +118 \text{ kJ/mol} = 118000 \text{ J/mol}\)
• \(R = 8.314 \text{ J/K mol}\)
• \(T = 2500 \text{ K}\)

Substitute the values into the formula:

\(118000 = - (8.314 \times 2500) \ln K\)

Solve for \(\ln K\):

\(\ln K = - \frac{118000}{8.314 \times 2500}\) \(\ln K = -5.674\)

Now, to find \(K\), take the exponential of both sides:

\(K = e^{-5.674}\) \(K \approx 3.42 \times 10^{-3}\)

Thus, the equilibrium constant for the reaction at 2500 K is \(3.42 \times 10^{-3}\). Therefore, the correct answer is \(3.42 \times 10^{-3}\).