What is the equilibrium constant (K) for a reaction with ΔG° = -40 kJ at 298 K?

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Question: What is the equilibrium constant (K) for a reaction with ΔG° = -40 kJ at 298 K?

Options:

Correct Answer: 1000

Solution:

ΔG° = -RTln(K); K = e^(-ΔG°/RT) = e^(40,000/(8.314*298)) ≈ 1000.

What is the equilibrium constant (K) for a reaction with ΔG° = -40 kJ at 298 K?

What is the equilibrium constant (K) for a reaction with ΔG° = -40 kJ at 298 K?

Step 1: Understand the formula for the equilibrium constant (K) in relation to Gibbs free energy (ΔG°). The formula is ΔG° = -RTln(K).
Step 2: Rearrange the formula to solve for K. This gives us K = e^(-ΔG°/RT).
Step 3: Identify the values needed for the calculation: ΔG° = -40 kJ, R (the gas constant) = 8.314 J/(mol·K), and T (temperature) = 298 K.
Step 4: Convert ΔG° from kJ to J. Since 1 kJ = 1000 J, -40 kJ = -40,000 J.
Step 5: Plug the values into the rearranged formula: K = e^(40,000 / (8.314 * 298)).
Step 6: Calculate the denominator: 8.314 * 298 = 2477.572.
Step 7: Divide 40,000 by 2477.572 to get approximately 16.14.
Step 8: Calculate e^(16.14) using a calculator or mathematical software to find K.
Step 9: The result is approximately K ≈ 1000.

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