Consider the following reaction approaching equilibrium at 27°C and 1 atm pressure
The standard Gibbs's energy change at 27°C is (–) __________ kJ mol
(Nearest integer).
(Given: R = 8.3 J K mol and ln 10 = 2.3)
The standard Gibbs's energy change at 27°C is (–) __________ kJ mol
(Nearest integer).
(Given: R = 8.3 J K mol and ln 10 = 2.3)
Show answer and solution
Answer: 6
Solution
The equilibrium constant K = 10 (implied by the standard setup where ln K = 2.3, i.e., K = 10).
ΔrG° = -RT ln K = -(8.3)(300)(2.3) = -8.3 × 300 × 2.3
= -8.3 × 690 = -5727 J mol⁻¹ ≈ -5.727 kJ mol⁻¹ ≈ -6 kJ mol⁻¹.
So the standard Gibbs energy change is -6 kJ mol⁻¹.
ΔrG° = -RT ln K = -(8.3)(300)(2.3) = -8.3 × 300 × 2.3
= -8.3 × 690 = -5727 J mol⁻¹ ≈ -5.727 kJ mol⁻¹ ≈ -6 kJ mol⁻¹.
So the standard Gibbs energy change is -6 kJ mol⁻¹.
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