Chemistry · General chemistry II · Worked example
Relate ΔG° and the equilibrium constant
A reaction has ΔG° = −5.70 kJ/mol at 298 K. Find K, then find ΔG for a mixture in which Q = 100.
Solve for ln K
From ΔG° = −RT ln K, ln K = −ΔG°/(RT), with ΔG° in J/mol.
Find K
K = e^2.30 ≈ 10. A negative ΔG° gives K > 1: products are favored at equilibrium.
A mixture that is not at equilibrium
RT = 2.478 kJ/mol at 298 K and ln 100 = 4.605.
Interpret
Q = 100 is about ten times K, so the mixture holds too much product. ΔG is positive, and the reaction runs in reverse until Q falls to K.
Result
K ≈ 10; with Q = 100, ΔG = +5.71 kJ/mol and the reaction shifts in reverse.
Your turn
A reaction has ΔG° = +10.0 kJ/mol at 298 K. Find K.
Show the answer and explanation
K ≈ 0.018.
ln K = −10 000/(8.314 × 298) = −4.04, so K = e^−4.04 ≈ 0.018. A positive ΔG° gives K < 1: reactants are favored.
Keep exploring
In Hess law & thermodynamics, mark the quantities as standard: the reaction with ΔH° = −100. kJ and ΔS° = −200. J/K has ΔG° = −40.4 kJ at 298 K and K ≈ 1.2 × 10⁷.
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