1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant of a reaction is 0.008 at 298 K. The standard free energy change of the reaction at the same temperature is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The equilibrium constant of a reaction is $0.008$ at $298\, K$. The standard free energy change of the reaction at the same temperature is

KCETKCET 2012Thermodynamics

Solution:

Standard free energy change,

$\Delta G^{\circ}=-2.303 \,RT \log\, K$

$=-2.303 \times 8.314 \times 298 \times \log 0.008$

$=-2.303 \times 8.314 \times 298 \times(3 \log 2-3 \log 10)$

$=-2.303 \times 8.314 \times 298 \times(0.903-3.0)$

$=-2.303 \times 8.314 \times 298 \times(-2.097)$

$=+11965.16\, J =+11.96\, kJ$