Question

At $300\,\mathrm{K}$, a certain reaction has an equilibrium constant equal to 10. The value of $\Delta G^\circ$ for this reaction (in $\mathrm{kJ\,mol^{-1}}$) is:} \[ (R=8.314\,\mathrm{J\,K^{-1}mol^{-1}}) \]

• A. $-5.74$
• B. $+5.74$
• C. $-2.49$
• D. $+2.49$

Hint:

Remember: \[ \Delta G^\circ=-RT\ln K. \] If $K>1$, then $\Delta G^\circ$ is negative.

Answer 1

The Correct Option is A

Concept: The standard Gibbs free energy change and equilibrium constant are related by \[ \Delta G^\circ=-RT\ln K. \] A large equilibrium constant corresponds to a negative value of $\Delta G^\circ$.

Step 1: Substitute the given values. \[ K=10 \] \[ T=300\,\mathrm{K} \] \[ R=8.314\,\mathrm{J\,K^{-1}mol^{-1}}. \] Therefore, \[ \Delta G^\circ = -(8.314)(300)\ln(10). \]

Step 2: Use $\ln 10 = 2.303$. \[ \Delta G^\circ = -(8.314)(300)(2.303). \] \[ =-5743.6\ \mathrm{J\,mol^{-1}}. \]

Step 3: Convert into kJ mol$^{-1}$. \[ \Delta G^\circ = -5.74\ \mathrm{kJ\,mol^{-1}}. \] Hence, \[ \boxed{-5.74\ \mathrm{kJ\,mol^{-1}}} \]