Question

At a temperature of 298.15 Kelvin, the free energy change of a reaction (\(\Delta G^{\circ}\)) is 19.737 kcal/mole. If the universal gas constant (R) = 1.98717 Calorie/degree/mole, the \(\log_{10}\) of the equilibrium constant \(K\) is _________ (rounded off to two decimal places).

Hint:

The Gibbs free energy equation can be used to calculate the equilibrium constant \(K\) for a reaction at a given temperature.

Answer 1

Correct Answer: -14.4

Step 1: Use the Gibbs Free Energy Equation.
The relationship between the free energy change \(\Delta G^{\circ}\), temperature \(T\), the gas constant \(R\), and the equilibrium constant \(K\) is given by: \[ \Delta G^{\circ} = -RT \ln K, \] where \(T\) is the temperature in Kelvin, \(R\) is the gas constant in appropriate units, and \(K\) is the equilibrium constant. Rearranging to solve for \(K\): \[ \log_{10} K = \frac{-\Delta G^{\circ}}{2.303 \cdot RT}. \] Substituting the given values: \[ \log_{10} K = \frac{-19.737 \times 1000}{2.303 \cdot 1.98717 \cdot 298.15} = 32.0. \] Step 2: Conclusion.
The \(\log_{10}\) of the equilibrium constant \(K\) is 32.0.