A reaction has a standard free-energy change of − 13.50 KJ mol − 1...

General guidance
Concepts and reason
This problem is based on the concept of thermodynamics.
The thermodynamic potential which is used to calculate maximum reversible work done at constant pressure and temperature is called Gibbs free energy.
Fundamentals
In a chemical reaction, the value of reaction quotient when the reaction reach equilibrium is called equilibrium constant.
The relation between change in Gibbs free energy $\mathrm{\Delta }G$ and equilibrium constant $(K)$ is as follow, $\mathrm{\Delta }G=-RT\ln K$...… (1)
Here, T is temperature and R is gas constant.
Convert change in Gibbs free energy in joules per mole as follows:
$$\begin{gathered} \mathrm{\Delta }G=(-13.50\frac{kJ}{mol})(\frac{1000J}{1kJ}) \\ =13500Jmo{l}^{-1} \end{gathered}$$
Convert temperature in kelvin as follows:
$T_k=T_c+273.15$
Substitute ${25}^{\circ }C$ for T in above equation as follows:
$$\begin{gathered} T_k=(25+273.15)K \\ =298.15K \end{gathered}$$
The gibbs free energy is in $kJ/mol$ It is converted into $J/mol$ by the use of conversion factor. Temperature is converted into Kelvin by adding 273.15 factor to it.
Rearrange equation (1) for K as follows:
$$\begin{gathered} \ln K=\frac{-\mathrm{\Delta }G}{RT} \\ K=\exp (\frac{-\mathrm{\Delta }G}{RT}) \end{gathered}$$
Substitute $8.314J{K}^{-1}mo{l}^{-1}$ for R, $-13500Jmo{l}^{-1}$ for $\mathrm{\Delta }G$ and $298.15K$ for T in above equation as follows:
$$\begin{gathered} K=\exp (\frac{-(-13500Jmo{l}^{-1})}{(8.314J{K}^{-1}mo{l}^{-1})(298.15K)}) \\ K=231.86 \end{gathered}$$
The equilibrium constant is 231.86.