Question

A reaction is in equilibrium at ${100}^{0}C$ and the enthalpy change for the reaction is 42.6 kJ $mo{l}^{-1}$. What will be the value of $\Delta S$ in J $K^{-1}mo{l}^{-1}$?

• A. 120
• B. 426.2
• C. 373.1
• D. 114.2

Answer 1

Answer: (D)

114.2

$\Delta G=\Delta H-T\Delta S$

At equlibrium, $\Delta G=0=\Delta H-T\Delta S$

Hence, $\Delta S=\frac{\Delta H}{T}$ = $\frac{42600Jmo{l}^{-1}}{373K}=114.2J{K}^{-1}mo{l}^{-1}$