Question

Calculate the value of change in entropy for a reaction if it is at equilibrium at $253{}^{o}C$ .The enthalpy change for the reaction is $52.6kJmo{l}^{-1}.$

• A. $500J{K}^{-1}mo{l}^{-1}$
• B. $50J{K}^{-1}mo{l}^{-1}$
• C. $1000J{K}^{-1}mo{l}^{-1}$
• D. $100J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is D $100J{K}^{-1}mo{l}^{-1}$

We know that ,
$\Delta G=\Delta H-T\Delta S$
At equilibrium, $\Delta G=0$
So, $\Delta H=T\Delta S$
$\Delta S=\frac{\Delta H}{T}$
$T=253+273=526K$
putting the values we get :
$\Delta S=\frac{52600Jmo{l}^{-1}}{526K}=100J{K}^{-1}mo{l}^{-1}$