Question

A reaction is at equilibrium at $153{}^{o}C$ and the enthalpy change for the reaction is $42.6kJmo{l}^{-1}.$ What will be the value of $△S$ ?

• A. 120 $J{K}^{-1}mo{l}^{-1}$
• B. 426.2 $J{K}^{-1}mo{l}^{-1}$
• C. 373.1 $J{K}^{-1}mo{l}^{-1}$
• D. 100 $J{K}^{-1}mo{l}^{-1}$

Answer 1

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

We know that ,
$△G=△H-T△S$
At equilibrium, $△G=0$
So, $△H=T△S$
$△S=\frac{△H}{T}$
$T=153+273=426K$
putting the values we get :
$△S=\frac{42600Jmo{l}^{-1}}{426K}=100J{K}^{-1}mo{l}^{-1}$