Question

For a chemical reaction $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$ ($∆{\mathrm{rH}}^{\mathrm{\psi }}=80\mathrm{kJ}{\mathrm{mol}}^{-1}$) the entropy change $∆{\mathrm{rS}}^{\mathrm{\psi }}$depends on the temperature $\mathrm{T}$(in $\mathrm{K}$) as $∆\mathrm{rS}=2\mathrm{T}(\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1})$. Minimum temperature at which it will become spontaneous is $\mathrm{K}$,

Answer 1

The solution of the above question can be solved as:

Step1: $∆{\mathrm{rH}}^{\mathrm{\psi }}=80\mathrm{kJ}{\mathrm{mol}}^{-1}$, $∆{\mathrm{rS}}^{\mathrm{\psi }}=2\mathrm{T}\left({\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)$

Step 2: According to thermodynamics, $∆G=∆H-T∆S$

Spontaneity criteria for a reaction, $∆G<0$

Thus,$∆H-T∆S<0$

$∆H<T∆S$; $∆H/∆S<T$

Step3: $∆{\mathrm{rH}}^{\mathrm{\psi }}/∆{\mathrm{rS}}^{\mathrm{\psi }}<\mathrm{T}$

$80000/2T<T$$=$ $40000$ $<{\mathrm{T}}^2$

$\mathrm{T}=200\mathrm{K}$

Therefore, The minimum temperature at which the reaction become spontaneous is $200\mathrm{K}$