Question

In the reaction; $A_2\left(g\right)+3B_2\left(g\right)⟶2A{B}_3\left(g\right)$ the standard entropies in $\left(J{K}^{-1}{mol}^{-1}\right)$ of $A_2\left(g\right),$ $B_2\left(g\right)$ and $A{B}_3\left(g\right)$ are respectively $190,$ $130$ and $195$ and the standard enthalpy change for the reaction is $-95$ $kJ$ ${mol}^{-1}$. The temperature (in $K$) at which the reaction attains equilibrium is (assuming both the standard entropy change and standard enthalpy change for this reaction are constant over a wide range of temperature):

• A. $500$
• B. $400$
• C. $300$
• D. $600$
• E. $700$

Answer 1

The correct option is A $500$

For the reaction,
$A_2\left(g\right)+3B_2\left(g\right)⟶2A{B}_3\left(g\right)$

Entropy change, $\Delta S^o=\sum S_{product}-\sum S_{reactants}$
$=2\times S_{A{B}_3}-[S_{A_2}+3S_{B_2}]$
$=2\times 195-[190+3\times 130]$
$=390-[190+390]$
$=-190$ $J{K}^{-1}{mol}^{-1}$

Given, $\Delta H^o=-95$ $kJ$ ${mol}^{-1}$
At equilibrium, $\Delta G^o=0$

We know that,
$\Delta G^o=\Delta H^o-T\Delta S^o$
$0=\Delta H^o-T\Delta S^o$
or $\Delta H^o=T\Delta S^o$

$T=\frac{\Delta H^o}{\Delta S^o}=\frac{-95\times {10}^{3}J{mol}^{-1}}{-190J{K}^{-1}{mol}^{-1}}$
$=500K$