Question

In the reaction: $A_{2\left(g\right)}+3B_{2\left(g\right)}\to 2A{B}_{3\left(g\right)}$
the standard entropies (in $J{K}^{-1}$ $mo{l}^{-1}$) of $A_{2(g)}$, $B_{2(g)}$ and $A{B}_{3(g)}$ are respectively $190,130$ and $195$ and the standard enthalpy change for the reaction is $-95kJmo{l}^{-1}$. The temperature ($inK$) 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

Answer: (A)

For the reaction,

$A_2(g)+3B_2(g)⟶2A{B}_3(g)$

Entropy change, $\Delta S^{\circ }=\Sigma S_{\text{product }}-\Sigma S_{\text{reactants }}$

$=2\times S_{A{B}_3}-\left[S_{A_2}+3S_{B_2}\right]$

$=2\times 195-[190+3\times 130]$

$=390-[190+390]$

$=-190J{K}^{-1}mo{l}^{-1}$

Given, $\Delta H^{\circ }=-95kJmo{l}^{-1}$

At equilibrium, $\Delta G^{\circ }=0$

We know that,

Or $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$

$0=\Delta H^{\circ }-T\Delta S^{\circ }$

or $\Delta H^{\circ }=T\Delta S^{\circ }$

$T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}=\frac{-95\times 10^{3}Jmo{l}^{-1}}{-190J{K}^{-1}mo{l}^{-1}}$ $=500K$