Question

For hypothetical reversible reaction:
$\frac{1}{2}{A}_2\left(g\right)+\frac{3}{2}{B}_2\left(g\right)⟶{AB}_3\left(g\right);$

$\Delta H=-20kJ$ if standard entropies of $A_2,B_2$ and ${AB}_3$ are 60, 40 and $50J.K^{-1}.mo{l}^{-1}$, respectively. The above reaction will be equilibrium at:

• A. 400 K
• B. 500 K
• C. 250 K
• D. 200 K

Answer 1

The correct option is B 500 K

For hypothetical reversible reaction,

$\frac{1}{2}{A}_{2\left(g\right)}+\frac{3}{2}{B}_{2\left(g\right)}\to A{B}_{2\left(g\right)}$

$△H=-20kJ=-20\times {10}^{3}J$

${△}_{r}S=\sum S_{product}^o-\sum S_{reactant}^o$

$=50J{K}^{-1}-(\frac{1}{2}\times 60J{K}^{-1}+\frac{3}{2}\times 40J{K}^{-1})$

$=50J{K}^{-1}-90J{K}^{-1}$

$\therefore {△}_{r}S=-40J{K}^{-1}$

At equilibrium, ${△}_{r}G=△H-T{△}_{r}S=0$

$\Rightarrow -20\times {10}^{3}J-T(-40J{K}^{-1})=0$

$\Rightarrow T=\frac{20\times {10}^{3}J}{40J{K}^{-1}}$

$\Rightarrow T=500K$