Question

The rate of the reaction $H_2+I_2\rightleftharpoons 2HI$ is given by:
$rate=1.7\times {10}^{-19}\left[H_2\right]\left[I_2\right]$ at ${25}^{\circ }C$
The rate of decomposition of gaseous $HI$ to $H_2$ and $I_2$ is given by:
$rate=2.4\times {10}^{-21}[HI{]}^2$ at ${25}^{\circ }C$.
Calculate the equilibrium constant for the formation of $HI$ from $H_2$ and $I_2$ at ${25}^{\circ }C$?

Answer 1

For the reversible reaction
$H_2+I_2\rightleftharpoons 2HI$
Equilibrium constant, $K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
At equilibrium $r_f=r_b$
given $r_f=1.7\times {10}^{=19}\left[H_2\right]\left[I_2\right]$
$r_b=2.4\times {10}^{-21}[HI{]}^2$
$\therefore$ at equilibrium
$1.7\times {10}^{-19}\left[H_2\right]\left[I_2\right]=2.4\times {10}^{-21}[HI{]}^2$
or $\frac{1.7\times {10}^{-19}}{2.4\times {10}^{-21}}=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=K_c$
$\therefore K_c=70.83$ at ${25}^{\circ }C$.