$HI \rightleftharpoons \frac {1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$
is $8.0$. The equilibrium constant of the reaction
$H_{2( g )} + I_{2(g)} \rightleftharpoons 2HI_{ ( g )}$ will be
$\mathrm{K}_{1}=\frac{\left[\mathrm{H}_{2}\right]^{1 / 2}\left[\mathrm{I}_{2}\right]^{1 / 2}}{[\mathrm{HI}]}$
$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) 2 \mathrm{H}(\mathrm{g})$
$\mathrm{K}_{2}=\frac{[\mathrm{H}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}$
From eqs $(i)$ and $(ii)$ $\mathrm{K}_{1}^{2}=\frac{1}{\mathrm{K}_{2}}$
$\mathrm{K}_{1}=8.0$
$\mathrm{k}_{2}=\frac{1}{\mathrm{k}_{1}^{2}}=\frac{1}{8^{2}}=\frac{1}{64}$
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