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1 mole of $\mathrm{HI}$ is heated in a closed container of capacity of $2 \mathrm{~L}$. At equilibrium half a mole of $\mathrm{HI}$ is dissociated.
The equilibrium constant of the reaction is
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The equilibrium constant of the reaction is
Solution:
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Verified Answer
The correct answer is:
0.25
$2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2$
Initial mole : 1 x $x$
Final mole: $(1-x) \quad x / 2 \quad x / 2$
$$
K_C=\frac{x^2}{4(1-x)^2}
$$
By putting the value of $x=1 / 2$, we get,
$$
K_C=\frac{1 / 4}{4(1-1 / 2)^2}=0.25
$$
Initial mole : 1 x $x$
Final mole: $(1-x) \quad x / 2 \quad x / 2$
$$
K_C=\frac{x^2}{4(1-x)^2}
$$
By putting the value of $x=1 / 2$, we get,
$$
K_C=\frac{1 / 4}{4(1-1 / 2)^2}=0.25
$$
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