A mixture of 0.3 mole of $H_{2}$ and 0.3 mole of $I_{2}$ is allowed to react in a 10 L evacuated flask at $500℃$. The reaction is $H_{2} + I_{2} \rightleftharpoons 2HI,$ the $K$ is found to be 64. The amount of unreacted $I_{2}$ at equilibrium is

Question

Dr. Abubacker · Accepted Answer

(b)
$K_{c} = \frac{\lbrack HIbrack^{2}}{\left\lbrack H_{2} ightbrack\lbrack I_{2}brack}$
$	herefore 64 = \frac{x^{2}}{0.03 	imes 0.03}$
$	herefore x^{2} = 64 	imes 9 	imes 10^{- 4}$
or, $x = 8 	imes 3 	imes 10^{- 2}$
$x$ is the amount of HI
At equilibrium, amount of $I_{2}$ will be
$0.30 - 0.24 = 0.06$

Q. A mixture of 0.3 mole of $H_{2}$ and 0.3 mole of $I_{2}$ is allowed to react in a 10 L evacuated flask at $500℃$. The reaction is $H_{2} + I_{2} \rightleftharpoons 2HI,$ the $K$ is found to be 64. The amount of unreacted $I_{2}$ at equilibrium is

Explanation

Reviewed by Dr. Abubacker

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