A mixture of 0-3 mole of H2 and 0-3 mole of I2 is allowed to react in a 10 litre evacuated flask at 500-C- The reaction is H2 - I2- 2HI- the K is found to be 64- The amount of unreacted I2 at equilibrium is

The correct option is B 0.06 mole The equilibrium reaction is H_2(g) + I_2(g) ⇔ 2HI(g) .The number of moles of various species present initially and a ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

A mixture of ...

Question

A mixture of $0.3$ mole of $H_{2}$ and $0.3$ mole of $I_{2}$ is allowed to react in a $10 l i t r e$ evacuated flask at $500^{\circ} C$ . The reaction is $H_{2} + I_{2} ⇌ 2 H I$ , the $K$ is found to be $64$ . The amount of unreacted $I_{2}$ at equilibrium is

0.15 m o l e

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0.06 m o l e

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0.03 m o l e

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0.2 m o l e

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Solution

The correct option is B $0.06 m o l e$
The equilibrium reaction is $H_{2} (g) + I_{2} (g) \Leftrightarrow 2 H I (g)$ .
The number of moles of various species present initially and at equilibrium are given in the following table.

$H_{2}$
$I_{2}$ $H I$
Initial
$0.3$
$0.3$
$0$
change
$x$
$x$
$2 x$
equilibrium
$0.3 - x$
$0.3 - x$
$2 x$
The expression for the equilibrium constant is $K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]}$
Substitute values in the above expression.

$64 = \frac{(\frac{2 x}{10})^{2}}{(\frac{0.3 - x}{10}) \times (\frac{0.3 - x}{10})}$

or $x = 0.24$ .
The amount of unreacted iodine is $0.3 - x = 0.3 - 0.24 = 0.06$ .

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	$H_{2}$	$I_{2}$	$H I$
Initial	$0.3$	$0.3$	$0$
change	$x$	$x$	$2 x$
equilibrium	$0.3 - x$	$0.3 - x$	$2 x$

A mixture of 0.3 mole of H2 and 0.3 mole of I2 is allowed to react in a 10 litre evacuated flask at 500∘C. The reaction is H2+I2⇌2HI, the K is found to be 64. The amount of unreacted I2 at equilibrium is

A mixture of $0.3$ mole of $H_{2}$ and $0.3$ mole of $I_{2}$ is allowed to react in a $10 l i t r e$ evacuated flask at $500^{\circ} C$ . The reaction is $H_{2} + I_{2} ⇌ 2 H I$ , the $K$ is found to be $64$ . The amount of unreacted $I_{2}$ at equilibrium is