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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

In a study of...

Question

In a study of equilibrium
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
1mol of $H_{2}$ and 3 mol of $I_{2}$ give rise at equilibrium to x mol of HI.
Addition of a further 2 mol of $H_{2}$ gave an additional x mol of HI. What is x?

0.5

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1.5

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None of these

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Solution

The correct option is B 1.5

$H_{2}$
$I_{2}$
HI
Initial
1
3
0
Equilibrium 1
1-0.5x
3-0.5x
x
Equilibrium 2
3-x
3-x
2x
The expression for the equilibrium constant for two equilibria will be $K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]} = \frac{x^{2}}{(1 - 0.5 x) (3 - 0.5 x)} = \frac{(2 x)^{2}}{(3 - x) (3 - x)}$ .
Hence, $x = 1.5$ .

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	$H_{2}$	$I_{2}$	HI
Initial	1	3	0
Equilibrium 1	1-0.5x	3-0.5x	x
Equilibrium 2	3-x	3-x	2x

In a study of equilibrium H2(g)+I2(g)⇌2HI(g)1mol of H2 and 3 mol of I2 give rise at equilibrium to x mol of HI.Addition of a further 2 mol of H2 gave an additional x mol of HI. What is x?

The correct option is B 1.5H2I2HIInitial130Equilibrium 11-0.5x3-0.5xxEquilibrium 23-x3-x2xThe expression for the equilibrium constant for two equilibria will be Kc=[HI]2[H2][I2]=x2(1−0.5x)(3−0.5x)=(2x)2(3−x)(3−x).Hence, x=1.5.

In a study of equilibrium
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
1mol of $H_{2}$ and 3 mol of $I_{2}$ give rise at equilibrium to x mol of HI.
Addition of a further 2 mol of $H_{2}$ gave an additional x mol of HI. What is x?