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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

4 moles of HI...

Question

4 moles of $H I$ are taken in a 1 litre closed vessel and heated till equilibrium is attained. At equilibrium, the concentration of $H_{2}$ is $1 {mol L}^{- 1}$ . What is the equilibrium constant?

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0.5

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0.25

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Solution

The correct option is D 0.25
$\begin{matrix} H I & ⇌ & H_{2} & + & I_{2} Initial & 4 & 0 & 0 Equilibrium & (4 - 2) & 1.0 & 1.0 \end{matrix}$

$K_{e q} = \frac{[H_{2}] [I_{2}]}{[H I]^{2}} = \frac{1 \times 1}{(2)^{2}} = \frac{1}{4} = 0.25$

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4 moles of HI are taken in a 1 litre closed vessel and heated till equilibrium is attained. At equilibrium, the concentration of H2 is 1 mol L−1. What is the equilibrium constant?

The correct option is D 0.25HI⇌H2+I2Initial400Equilibrium(4−2)1.01.0 Keq=[H2][I2][HI]2=1×1(2)2=14=0.25

4 moles of $H I$ are taken in a 1 litre closed vessel and heated till equilibrium is attained. At equilibrium, the concentration of $H_{2}$ is $1 {mol L}^{- 1}$ . What is the equilibrium constant?

The correct option is D 0.25
$\begin{matrix} H I & ⇌ & H_{2} & + & I_{2} Initial & 4 & 0 & 0 Equilibrium & (4 - 2) & 1.0 & 1.0 \end{matrix}$

$K_{e q} = \frac{[H_{2}] [I_{2}]}{[H I]^{2}} = \frac{1 \times 1}{(2)^{2}} = \frac{1}{4} = 0.25$