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Standard XII

Chemistry

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

Two moles of ...

Question

Two moles of $N_{2}$ and 2 moles of $H_{2}$ are taken in a closed vessel of 5 L capacity and suitable conditions are provided. When the equilibrium is reached, it is found that 0.6 mole of $H_{2}$ is used up. The equilibrium concentration of $N H_{3}$ is:

0.4

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0.08

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0.8

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0.24

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Solution

The correct option is A 0.4
$N_{2} + 3 H_{2} ⇋ 2 N H_{3}$
At $t = 0$ - $\frac{2}{5}$ $\frac{2}{5}$ $0$
At equilibrium- $(2 / 5 - x)$ $(2 / 5 - 3 x)$ $2 x$
But $\frac{2}{5} - (\frac{2}{5} - 3 x) = 0.6$
$∴ 3 x = 0.6$
$∴ x = 0.2$
$∴$ Equilibrium concentration of $N H_{3} = 2 x = 2 \times 0.2 = 0.4$

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Two moles of N2 and 2 moles of H2 are taken in a closed vessel of 5 L capacity and suitable conditions are provided. When the equilibrium is reached, it is found that 0.6 mole of H2 is used up. The equilibrium concentration of NH3 is:

The correct option is A 0.4 N2+3H2⇋2NH3At t=0- 25 25 0At equilibrium- (2/5−x) (2/5−3x) 2xBut 25−(25−3x)=0.6∴3x=0.6∴x=0.2∴ Equilibrium concentration of NH3=2x=2×0.2=0.4

Two moles of $N_{2}$ and 2 moles of $H_{2}$ are taken in a closed vessel of 5 L capacity and suitable conditions are provided. When the equilibrium is reached, it is found that 0.6 mole of $H_{2}$ is used up. The equilibrium concentration of $N H_{3}$ is: