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Q.

A one litre closed vessel contains 2 moles of nitrogen, 1 mole of hydrogen and 2 moles of ammonia at equilibrium. Equilibrium constant for the reaction, $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {NH}_{3} is$

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a

2 M

b

1.414 M

c

1.414 M^-1

d

2 M^-1

answer is C.

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Detailed Solution

$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3} K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}; K_{c} = \frac{{(\frac{2}{1})}^{2}}{(\frac{2}{1}) {(\frac{1}{1})}^{3}}; K_{c} = 2 \times 2 \times \frac{1}{2} = 2 M^{- 2} For \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {NH}_{3}, K_{c} = {(2 M^{- 2})}^{\frac{1}{2}}$

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