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3 moles of ${NH}_{3}$ are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of $N_{2}$ is $0.2$ mole / lit. Then the total number of moles at equilibrium is

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$1.5$

$7.5$

$2.5$

answer is B.

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Detailed Solution

$2 {NS}_{3} ⇌ N_{2} + 3 H_{2}$

3 0 0

3-2x x 3x

x = 0.1 x 5 =1 mole

Total no of moles= $3 - 2 x + x + 3 x$

$= 3 - 2 + 1 + 3 = 5$ moles

Hence the correct option is (B) 5.

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