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$3 C_{2} H_{2} \Leftrightarrow C_{6} H_{6}$ he above reaction is performed in a 1 lit vessel. Equilibrium is established when 0.5 mole of benzene is present at certain tempe-rature. If equilibrium constant is $4 {lit}^{2} {mole}^{- 2} .$ The total number of moles of the substances present at equilibrium.

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0.5

1.5

answer is B.

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Detailed Solution

$3 C_{2} H_{2} ⇌ C_{6} H_{6}$

$x 0.5$

$K_{C} = \frac{[C_{6} H_{6}]}{{[C_{2} H_{2}]}^{3}}$

$\Rightarrow {[C_{2} H_{2}]}^{3} = \frac{[C_{6} H_{6}]}{K_{C}} = \frac{0.5}{4}$

$\Rightarrow {[C_{2} H_{2}]}^{3} = 0.125$

$\Rightarrow [C_{2} H_{2}] = 0.5$

So no. of moles of $C_{2} H_{2} = 0.5 mole$

Hence total no. of moles=0.5+0.5=1

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