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$3 C_{2} H_{2} \Leftrightarrow C_{6} H_{6}$ the above reaction is performed initially with 2 moles of C₂H₂ in a litre vessel. Equilibrium is established when 0.5 mole of benzene is present at certain temperature. If equilibrium constant is $4 {lit}^{2} {mole}^{- 2}$ . The total number of moles of the substance present at equilibrium.

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1.5

0.5

answer is B.

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Detailed Solution

$\begin{array}{l} K_{C} = \frac{|C_{6} H_{6}|}{{(C_{2} H_{2})}^{3}}, 4 = \frac{0 .5}{x^{3}}, x^{3} = \frac{0 .5}{4} = 0 .125 \\ x = \sqrt[3]{0 .125} \end{array}$
Total = 0.5+0.5 =1

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