Question

The equilibrium constant, $k_{c}$ for
$3 C _{2} H _{2}(g) \rightleftharpoons C _{6} H _{6}(g)$
is $4 \,L ^{2} mol ^{-2}$. If the equilibrium concentration of benzene is $0.5\, mol\, L ^{-1}$, that of acetylene in $mol \m,L ^{-1}$ must be

• A. $0.025$
• B. $0.25$
• C. $0.05$
• D. $0.5$

Answer 1

Answer: (D)

For the reaction,
$3 C _{2} H _{2}(g) \rightleftharpoons C _{6} H _{6}(g)$
$K_{c} =\frac{\left[ C _{6} H _{6}\right]}{\left[ C _{2} H _{2}\right]^{3}}$
$ \Rightarrow 4=\frac{0.5}{\left[ C _{2} H _{2}\right]^{3}} $
${\left[ C _{2} H _{2}\right]^{3} } =\frac{0.5}{4}=\frac{1}{8}$
${\left[ C _{2} H _{2}\right] } =\frac{1}{2}=0.5 \,mol / L$