1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant, kc for 3 C 2 H 2(g) leftharpoons 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

Q. 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

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Solution:

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}} $
${\left[ C _{2} H _{2}\right]^{3} } =\frac{0.5}{4}=\frac{1}{8}$
${\left[ C _{2} H _{2}\right] } =\frac{1}{2}=0.5 \,mol / L$

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