Question

For the system $A\left(g\right)+2B\left(g\right)\rightleftharpoons C\left(g\right)$ the equilibrium concentration is

$A=0.06mol{L}^{-1};B=0.12mol{L}^{-1}$

$C=0.216mol{L}^{-1}$. The $K_{eq}$ for the reaction is :

• A. 250
• B. 416
• C. $4\times {10}^{-3}$
• D. 125

Answer 1

The correct option is A 250

$\underset{0.06M}{A\left(g\right)}+\underset{0.12M}{2B\left(g\right)}\rightleftharpoons \underset{0.216}{C\left(g\right)}$

$K=\frac{\left[C\right]}{\left[B^2\right]\left[A\right]}=\frac{0.216}{0.12\times 0.12\times 0.06}=250$