Question

For the reaction, $A\left(g\right)+2B\left(g\right)\leftrightharpoons 2C\left(g\right)$, one mole of A and $1.5$ moles of B are taken in a $2.0$ litre vessel. At equilibrium, the concentration of C was found to be a $0.35mol{L}^{-1}$. The equilibrium constant $K_C$ of the reaction would be:

• A. $0.295M^{-1}$
• B. $2.36M^{-1}$
• C. $1.178M^{-1}$
• D. $0.673M^{-1}$

Answer 1

Answer: (D)

$0.673M^{-1}$

$A_{\left(g\right)}+2B_{\left(g\right)}\rightleftharpoons 2C_{\left(g\right)}$

$1.5$ $1.5$ $-$ $x=0.35mol{L}^{-1}$

$\frac{1.5-x}{2}$ $\frac{1.5-x}{2}$ $x$

$K_C=\frac{{\left[C\right]}^2}{\left[A\right]{\left[B\right]}^2}=\frac{{\left[0.35\right]}^2}{\left[0.575\right]{\left[0.575\right]}^2}=\frac{0.1225}{0.575\times 0.331}=0.673$