Question

A container contains three gases: $A,B,$ and $C$ in equilibrium $A\rightleftharpoons 2B+C$At equilibrium the concentration of $A$ was 3M, and of $B$ was 4M. On doubling the volume of container, the new equilibrium concentration of $B$ was 3M. Calculate $K_c$ and initial equilibrium concentration of $C$.

Answer 1

$A\rightleftharpoons 2B+CInitial:a_{\circ }00Equilibrium:a_{\circ }-x2xx$

$\left[A\right]=\frac{a_{\circ }-x}{V}=3,\left[B\right]=\frac{2x}{V}=4M$

Where $V=$ Volume of the container.

$\therefore \frac{x}{V}=2M,\frac{a_{\circ }}{V}=5M$

$\left[C\right]=\frac{x}{V}=2M=$ Initial equilibrium concentration of $C.$

$\therefore K_C=\frac{\left[B{]}^2\right[C]}{\left[A\right]}=\frac{4^2\times 2}{3}=10.67$

On doubling the value, $\left[B\right]=\frac{2x^1}{2V}=4M⟹\frac{x^1}{V}=4M$

$\therefore \frac{a_{\circ }}{V}=3+4=7M$ $\left[C\right]=\frac{x^1}{V}=4M,\left[A\right]=3M$