Question

For $A+B\rightleftharpoons C$, the equilibrium concentration of $A$ and $B$ at a temperature are $15mol{L}^{-1}$. When volume is doubled the reaction has equilibrium concentration of $A$ as calculate $K_C$.

• A. $K_C=0.04{Mol}^{-1}L$
• B. $K_C=3{Mol}^{-1}L$
• C. $K_C=0.2{Mol}^{-1}L$
• D. $K_C=1.2{Mol}^{-1}L$

Answer 1

The correct option is C $K_C=0.2{Mol}^{-1}L$

$A+B\rightleftharpoons C$
$1515a$ Conc at 1 equilibrium (Volume in $VL$)

$\frac{15}{2}$ ($\frac{15}{2}$) $\frac{a}{2}$ (Conc when $V$ is doubled)
Since. on increasing volume, pressure decreases and the reaction proceeds in the direction where it shows an increase in moles, i.e. backward reaction.
Conc at II equilibrium
$\left[\frac{15}{2}+x\right]\left[\frac{15}{2}+x\right]\left[\frac{a}{2}-x\right]$
$\because \frac{15}{2}+x=10$
$\because x=\frac{5}{2}$
Now,
$K_C=\frac{\left[C\right]}{\left[A\right]\left[B\right]}=\frac{\left[\frac{a}{2}-\frac{5}{2}\right]}{\left[\frac{15}{2}+\frac{15}{2}\right]\left[\frac{15}{2}+\frac{5}{2}\right]}$
$K_C=\frac{\left[C\right]}{\left[A\right]\left[B\right]}=\frac{a}{15\times 15}$
$\because K_{C}aresame$
$\therefore \frac{(a-5)/2}{10\times 10}=\frac{a}{15\times 15}$
$\therefore a=45M$
Now $K_C=\frac{45}{15\times 15}=0.2{Mol}^{-1}{L}^{+1}$