Question

The equilibrium constant at $298K$ for a reaction $A+B\rightleftharpoons C+D$ is $100$. If the initial concentration of all the four species were $1M$ each, then equilibrium concentration of $D$ (in $mol{L}^{-1}$) will be:

• A. $0.182$
• B. $0.818$
• C. $1.818$
• D. $1.182$

Answer 1

The correct option is C $1.818$

Solution:

$A+B\rightleftharpoons C+D$

$1111$

$1-x1-x1+x1+x$

$K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

$=\frac{(1+x){}^2}{(1-x){}^2}$

$=100$

$\Rightarrow \frac{1+x}{1-x}=10$

$\Rightarrow 1+x=10-10x$

$\Rightarrow 11x=9$

$x=\frac{9}{11}$

$\left[D\right]=1+x=1+\frac{9}{11}$

$=1.818M$ is the answer.