Question

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

Answer 1

$\begin{array}{cccccccc}& A& +& B& \rightleftharpoons & C& +& D\\ \text{Initially at t = 0}& 1& & 1& & 1& & 1\\ \text{At equilibrium}& 1-x& & 1-x& & 1+x& & 1+x\end{array}$

$K_{eq}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{(1+x)(1+x)}{(1-x)(1-x)}=\frac{(1+x{)}^2}{(1-x{)}^2}$
$\text{or},100={\left(\frac{1+x}{1-x}\right)}^2\text{or},10=\frac{1+x}{1-x}$
$\text{or},10-10x=1+x$
$\text{or},10-1=x+10x$
$\text{or},9=11x$
$\text{or},x=\frac{9}{11}=0.818$
$\therefore \left[D\right]=1+x=1+0.818=1.818$