Discussion Forum

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

Answer:

Explanation:

                      $A + B \rightleftharpoons  C  +  D$

initially at t=0,  1      1                             1      1

At equilibrium  1+x   1+x                          1-x    1-x

$K_{eq}=\frac{[C][D]}{[A][B]}$

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

$100= \left(\frac{1+x}{1-x}\right)^2$

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

$10-10x=1+x$

$9=11x$

$x=\frac{9}{11}=0.818$

D=1+x=1+0.818=1.818