Chemistry - Chemical Equilibrium Question with Solution | TestHub

The reaction is $A+B\rightleftharpoons C+D$.

Given $K_c=100$ and initial concentrations $[A{]}_0=[B{]}_0=[C{]}_0=[D{]}_0=1$ M.

The reaction quotient $Q_c=\frac{[C{]}_0[D{]}_0}{[A{]}_0[B{]}_0}=\frac{(1)(1)}{(1)(1)}=1$.

Since $Q_c<K_c$, the reaction will proceed in the forward direction to reach equilibrium.

Let $x$ be the change in concentration.

Initial: $[A]=1,[B]=1,[C]=1,[D]=1$

Change: $[A]=-x,[B]=-x,[C]=+x,[D]=+x$

Equilibrium: $[A]=1-x,[B]=1-x,[C]=1+x,[D]=1+x$

$K_c=\frac{[C][D]}{[A][B]}=\frac{(1+x)(1+x)}{(1-x)(1-x)}={\left(\frac{1+x}{1-x}\right)}^2=100$

Taking the square root of both sides:

$\frac{1+x}{1-x}=\sqrt{100}=10$

$1+x=10(1-x)$

$1+x=10-10x$

$11x=9$

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

Equilibrium concentration of D:

$[D]=1+x=1+0.818=1.818$ M