Question

An equilibrium mixture of $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons {CO}_2\left(g\right)+H_2$ present in a vessel of one litre capacity at ${815}^0$C was found by analysis to contain 0.4 mol of $CO$, 0.3 mol of $H_{2}O$, 0.2 mol of ${CO}_2$ and 0.6 mol of $H_2$.

a) Calculate $K_C$.
b) If it is derived to increase the concentration of $CO$ to 0.6 mol by adding ${CO}_2$ to the vessel, how many moles must be added into equilibrium mixture at a constant temperature in order to get this change?

• A. a) $K_C=1.25$
  b) $a=1.5moles$
• B. a) $K_C=1$
  b) $a=0.5moles$
• C. a) $K_C=1.5$
  b) $a=0.75moles$
• D. a) $K_C=1$
  b) $a=0.75moles$

Answer 1

Answer: (D)

a) $K_C=1$

b) $a=0.75moles$
$CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons {CO}_2\left(g\right)+H_2\left(g\right)$
Moles at equilibrium 0.4 0.3 0.2 0.6

a) $K_C=\frac{{[CO}_2\left]{[H}_2\right]}{\left[CO\right]{[H}_{2}O]}=\frac{0.2\times 0.6}{0.4\times 0.3}=1$
$(\because △n=0,\therefore Volumetermsarenotneeded)$

b) Now it is desired to increase the concentration of $CO$ by 0.2 at equilibrium by forcing ${CO}_2$ into equilibrium mixture.
Suppose $a$ mol of ${CO}_2$ are forced in vessel at equilibrium; by doing so reaction proceeds in backward direction,i.e,

${CO}_2+H_2\left(g\right)\rightleftharpoons CO+H_{2}O$
(0.2+a) 0.6 0.4 0.3
(0.2+a-0.2) (0.6-0.2) (0.4+0.2) (0.3+0.2)
a 0.4 0.6 0.5
$\frac{1}{K_C}=\frac{\left[CO\right]{[H}_{2}O]}{{[CO}_2\left]\right[H_2]}=\frac{0.6\times 0.5}{a\times 0.4}$

$\therefore a=0.75moles$