Question

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

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

• A. 2 and 7.5 moles
• B. 1 and 0.75 moles
• C. 1 and 75 moles
• D. 1 and 0.075 mole

Answer 1

Answer: (B)

1 and 0.75 moles

$K_p=\frac{\left[C{O}_2\right]\left[H_2\right]}{\left[CO\right]\left[H_{2}O\right]}$

$K_p=\frac{0.2\times 0.6}{0.4\times 0.3}$

$K_p=1$
Now we wish to find the number of moles of $C{O}_2$ added so that the equilibrium shifts backward and finally there are 0.6 moles of $C{O}_2$.
Since, $\Delta n_g=0$
$K_p=K_c=1$
Suppose 'x' mole $C{O}_2$ is added. then,
$CO\left(g\right)+H_{2}O\left(g\right)\to C{O}_2\left(g\right)+H_2\left(g\right)$
Initially (0.4) (0.3) (0.2 + x) (0.6)

At equilibrium (0.4+0.2) (0.3+0.2) (0.2+x - 0.2) (0.6-0.2)
$K_p=\frac{0.6\times 0.5}{x\times 0.4}=1$

$⟹x=0.75$

Option B is correct.