Question

At certain temperature(T) for the gas phase reaction $2H_{2}O\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 4HCl\left(g\right)+O_2\left(g\right)$ $K_P=12\times {10}^8$atm. If $C{l}_2,HCl$ & $O_2$ are mixed in such a manner that the partial pressure of each is $2$atm and the mixture is brought into contact with excess of liquid water. What would be approximate partial pressure of $C{l}_2$ when equilibrium is attained at temperature (T)?

[Given: Vapour pressure of water is $380$mm Hg at temperature (T)].

• A. $3.6\times {10}^{-5}$ atm
• B. ${10}^{-4}$ atm
• C. $3.6\times {10}^{-3}$ atm
• D. $0.01$ atm

Answer 1

The correct option is C $3.6\times {10}^{-3}$ atm

$K_p$, the pressure constant at equilibrium and $K_c$, the equilibrium constant only changes with the change in temperature and no other factor.

$K_p=12\times {10}^{8}atm;PartialPressur{e}_{water}=0.5atm$

$2H_{2}O\left(g\right)+2C{l}_2\left(g\right)\leftrightharpoons 4HCl\left(g\right)+O_2\left(g\right)$

$0.5-2\alpha$ $2-2\alpha$ $2+4\alpha$ $2+\alpha$

$K_p=\frac{(P_{HCl}{)}^4\times P_{O_2}}{(P_{H_{2}O}{)}^2\times (P_{C{l}_2}{)}^2}$

$P_{C{l}_2}=3.6\times {10}^{-3}atm$

Hence, option (C) is correct.