Question

2H₂O(g) + 2Cl₂ (g) $\rightleftharpoons$ 4HCl(g) + O₂(g) Kₚ = 12 × 10⁸ atm

At certain temperature (T) for the gas phase reaction If Cl₂, HCl & O₂ are mixed in such a manner that the partial pressure of each is 2 atm and the mixture is brough into contact with excess of liquid water. What would be approximate partial pressure (in 10⁻³ atm) of Cl₂ when equilibrium is attained at temperature (T)?

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

• A. 3.6
• B. 7.2
• C. 1.8
• D. 0.9

Answer 1

Answer: (A)

3.6

Answer 2

The reaction is: $2\text{H}_2\text{O(g)} + 2\text{Cl}_2\text{(g)} \rightleftharpoons 4\text{HCl(g)} + \text{O}_2\text{(g)}$ $K_p = 12 \times 10^8$ atm.

Vapor pressure of water: 380 mm Hg. $P_{\text{H}_2\text{O}} = \frac{380}{760} \text{ atm} = 0.5 \text{ atm}$. Since there is excess liquid water, $P_{\text{H}_2\text{O}} = 0.5$ atm is constant.

Initial partial pressures: $P_{\text{Cl}_2} = 2$ atm, $P_{\text{HCl}} = 2$ atm, $P_{\text{O}_2} = 2$ atm. The reaction quotient $Q_p = \frac{(P_{\text{HCl}})^4 P_{\text{O}_2}}{(P_{\text{H}_2\text{O}})^2 (P_{\text{Cl}_2})^2} = \frac{(2)^4 \times 2}{(0.5)^2 (2)^2} = \frac{16 \times 2}{0.25 \times 4} = \frac{32}{1} = 32$. Since $Q_p \ll K_p$, the reaction proceeds to the right.

Given $K_p$ is very large, the reaction will proceed significantly to the right. Let the equilibrium partial pressure of $Cl_2$ be $p$. Since the reaction is strongly product-favored, we can assume that the reaction consumes most of the $Cl_2$. Let $P_{\text{Cl}_2} = p$. The change in $P_{\text{Cl}_2}$ is $p - 2$. The change in $P_{\text{HCl}}$ is $\frac{4}{2}(p - 2) = 2(p - 2)$. The change in $P_{\text{O}_2}$ is $\frac{1}{2}(p - 2)$.

Equilibrium partial pressures: $P_{\text{H}_2\text{O}} = 0.5$ atm (constant) $P_{\text{Cl}_2} = p$ $P_{\text{HCl}} = 2 + 2(p-2) = 2 + 2p - 4 = 2p - 2$ $P_{\text{O}_2} = 2 + \frac{1}{2}(p-2) = 2 + \frac{p}{2} - 1 = 1 + \frac{p}{2}$

Since $K_p$ is very large, $p$ must be very small. We can approximate: $P_{\text{HCl}} \approx 2(0) - 2 = -2$ (This approach is flawed as it assumes the change is relative to initial state, but $P_{H_2O}$ is fixed).

Let's assume the reaction proceeds until $P_{\text{Cl}_2}$ is very small, $p$. The amount of $Cl_2$ consumed is $2-p$. This implies that the amount of $H_2O$ consumed is $\frac{2}{2}(2-p) = 2-p$. The amount of $HCl$ produced is $\frac{4}{2}(2-p) = 2(2-p)$. The amount of $O_2$ produced is $\frac{1}{2}(2-p)$.

Equilibrium partial pressures: $P_{\text{H}_2\text{O}} = 0.5$ atm (constant due to excess liquid) $P_{\text{Cl}_2} = p$ $P_{\text{HCl}} = 2 + 2(2-p) = 2 + 4 - 2p = 6 - 2p$ $P_{\text{O}_2} = 2 + \frac{1}{2}(2-p) = 2 + 1 - \frac{p}{2} = 3 - \frac{p}{2}$

Since $p$ is very small, we can approximate: $P_{\text{HCl}} \approx 6$ atm $P_{\text{O}_2} \approx 3$ atm

Substitute these into the $K_p$ expression: $K_p = \frac{(P_{\text{HCl}})^4 P_{\text{O}_2}}{(P_{\text{H}_2\text{O}})^2 (P_{\text{Cl}_2})^2}$ $12 \times 10^8 = \frac{(6)^4 \times 3}{(0.5)^2 \times p^2}$ $12 \times 10^8 = \frac{1296 \times 3}{0.25 \times p^2}$ $12 \times 10^8 = \frac{3888}{0.25 \times p^2}$ $p^2 = \frac{3888}{0.25 \times 12 \times 10^8} = \frac{3888}{3 \times 10^8} = 1296 \times 10^{-8}$ $p = \sqrt{1296 \times 10^{-8}} = 36 \times 10^{-4}$ atm.

Convert to $10^{-3}$ atm: $p = 36 \times 10^{-4} \text{ atm} = 3.6 \times 10^{-3} \text{ atm}$. The approximate partial pressure of $Cl_2$ is $3.6 \times 10^{-3}$ atm. The value in $10^{-3}$ atm is 3.6.