Question

In a vessel of 1.0 L capacity, O$_2$(g) at 0.25 atm pressure and HCl(g) at 1.0 atm pressure are allowed to react in the presence of liquid water at 57oC.

4HCl(g) + O$_2$(g) $\rightleftharpoons$ 2Cl$_2$(g) + 2H$_2$O(g); K$_p$ = 5.0 × 10$^{12}$ atm$^{-1}$.

The volume occupied by liquid water is negligible, but it is sufficient at achieve equilibrium with water vapour. The vapour pressure of water at 57oC is 0.4 atm. Select the correct statement(s) regarding the equilibrium mixture.

• A. The partial pressure of water vapour at any stage of reaction is 0.4 atm.
• B. The partial pressure of Cl$_2$(g) at equilibrium is 0.5 atm.
• C. The partial pressure of O$_2$(g) at equilibrium is 5.0 × 10$^{-4}$ atm.
• D. The partial pressure of HCl(g) at equilibrium is 2.0 × 10$^{-3}$ atm.

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem describes an equilibrium reaction between HCl(g) and O$_2$(g) to form Cl$_2$(g) and H$_2$O(g) in the presence of liquid water. The temperature is 57°C.

The reaction is:

$4HCl(g) + O_2(g) \rightleftharpoons 2Cl_2(g) + 2H_2O(g)$

Given $K_p = 5.0 \times 10^{12} \text{ atm}^{-1}$ at 57°C. Initial pressures: $P_{HCl} = 1.0 \text{ atm}$, $P_{O_2} = 0.25 \text{ atm}$. Vapour pressure of water at 57°C is 0.4 atm.

Analysis of Statement (A): The partial pressure of water vapour at any stage of reaction is 0.4 atm.

Since liquid water is present and is sufficient to achieve equilibrium with water vapour, the partial pressure of water vapour ($P_{H_2O}$) in the vessel will be constant and equal to its vapour pressure at the given temperature, as long as liquid water is present. Thus, $P_{H_2O} = 0.4 \text{ atm}$ at equilibrium. This statement is correct.

Analysis of Equilibrium:

The equilibrium constant expression is:

$K_p = \frac{(P_{Cl_2})^2 (P_{H_2O})^2}{(P_{HCl})^4 (P_{O_2})}$

Since $P_{H_2O}$ is fixed at 0.4 atm, we can define an effective equilibrium constant $K_p'$ for the gas-phase species:

$K_p' = \frac{K_p}{(P_{H_2O})^2} = \frac{5.0 \times 10^{12}}{(0.4)^2} = \frac{5.0 \times 10^{12}}{0.16} = 31.25 \times 10^{12} = 3.125 \times 10^{13} \text{ atm}^{-3}$

This $K_p'$ value is very large, indicating that the reaction proceeds almost to completion in the forward direction.

Let's determine the limiting reactant if the reaction were to go to completion:

Initial moles are proportional to initial pressures for gases in the same volume.

For HCl: $1.0 \text{ atm}$

For O$_2$: $0.25 \text{ atm}$

Stoichiometric ratio of HCl to O$_2$ is 4:1.

Moles of HCl required to react with 0.25 atm of O$_2$ = $4 \times 0.25 = 1.0 \text{ atm}$.

Since the initial pressure of HCl (1.0 atm) is exactly what is required to react with 0.25 atm of O$_2$, both reactants (HCl and O$_2$) are consumed completely if the reaction goes to completion.

If the reaction goes to completion:

$P_{HCl, final} = 1.0 - 4(0.25) = 0 \text{ atm}$

$P_{O_2, final} = 0.25 - 0.25 = 0 \text{ atm}$

$P_{Cl_2, final} = 2(0.25) = 0.5 \text{ atm}$

$P_{H_2O, final} = 0.4 \text{ atm}$ (fixed)

Since $K_p'$ is very large, the equilibrium pressures of HCl and O$_2$ will be very small, and the equilibrium pressure of Cl$_2$ will be very close to 0.5 atm.

Let's assume a small amount of reverse reaction occurs from the "completion" state.

Let $P_{O_2, eq} = \xi$.

From stoichiometry of the reverse reaction ($2Cl_2 + 2H_2O \rightarrow 4HCl + O_2$), if $\xi$ atm of O$_2$ is formed, then $4\xi$ atm of HCl is formed, and $2\xi$ atm of Cl$_2$ is consumed.

Equilibrium pressures:

$P_{HCl, eq} = 4\xi$

$P_{O_2, eq} = \xi$

$P_{Cl_2, eq} = 0.5 - 2\xi$

$P_{H_2O, eq} = 0.4$ (fixed)

Substitute these into the $K_p'$ expression:

$K_p' = \frac{(P_{Cl_2})^2}{(P_{HCl})^4 (P_{O_2})}$

$3.125 \times 10^{13} = \frac{(0.5 - 2\xi)^2}{(4\xi)^4 (\xi)}$

Since $\xi$ is very small, we can approximate $0.5 - 2\xi \approx 0.5$:

$3.125 \times 10^{13} = \frac{(0.5)^2}{256 \xi^4 \cdot \xi}$

$3.125 \times 10^{13} = \frac{0.25}{256 \xi^5}$

$\xi^5 = \frac{0.25}{256 \times 3.125 \times 10^{13}}$

$\xi^5 = \frac{0.25}{800 \times 10^{13}} = \frac{0.25}{8 \times 10^{15}} = \frac{25 \times 10^{-2}}{8 \times 10^{15}} = 3.125 \times 10^{-17}$

$\xi = (3.125 \times 10^{-17})^{1/5} = (31.25 \times 10^{-18})^{1/5}$

$\xi = (31.25)^{1/5} \times 10^{-18/5} = (31.25)^{1/5} \times 10^{-3.6}$

We know $2^5 = 32$, so $(31.25)^{1/5}$ is slightly less than 2. Let's use $(31.25)^{1/5} \approx 1.99$.

$\xi \approx 1.99 \times 10^{-3.6} = 1.99 \times 10^{-4} \times 10^{0.4}$

$10^{0.4} \approx 2.51$

$\xi \approx 1.99 \times 2.51 \times 10^{-4} \approx 5.0 \times 10^{-4} \text{ atm}$

Now, let's calculate the equilibrium partial pressures:

(B) The partial pressure of Cl$_2$(g) at equilibrium is 0.5 atm.

$P_{Cl_2, eq} = 0.5 - 2\xi = 0.5 - 2(5.0 \times 10^{-4}) = 0.5 - 0.001 = 0.499 \text{ atm}$.

This is very close to 0.5 atm. So, statement (B) is correct.

(C) The partial pressure of O$_2$(g) at equilibrium is 5.0 × 10$^{-4}$ atm.

$P_{O_2, eq} = \xi = 5.0 \times 10^{-4} \text{ atm}$.

This statement is correct.

(D) The partial pressure of HCl(g) at equilibrium is 2.0 × 10$^{-3}$ atm.

$P_{HCl, eq} = 4\xi = 4 \times (5.0 \times 10^{-4}) = 20.0 \times 10^{-4} = 2.0 \times 10^{-3} \text{ atm}$.

This statement is correct.

All statements (A), (B), (C), and (D) are correct.