Chemistry - Chemical Equilibrium Question with Solution | TestHub

A. Question:

In the system LaCl₃(s) + H₂O(g) + heat ⇌ LaClO(s) + 2HCl(g), equilibrium is established. More water vapor is added to re-establish the equilibrium. The pressure of water vapor is doubled at the new equilibrium compared to the old equilibrium. What is the factor by which the pressure of HCl is changed?

B. Given Data:

Reaction: LaCl₃(s) + H₂O(g) + heat ⇌ LaClO(s) + 2HCl(g)

Pressure of H₂O at new equilibrium = $2\times$ Pressure of H₂O at old equilibrium

C. Concept: Le Chatelier's Principle.

D. Mathematical Calculation:

1. Initial Equilibrium:

Let $P_{H_{2}O}=P_1$ and $P_{HCl}=P_2$.

Then, the equilibrium constant $K_p$ is given by:

$K_p=\frac{(P_2{)}^2}{P_1}$

2. New Equilibrium:

Let $P_{H_{2}O}=2P_1$ and $P_{HCl}=x{P}_2$ (where $x$ is the factor by which the pressure of HCl changes).

Then, $K_p$ is given by:

$K_p=\frac{(x{P}_2{)}^2}{2P_1}$

3. Equating $K_p$:

Since the equilibrium constant remains the same at a given temperature:

$\frac{(P_2{)}^2}{P_1}=\frac{(x{P}_2{)}^2}{2P_1}$

$1=\frac{x^2}{2}$

$x^2=2$

$x=\sqrt{2}$

E. Final Answer:

The pressure of HCl is changed by a factor of $\sqrt{2}$. The closest option to $\sqrt{2}$ is $1.414$. The factor by which the pressure of HCl changes is $\sqrt{2}$.