Question

Question: The equilibrium constant is 0.0900 at 25^oC for the reaction

${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{HOCl}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift?

a. ${\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{atm}$,${\mathbf{P}}_{{\mathbf{Cl}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{atm}$,${\mathbf{P}}_{\mathbf{HOCl}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{atm}$.

b. ${\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{200}\mathbf{.}\mathbf{ }\mathbf{torr}$, ${\mathbf{P}}_{{\mathbf{Cl}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{49}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{torr}$,${\mathbf{P}}_{\mathbf{HOCl}}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{torr}$

c. ${\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{296}\mathbf{ }\mathbf{torr}$,${\mathbf{P}}_{{\mathbf{Cl}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{torr}$,${\mathbf{P}}_{\mathbf{HOCl}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{torr}$

Short answer

a. It is not at equilibrium because Q>Kp. The system will shift to the left to re-establish the equilibrium.

b. It is not at equilibrium because Q<Kp. The system will shift to the right.

c. It is at equilibrium because Q=K_p.

Step-by-step solution

Find the Q value for the given set of conditions in part a.

The expression for the reaction quotient, Q for the given reaction, is shown below:

$\mathrm{Q}=\frac{{\mathrm{P}}_{{\mathrm{HOCl}}^2}}{\left({\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}\right)\left({\mathrm{P}}_{{\mathrm{Cl}}_2\mathrm{O}}\right)}...\left(1\right)$

The given partial pressures are :localid="1648814565409" ${\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}=1.00\mathrm{atm},{\mathrm{P}}_{{\mathrm{Cl}}_2\mathrm{O}}=1.00\mathrm{atm},{\mathrm{P}}_{\mathrm{HOCl}}=1.00\mathrm{atm}.$

Substitute these values into Equation (1).

$$\begin{gathered} Q=\frac{{\left(1.00\right)}^2}{\left(1.00\right)\left(1.00\right)} \\ =1.00 \end{gathered}$$

The equilibrium constant value is given as 0.0900.

The obtained Q value is greater than K.So it is not at equilibrium. Hence, the system will shift to the left.

Calculate the Q value for the given set of conditions in part b.

The given partial pressures are ${\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}=200.\mathrm{torr},{\mathrm{P}}_{{\mathrm{Cl}}_2\mathrm{O}}=49.8\mathrm{torr},{\mathrm{P}}_{\mathrm{HOCl}}=21.0\mathrm{torr}$.

Substitute these values into Equation (1).

$$\begin{gathered} Q=\frac{{\left(21.0\right)}^2}{\left(200\right)\left(49.8\right)} \\ =4.43\times {10}^{-2} \end{gathered}$$

The calculated Q value is less than K_p. So, it is not at equilibrium. Hence, the system will shift to the right to reach equilibrium again.

Calculate the Q value for the given conditions in part c.

The given partial pressures are ${\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}=296 \mathrm{torr}$,${\mathrm{P}}_{{\mathrm{Cl}}_2\mathrm{O}}=15.0 \mathrm{torr}$,${\mathrm{P}}_{\mathrm{HOCl}}=20.0 \mathrm{torr}$

Substitute these values into Equation (1).

$$\begin{gathered} \text{Q =}\frac{{\left(\text{20.0}\right)}^{\text{2}}}{\left(\text{296}\right)\left(\text{15.0}\right)} \\ \text{= 0.0901} \end{gathered}$$

The calculated Q value is approximately equal to the K_p value. Hence, it is at equilibrium.