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. A 1.0-L flask contains 1.0 mole of HOCl, 0.10 mole of Cl₂O, and 0.10 mole of H₂O.

b. A 2.0-L flask contains 0.084 mole of HOCl, 0.080 mole of Cl₂O, and 0.98 mole of H₂O.

c. A 3.0-L flask contains 0.25 mole of HOCl, 0.0010 mole of Cl₂O, and 0.56 mole of H₂O.

Short answer

a. It is not at equilibrium because Q>K. The system will shift to the left.

b. It is at equilibrium because Q=K.

c. It is not at equilibrium for the given conditions because Q>K. The system will shift to the left.

Step-by-step solution

Calculate the Q value for the given set of conditions.

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

$\text{Q =}\frac{{\left[\text{HOCl}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\text{O}\right]\left[{\text{Cl}}_{\text{2}}\text{O}\right]}$ …(1)

The concentrations of all chemical species can be calculated as follows:

$$\begin{gathered} \left[\text{HOCl}\right]\text{=}\frac{\text{1.0} \text{mol}}{\text{1.0} \text{L}} \\ \text{= 1.0} \text{M} \end{gathered}$$

$$\begin{gathered} \left[{\text{H}}_{\text{2}}\text{O}\right]\text{=}\frac{\text{0.10} \text{mol}}{\text{1.0} \text{L}} \\ \text{= 0.10} \text{M} \end{gathered}$$

$$\begin{gathered} \left[{\text{Cl}}_{\text{2}}\text{O}\right]\text{=}\frac{\text{0.10} \text{mol}}{\text{1.0} \text{L}} \\ \text{= 0.10} \text{M} \end{gathered}$$

Substitute these values into Equation (1).

$$\begin{gathered} Q=\frac{{\left(1.0\right)}^2}{\left(0.10\right)\left(0.10\right)} \\ =1.0\times {10}^2 \end{gathered}$$

The equilibrium constant, K value, is given as 0.0900.

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

Find the Q value for the given set of conditions.

The concentrations of all chemical species can be calculated as follows:

$$\begin{gathered} \left[\text{HOCl}\right]\text{=}\frac{\text{0.084} \text{mol}}{\text{2.0} \text{L}} \\ \text{= 0.042} \text{mol} \end{gathered}$$

$$\begin{gathered} \left[{\text{H}}_{\text{2}}\text{O}\right]\text{=}\frac{\text{0.98} \text{mol}}{\text{2.0} \text{L}} \\ \text{= 0.49} \text{M} \end{gathered}$$

$$\begin{gathered} \left[{\text{Cl}}_{\text{2}}\text{O}\right]\text{=}\frac{\text{0.080} \text{mol}}{\text{2.0} \text{L}} \\ \text{= 0.040} \text{M} \end{gathered}$$

Now insert these values into Equation (1).

$$\begin{gathered} \text{Q =}\frac{{\left(\text{0.042}\right)}^{\text{2}}}{\left(\text{0.49}\right)\left(\text{0.040}\right)} \\ \text{= 0.090} \end{gathered}$$

The calculated Q value is exactly equal to K. Hence, it is at equilibrium.

$$\begin{gathered} \left[{\text{Cl}}_{\text{2}}\text{O}\right]\text{=}\frac{\text{0.080} \text{mol}}{\text{2.0} \text{L}} \\ \text{= 0.040} \text{M} \end{gathered}$$

Determine the Q value for the given set of conditions.

The concentrations of all chemical species can be calculated as follows:

$$\begin{gathered} \left[\text{HOCl}\right]\text{=}\frac{\text{0.25} \text{mol}}{\text{3.0} \text{L}} \\ \text{= 0.083} \text{M} \end{gathered}$$

$$\begin{gathered} \left[{\text{H}}_{\text{2}}\text{O}\right]\text{=}\frac{\text{0.56} \text{mol}}{\text{3.0} \text{L}} \\ \text{= 0.19} \text{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{Cl}}_2\mathrm{O}\right]=\frac{0.0010\mathrm{mol}}{3.0\mathrm{L}} \\ =3.3\times {10}^{-4}\mathrm{M} \end{gathered}$$

Substitute these values into Equation (1).

$$\begin{gathered} Q=\frac{{\left(0.083\right)}^2}{\left(0.19\right)\left(3.3\times {10}^{-4}\right)} \\ =110 \end{gathered}$$

The determined Q value is larger than the K value. Thus, it is not at equilibrium. Hence, the system will shift to the left.