Question

At 25^oC, K = 0.090 for the reaction

${\mathit{H}}_{\mathbf{2}}\mathit{O}\left(\mathit{g}\right)\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathit{O}\left(\mathit{g}\right)\mathbf{\rightleftharpoons }\mathbf{2}\mathit{H}\mathit{O}\mathit{C}\mathit{l}\left(\mathit{g}\right)$

Calculate the concentrations of all species at equilibrium for each of the following cases.

a. 1.0 g of H₂O and 2.0 g of Cl₂O are mixed in a 1.0-L flask.

b. 1.0 mole of pure HOCl is placed in a 2.0-L flask.

Short answer

a. The equilibrium concentrations of HOCI, H₂O, Cl₂O are 9.2 x 10^-3M, 5.1 x 10^-2M , and 1.8 x 10^-2M, respectively.

b. The equilibrium concentrations of HOCI, H₂O, Cl₂O are 0.07M, 0.22M, and 0.22M, respectively.

Step-by-step solution

Find the initial concentrations of chlorine monoxide and water.

The expression for the equilibrium constant, K, for this reaction is shown below:

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

The given equilibrium constant, K value, is 0.090.

The initial concentrations ofCl₂O and H₂O can be calculated as follows:

For Cl₂O :

$\frac{2.0g}{1.0g}\times \frac{1mol}{86.9g}=2.3\times {10}^{-2}mol/L$

For H₂O :

$\frac{1.0g}{1.0g}\times \frac{1mol}{18.0g}=5.6\times {10}^{-2}mol/L$

Find the value of x.

Now construct the ICE table:

Substitute these equilibrium concentrations in terms of x into equation (1).

$0.090=\frac{{\left(2x\right)}^2}{\left(5.6\times {10}^{-2}-x\right)\left(2.3\times {10}^{-2}-x\right)}$

Solve for x.

x = 4.6 x 10^-3

Calculate the equilibrium concentrations of all chemical species.

The equilibrium concentrations of all chemical speciescan be calculated as follows:

$$\begin{gathered} \left[\text{HOCl}\right]\text{= 2x} \\ =2\times 4.6x{10}^{-3} \\ =9.2x{10}^{-3}M \\ {H}_{2}O=5.6\times {10}^{-2}\times x \\ =5.6\times {10}^{-2}\times 4.6x{10}^{-3} \\ =5.1\times {10}^{-2}M \\ C{l}_{2}O=2.3\times {10}^{-2}\times x \\ =2.3\times {10}^{-2}\times 4.6x{10}^{-3} \\ =1.8\times {10}^{-2}M \end{gathered}$$

Therefore, the equilibrium concentrations ofHOCI, H₂O, and Cl₂O are 9.2 x 10^-3M, 5.1 x 10^-2M, and 1.8 x 10^-2M, respectively.

Find the value of x.

Setup the ICE table:

Substitute the K_p value and these equilibrium concentrations in terms of x into equation (1).

$0.090=\frac{{\left(2x\right)}^2}{\left(5.6\times {10}^{-2}-x\right)\left(2.3\times {10}^{-2}-x\right)}$

Solve for x.

x = 0.217

Calculate the equilibrium concentrations of all chemical species.

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

$$\begin{gathered} \left[\text{HOCl}\right]\text{= 0.50 - 2x} \\ =0.50-2\times 0.217 \\ =0.07M \\ {H}_{2}O=x \\ =0.22M \\ C{l}_{2}O=x \\ =0.22M \end{gathered}$$

Hence, the equilibrium concentrations ofHOCI, H₂O, and Cl₂O are 0.07M, 0.22M, and 0.22M, respectively.