Question

At 35^oC, $\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$for the reaction

role="math" localid="1662208995211" $\mathbf{2}\mathit{N}\mathbf{0}\mathit{C}\mathit{l}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\iff }\mathbf{2}\mathit{N}\mathit{O}\left(g\right)\mathbf{+}\mathit{c}{\mathit{l}}_{\mathbf{2}}\left(g\right)$

Calculate the concentrations of all species at equilibriumfor each of the following original mixtures.

a. 2.0 moles of pure NOCl in a 2.0-L flask

b. 2.0 moles of NO and 1.0 mole of Cl₂ in a 1.0-L flask

c. 1.0 mole of NOCl and 1.0 mole of NO in a 1.0-L flask

d. 3.0 moles of NO and 1.0 mole of Cl₂ in a 1.0-L flask

e. 2.0 moles of NOCl, 2.0 moles of NO, and 1.0 mole of Cl₂ in a 1.0-L flask

f. 1.00 mol/L concentration of all three gases

Short answer

a. The concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are $1.0\mathrm{M},0.032\mathrm{M}\mathrm{and}0.016\mathrm{M}$respectively.

b. The concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are $2.0\mathrm{M},0.050\mathrm{M}\mathrm{and}0.025\mathrm{M}$respectively.

c. The concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are $1.0\mathrm{M},1.0\mathrm{M}\mathrm{and}1.6\times {10}^{-5}\mathrm{M}$respectively.

d. The concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are $2.0\mathrm{M},1.0\mathrm{M}\mathrm{and}6.4\times {10}^{-5}\mathrm{M}$respectively.

e. The concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are $3.9\mathrm{M},0.080\mathrm{M}\mathrm{and}0.040\mathrm{M}$respectively.

f. The concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are$1.99\mathrm{M},1.1\times {10}^{-2}\mathrm{M}\mathrm{and}0.51\mathrm{M}$

respectively.

Step-by-step solution

Find the concentrations of all species at equilibrium.

The expression for the equilibrium constant, K in terms of concentrations is written below:

$\mathrm{K}=\frac{{\left[\mathrm{NO}\right]}^2\left[{\mathrm{Cl}}_2\right]}{{\left[\mathrm{NOCl}\right]}^2}...\left(1\right)$

The given K value is $1.6\times {10}^{-5}$

The initial concentration of NOCl is,

$$\begin{gathered} \left[\mathrm{NOCl}\right]=\frac{2.0\mathrm{mol}}{2.0\mathrm{L}} \\ =1.0\mathrm{M} \end{gathered}$$

Setup the ICE table:

$$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{InitiaL}\left(\mathrm{M}\right)1.000 \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ ----------------------- \\ \mathrm{Equlibrium}\left(\mathrm{M}\right)1.0-2\mathrm{x}+2\mathrm{x}\mathrm{x} \end{gathered}$$

Substitute these values into Equation (1).

$1.6\times {10}^{-5}=\frac{{\left(2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{\left(1.0-2\mathrm{x}\right)}^2}$

Solve for x.

x=0.016

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

$$\begin{gathered} \mathrm{NOCl}=1.0-2\mathrm{x} \\ =1.0-2\times 0.016 \\ =0.97 \\ \approx 1.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=2\mathrm{x} \\ =2\times 0.016 \\ =0.032\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{CL}}_2\right]=\mathrm{x} \\ =0.016\mathrm{M} \end{gathered}$$

Therefore, the concentrations of$\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are$1.0\mathrm{M},0.032\mathrm{M}\mathrm{and}0.016\mathrm{M}$respectively.

Calculate the concentrations of all species at equilibrium.

The initial concentrations of NO and ${\mathrm{Cl}}_2$are,

$$\begin{gathered} \left[\mathrm{NO}\right]=\frac{2.0\mathrm{mol}}{1.0\mathrm{L}} \\ =2.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{Cl}}_2\right]=\frac{1.0\mathrm{mol}}{1.0\mathrm{L}} \\ =1.0\mathrm{M} \end{gathered}$$

Setup the ICE table:

localid="1651129344446" $$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{InitiaL}\left(\mathrm{M}\right)02.01.0 \\ \mathrm{Change}\left(\mathrm{M}\right)+2.0-2.0-1.0\mathrm{Reacts}\mathrm{fully} \\ ------------------------------- \\ \mathrm{Final}\left(\mathrm{M}\right)2.000\mathrm{New}\mathrm{initial} \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \mathrm{Equlibrium}\left(\mathrm{M}\right)2.0-2\mathrm{x}+2\mathrm{x}\mathrm{x} \end{gathered}$$

Substitute these values into Equation (1).

$1.6\times {10}^{-5}=\frac{{\left(2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{\left(2.0-2\mathrm{x}\right)}^2}$

Solve for x.

x=0.025

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

$$\begin{gathered} \mathrm{NOCl}=2.0-2\mathrm{x} \\ =2.0-2\times 0.025 \\ =1.95\mathrm{M} \\ \approx 2.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=2\mathrm{x} \\ =0.050\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{CL}}_2\right]=\mathrm{x} \\ =0.025\mathrm{M} \end{gathered}$$

Hence, the concentrations of$\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are$2.0\mathrm{M},0.050\mathrm{M}\mathrm{and}0.025\mathrm{M}$respectively.

Determine the concentrations of all species at equilibrium.

The initial concentrations of NOCl and NO are,

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

$$\begin{gathered} \left[\mathrm{NO}\right]=\frac{1.0\mathrm{mol}}{1.0\mathrm{L}} \\ =1.0\mathrm{M} \end{gathered}$$

Setup the ICE table:

$$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{InitiaL}\left(\mathrm{M}\right)1.000 \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ ----------------------- \\ \mathrm{Equlibrium}\left(\mathrm{M}\right)1.0-2\mathrm{x}+2\mathrm{x}\mathrm{x} \end{gathered}$$

Substitute these values into Equation (1).

$1.6\times {10}^{-5}=\frac{{\left(1.0+2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{\left(1.0-2\mathrm{x}\right)}^2}$

Solve for x.

$\mathrm{x}=1.6\times {10}^{-5}$

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

$$\begin{gathered} \mathrm{NOCl}=1.0-2\mathrm{x} \\ =1.0-2\times 1.6\times {10}^{-5} \\ =1.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=1.0+2\mathrm{x} \\ =1.0+2\times 1.6\times {10}^{-5} \\ =1.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{CL}}_2\right]=\mathrm{x} \\ =1.6\times {10}^{-5}\mathrm{M} \end{gathered}$$

Hence, the concentrations of$\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are $1.0\mathrm{M},1.0\mathrm{M}\mathrm{and}1.6\times {10}^{-5}\mathrm{M}$respectively.

Find the concentrations of all species at equilibrium.

The initial concentrations of NO and ${\mathrm{Cl}}_2$are,

$$\begin{gathered} \left[\mathrm{NO}\right]=\frac{3.0\mathrm{mol}}{1.0\mathrm{L}} \\ =3.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{Cl}}_2\right]=\frac{1.0\mathrm{mol}}{1.0\mathrm{L}} \\ =1.0\mathrm{M} \end{gathered}$$

Setup the ICE table:

localid="1651130668418" $$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{InitiaL}\left(\mathrm{M}\right)03.01.0 \\ \mathrm{Change}\left(\mathrm{M}\right)+2.0-2.0-1.0\mathrm{Reacts}\mathrm{fully} \\ ------------------------------- \\ \mathrm{Final}\left(\mathrm{M}\right)2.01.00\mathrm{New}\mathrm{initial} \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \mathrm{Equlibrium}\left(\mathrm{M}\right)2.0-2\mathrm{x}1.0+2\mathrm{x}\mathrm{x} \end{gathered}$$

Substitute these values into Equation (1).

$1.6\times {10}^{-5}=\frac{{\left(1.0+2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{\left(2.0-2\mathrm{x}\right)}^2}$

Solve for x.

$\mathrm{x}=6.4\times {10}^{-5}$

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

$$\begin{gathered} \mathrm{NOCl}=2.0-2\mathrm{x} \\ =2.0-2\times 6.4\times {10}^{-5} \\ =2.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=1.0+2\mathrm{x} \\ =1.0+2\times 6.4\times {10}^{-5} \\ =1.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{CL}}_2\right]=\mathrm{x} \\ =6.4\times {10}^{-5}\mathrm{M} \end{gathered}$$

Hence, the concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are$2.0\mathrm{M},1.0\mathrm{M}\mathrm{and}6.4\times {10}^{-5}\mathrm{M}$respectively.

Find the concentrations of all species at equilibrium.

The initial concentrations of $\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$are,

$$\begin{gathered} \left[\mathrm{NOCl}\right]=\frac{2.0\mathrm{mol}}{1.0\mathrm{L}} \\ =2.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=\frac{2.0\mathrm{mol}}{1.0\mathrm{L}} \\ =2.0\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{Cl}}_2\right]=\frac{1.0\mathrm{mol}}{1.0\mathrm{L}} \\ =1.0\mathrm{M} \end{gathered}$$

Setup the ICE table:

$$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{InitiaL}\left(\mathrm{M}\right)2.02.01.0 \\ \mathrm{Change}\left(\mathrm{M}\right)+2.0-2.0-1.0\mathrm{Reacts}\mathrm{fully} \\ ------------------------------- \\ \mathrm{Final}\left(\mathrm{M}\right)4.000\mathrm{New}\mathrm{initial} \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \mathrm{Equlibrium}\left(\mathrm{M}\right)4.0-2\mathrm{x}2\mathrm{x}\mathrm{x} \end{gathered}$$

Substitute these values into Equation (1).

$1.6\times {10}^{-5}=\frac{{\left(2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{\left(4.0-2\mathrm{x}\right)}^2}$

Solve for x.

x=0.040

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

$$\begin{gathered} \mathrm{NOCl}=4.0-2\mathrm{x} \\ =4.0-2\times 0.040 \\ =3.92\mathrm{M} \\ \approx 3.9\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=2\mathrm{x} \\ =0.080\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{CL}}_2\right]=\mathrm{x} \\ =0.040\mathrm{M} \end{gathered}$$

Hence, the concentrations of$\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are$3.9\mathrm{M},0.080\mathrm{M}\mathrm{and}0.040\mathrm{M}$respectively.

Find the concentrations of all species at equilibrium.

Setup the ICE table:

$$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{InitiaL}\left(\mathrm{M}\right)1.01.01.0 \\ \mathrm{Change}\left(\mathrm{M}\right)+1.0-1.0-0.500\mathrm{Reacts}\mathrm{fully} \\ ------------------------------- \\ \mathrm{Final}\left(\mathrm{M}\right)2.000.50\mathrm{New}\mathrm{initial} \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \mathrm{Equlibrium}\left(\mathrm{M}\right)2.0-2\mathrm{x}2\mathrm{x}0.50+\mathrm{x} \end{gathered}$$

Substitute these values into Equation (1).

$1.6\times {10}^{-5}=\frac{{\left(\right)}^2\left(0.5+\mathrm{x}\right)}{{\left(2.0-2\mathrm{x}\right)}^2}$

Solve for x.

x=0.0057

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

$$\begin{gathered} \mathrm{NOCl}=2.00-2\mathrm{x} \\ =2.00-2\times 0.0057 \\ =1.99\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[\mathrm{NO}\right]=2\mathrm{x} \\ =2\times 0.0057 \\ =1.1\times {10}^{-2}\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{CL}}_2\right]=\mathrm{x} \\ =0.0057\mathrm{M} \end{gathered}$$

Hence, the concentrations of$\mathrm{NOCl},\mathrm{NO}\mathrm{and}{\mathrm{Cl}}_2$at equilibrium are$1.99\mathrm{M},1.1\times {10}^{-2}\mathrm{M}\mathrm{and}0.51\mathrm{M}$respectively.