Question

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty 2.0-L container. Equilibrium was established by the following reaction:

$\mathbf{2}\mathbf{NOCl}\left(\mathbf{g}\right)\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$

(a) Using numerical values for the concentrations in the Initial row and expressions containing the variable x in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let x 5 the concentration of ${\mathbf{Cl}}_{\mathbf{2}}$ that is present at equilibrium.

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{NOCl}\left(\mathbf{g}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Initial}\mathbf{\left(}\mathbf{M}\mathbf{\right)} \\ \mathbf{Change}\mathbf{\left(}\mathbf{M}\mathbf{\right)} \\ \mathbf{At}\mathbf{}\mathbf{equilibrium}\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{-}\mathbf{2}\mathbf{x} \end{gathered}$$

(b) Calculate the equilibrium concentrations for all species

Short answer

a.

The equilibrium reaction is as shown:

$2\mathrm{NOCl}\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

The number of moles of NOCl is 5.2 mol and the volume is 2.0 L. The molarity of NOCl as follows:

$\begin{array}{c}Molarity=\frac{n}{V}\\ =\frac{5.2mol}{2.0L}\\ =2.6M\end{array}$

The concentration of NOCl initially is 2.6 M. The ICE table for the reaction is as shown:

$$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{Initial}\left(\mathrm{M}\right)2.600 \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ \mathrm{At}\mathrm{equilibrium}\left(\mathrm{M}\right)2.6-2\mathrm{x}2\mathrm{x}\mathrm{x} \end{gathered}$$

b.

$\left[C{l}_2\right]$ is $3.0\times {10}^{-2}M$, $\left[NO\right]$ is $6.0\times {10}^{-2}M$ and $\left[NOCl\right]$ is 2.54 M.

Step-by-step solution

Equilibrium constant

The equilibrium constant (K) of a chemical reaction is reaction quotient at equilibrium stage.

For a general reaction $\mathbf{aA}\mathbf{+}\mathbf{bB}\to \mathbf{cC}\mathbf{+}\mathbf{dD}$

The equilibrium constant (K) is written as follows:

$\mathbf{K}\mathbf{=}\frac{{\left[\mathbf{C}\right]}^{\mathbf{c}}{\left[\mathbf{D}\right]}^{\mathbf{d}}}{{\left[\mathbf{A}\right]}^{\mathbf{a}}{\left[\mathbf{B}\right]}^{\mathbf{b}}}$

Where $\left[\mathbf{A}\right]\mathbf{,}\left[\mathbf{B}\right]\left[\mathbf{C}\right]\mathbf{}\mathbf{and}\mathbf{}\left[\mathbf{D}\right]$ are the equilibrium concentrations of A,B,C and D respectively. Only gaseous and aqueous entities are included in the above expression.

Subpart (a)

The equilibrium reaction is as shown:

$2\mathrm{NOCl}\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

The number of moles of NOCl is 5.2 mol and the volume is 2.0 L. The molarity of NOCl as follows:

$\begin{array}{c}Molarity=\frac{n}{V}\\ =\frac{5.2mol}{2.0L}\\ =2.6M\end{array}$

The concentration of NOCl initially is 2.6 M. The ICE table for the reaction is as shown:

$$\begin{gathered} 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{Initial}\left(\mathrm{M}\right)2.600 \\ \mathrm{Change}\left(\mathrm{M}\right)-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ \mathrm{At}\mathrm{equilibrium}\left(\mathrm{M}\right)2.6-2\mathrm{x}2\mathrm{x}\mathrm{x} \end{gathered}$$

Subpart (b)

The equilibrium expression for the reaction is as shown:

$\begin{array}{c}K=\frac{{\left[NO\right]}^2\left[C{l}_2\right]}{{\left[NOCl\right]}^2}\\ 1.6\times {10}^{-5}=\frac{{\left(2x\right)}^2\left(x\right)}{\left(2.6-x\right)}\\ 1.6\times {10}^{-5}=\frac{4x^3}{2.6-x}\end{array}$

Small value of K indicates that product concentration would be small at equilibrium. So, it is assumed that 2.6-x=2.6.

On simplifying the given equation, we get

$\begin{array}{c}1.6\times {10}^{-5}=\frac{4x^3}{2.6}\\ x=3.0\times {10}^{-2}M\end{array}$

The equilibrium concentration of data-custom-editor="chemistry" ${\mathrm{Cl}}_2$is x which is data-custom-editor="chemistry" $3.0\times {10}^{-2}M$.

The equilibrium concentration of NO is 2x which is as 2x.

data-custom-editor="chemistry" $\begin{array}{c}2x=2\times 3.0\times {10}^{-2}M\\ =6.0\times {10}^{-2}M\end{array}$

The equilibrium concentration of NOCl is 2.6-2x which is as follows:

data-custom-editor="chemistry" $\begin{array}{c}=2.6-2\left(3.0\times {10}^{-2}M\right)\\ =2.6-6.0\times {10}^{-2}M\\ =2.54M\end{array}$Thus, $\left[C{l}_2\right]$is $3.0\times {10}^{-2}M$, $\left[NO\right]$ is $6.0\times {10}^{-2}M$ and $\left[NOCl\right]$ is 2.54 M.