Question

At a particular temperature, 8.1 moles of gas is placed in a 3.0-L container. Over time the${\mathbf{NO}}_2$ decomposes to NOand${\mathbf{O}}_2$:

$\mathbf{2}{\mathbf{NO}}_2\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_2\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

At equilibrium, the concentration of $\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$was found to be 1.4 mol/L. Calculate the value of K for this reaction.

Short answer

The equilibrium constant of the above reaction is 0.82.

Step-by-step solution

Equilibrium Constant

The equilibrium of any reaction is based on the concentration of the reactant and products. The ratio of products of the concentration of the products and the product of the reactants’ concentration is the equilibriumconstant${\mathbf{K}}_c$. Theconcentration of the entities which are considered while depicting${\mathbf{K}}_c$ arein their gaseous states, always. The solid and liquid states are not considered.

Calculation

Molarity is the number of moles of solute per liter of the solution and it is expressed as follows:

$Molarity=\frac{n}{V}$

Where n is the number of moles of the solute and V is the volume of the solution in liters.

The given reaction is as follows:

${\text{2NO}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2NO}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)$

The number of moles of ${\text{NO}}_{\text{2}}\left(\text{g}\right)$ present in a 3.0 L container is 8.1 moles. The initial concentration of ${\text{NO}}_{\text{2}}$is calculated as follows:

$\begin{array}{c}Molarity=\frac{n}{V}\\ =\frac{8.1mol}{3.0L}\\ =2.7M\end{array}$

The ICE table for the equilibrium reaction is as follows:

$\begin{array}{ccccccc}& & {\text{2NO}}_{\text{2}}\left(\text{g}\right)& \rightleftharpoons & 2NO\left(g\right)& +& O_2\left(g\right)\\ Initial\left(M\right)& & 2.7& & 0& & 0\\ Change\left(M\right)& & -2x& & x& & x\\ Atequilibrium& & 2.7-2x& & 2x& & x\end{array}$

The concentration of NO at equilibrium is 1.4 mol/L.

So,

$\begin{array}{c}+2x=1.4mol/L\\ x=\frac{1.4M}{2}\\ =0.7M\end{array}$

Therefore, the concentrations are calculated as follows:

$\begin{array}{c}\left[N{O}_2\right]=2.7M-2x\\ =2.7-2\left(0.7\right)\\ =1.3M\end{array}$

And

$\begin{array}{c}\left[NO\right]=2x\\ =2\times 0.7M\\ =1.4M\end{array}$

On substituting the values of concentration in the equilibrium constant equation as follows:

$\begin{array}{c}k=\frac{{\left[NO\right]}^2\left[O_2\right]}{{\left[N{O}_2\right]}^2}\\ =\frac{{\left(1.4M\right)}^2\left(0.7M\right)}{{\left(1.3M\right)}^2}\\ =\frac{1.4}{1.7}\\ =0.82\end{array}$

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Hence, the equilibrium constant of the above reaction is 0.82.