Q42P - Chemistry: Molecular Nature Of Matter And Change

Question

Gaseous ammonia was introduced into a sealed container and heated to a certain temperature: 

$\mathbf{2}{\mathbf{NH}}_{\mathbf{3}\left(g\right)}\mathbf{ }\mathbf{֏}\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{N}}_{\mathbf{2}\left(g\right)}\mathbf{ }\mathbf{ }\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}\left(g\right)}$

At equilibrium, $\left[{\mathrm{NH}}_3\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0225}\mathbf{ }\mathbf{M}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\left[N_2\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{114}\mathbf{ }\mathbf{M}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{and}\mathbf{ }\left[H_2\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{342}\mathbf{ }\mathbf{M}$. Calculate Kc for the reaction at this temperature.

Step-by-Step Solution

Answer

The value of the equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$ for the reaction at a certain temperature is 9.01.

Step 1:Write the equilibrium constant

The equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$ is the constant that depends on the concentrations of reactants and products.

For a general chemical reaction,

$\mathrm{aA}+\mathrm{bB} \mathrm{֏} \mathrm{cC}+\mathrm{dD}$

The equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$is given as,

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

Here, a, b, c, and dare the stoichiometric coefficients.

Step2:Calculate Kc

Consider the following reaction;

$2{\mathrm{NH}}_{3\left(\mathrm{g}\right)} \mathrm{֏} {\mathrm{N}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2\left(\mathrm{g}\right)}$

The equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$ for this reaction is given as,

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2}$

From the given data,

$$\begin{gathered} \left[{\mathrm{NH}}_3\right]=0.0225 \mathrm{M} \\ \left[{\mathrm{N}}_2\right]=0.114 \mathrm{M} \\ \left[{\mathrm{H}}_2\right]=0.342 \mathrm{M} \end{gathered}$$

Substitute the concentration of ${\mathrm{N}}_2,{\mathrm{H}}_2, \mathrm{and} {\mathrm{NH}}_3$ to calculate the equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$as follows:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2} \\ =\frac{\left(0.114\right){\left(0.342\right)}^3}{{\left(0.0225\right)}^2} \\ =\frac{0.00456}{0.000506} \\ =9.01 \end{gathered}$$

Thus, the value of the equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$ for the reaction at a certain temperature is 9.01.