Question

At a given temperature, K = 1.3×10-2 for the reaction

N₂(g) + 3H₂(g)$\mathbf{\rightleftharpoons }$2NH₃(g)

Calculate values of K for the following reactions at this temperature.

a.N₂(g) +H₂(g)$\mathbf{\rightleftharpoons }$NH₃(g)

b. 2NH₃(g)$\mathbf{\rightleftharpoons }$N₂(g) + 3H₂(g)

c. NH₃(g)$\mathbf{\rightleftharpoons }$N₂(g) +H₂(g)

d. 2N₂(g) + 6H₂(g)$\mathbf{\rightleftharpoons }$4NH₃(g)

Short answer

a. The equilibrium constant, K value for the reactionN₂(g) +H₂(g)$\rightleftharpoons$NH₃(g)is 0.11.

b. The equilibrium constant, K value for the reaction 2NH₃(g)$\rightleftharpoons$N₂(g) + 3H₂(g)is 77.

c. The equilibrium constant, K value for the reaction NH₃(g)$\rightleftharpoons$ N₂(g) +H₂(g)is 8.8.

d. The equilibrium constant, K value for the reaction 2N₂(g) + 6H₂(g)$\rightleftharpoons$4NH₃(g) is $1.7\times {10}^{ - 4}$.

Step-by-step solution

Part a.

The equilibrium constant, K expression for the reaction N₂(g) +3H₂(g)$\rightleftharpoons$2NH₃(g)is shown below:

$\mathrm{K}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$ …(1)

The equilibrium constant, K expression in terms of concentrations, for the reaction reactionN₂(g) +H₂(g)$\rightleftharpoons$NH₃(g)is written down:

${\mathrm{K}}^{\mathrm{a}}=\frac{\left[{\mathrm{NH}}_3\right]}{{\left[{\mathrm{N}}_2\right]}^{1/2}{\left[{\mathrm{H}}_2\right]}^{3/2}}$ …(2)

Rewrite the above expression as follows:

${\mathrm{K}}^{\mathrm{a}}={\left(\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}\right)}^{1/2}$

Replace$\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$ with K into the above expression as shown below:

${\mathrm{K}}^{\mathrm{a}}={\left(\mathrm{K}\right)}^{1/2}$

Now substitute the value of K into the above expression.

$\begin{array}{c}{\mathrm{K}}^{\mathrm{a}}={\left(1.3\times {10}^{ -- 2}\right)}^{1/2}\\ =0.11\end{array}$

Therefore, the equilibrium constant, K value for this reaction is 0.11.

Part b.

The equilibrium constant expression for the reaction 2NH₃(g)$\rightleftharpoons$N₂(g) + 3H₂(g)can be written as follows:

$\begin{array}{c}{\mathrm{K}}^{\mathrm{b}}=\frac{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2}\\ =\frac{1}{\left(\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}\right)}\\ =\frac{1}{\mathrm{K}}\\ =\frac{1}{1.3\times {10}^{ -- 2}}\\ =77\end{array}$

Hence, the equilibrium constant, K value for this reaction is 77.

Part c.

The equilibrium constant expression for the reaction NH₃(g)$\rightleftharpoons$ N₂(g) +H₂(g)can be written as follows:

$\begin{array}{c}{\mathrm{K}}^{\mathrm{c}}=\frac{{\left[{\mathrm{N}}_2\right]}^{1/2}{\left[{\mathrm{H}}_2\right]}^{3/2}}{\left[{\mathrm{NH}}_3\right]}\\ =\frac{1}{\left(\frac{\left[{\mathrm{NH}}_3\right]}{{\left[{\mathrm{N}}_2\right]}^{1/2}{\left[{\mathrm{H}}_2\right]}^{3/2}}\right)}\\ =\frac{1}{{\left(\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}\right)}^{1/2}}\\ =\frac{1}{{\left(\mathrm{K}\right)}^{1/2}}\\ =\frac{1}{{\left(1.3\times {10}^{ -- 2}\right)}^{1/2}}\\ =8.8\end{array}$

Therefore, the equilibrium constant, K value for this reaction is 88.

Part d.

The equilibrium constant expression for the reaction 2N₂(g) + 6H₂(g)$\rightleftharpoons$4NH₃(g)can be written as follows:

$\begin{array}{c}{\mathrm{K}}^{\mathrm{d}}=\frac{{\left[{\mathrm{NH}}_3\right]}^4}{{\left[{\mathrm{N}}_2\right]}^2{\left[{\mathrm{H}}_2\right]}^6}\\ ={\left(\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}\right)}^2\\ ={\left(\mathrm{K}\right)}^2\\ ={\left(1.3\times {10}^{ -- 2}\right)}^2\\ =1.7\times {10}^{ -- 4}\end{array}$

Hence, the equilibrium constant, K value for this reaction is $1.7\times {10}^{ - 4}$.