Question

Analysis of the gases in a sealed reaction vessel containing \(N{H_3}\), \({N_2}\), and \({H_2}\) at equilibrium at \(40{0^0}C\) established the concentration of \({N_2}\) to be \(1.2M\) and the concentration of \({H_2}\) to be \(0.24M\).

\({N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)\)

\({K_c} = 0.50\,at\,40{0^o}C\)

Calculate the equilibrium molar concentration of \(N{H_3}\).

Short answer

The concentration of \(N{H_3}\) is \(0.091\,{\rm{M}}\) .

Step-by-step solution

Given information:

\({N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)\)

1. Value of equilibrium constant at \(40{0^0}C\)is \({K_c} = 0.50\)
2. The concentration of \({N_2}\)at equilibrium is \(1.2M\)
3. The concentration of \({H_2}\) at equilibrium is \(0.24M\)

The value of equilibrium molar concentration needs to be calculated using the equation relating equilibrium constant and molar concentrations.

Determine the concentration of \(N{H_3}\)at equilibrium

\(\begin{aligned}{}{K_c} &= \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right] \times {{\left[ {{H_2}} \right]}^3}}}\\{\left[ {N{H_3}} \right]^2} &= {K_c} \times \left[ {{N_2}} \right] \times {\left[ {{H_2}} \right]^3}\\{\left[ {N{H_3}} \right]^2} &= 0.50 \times (1.2M) \times {(0.24M)^3}\\{\left[ {N{H_3}} \right]^2} &= 0.0083\end{aligned}\)

\(\begin{aligned}{}\left[ {N{H_3}} \right] &= \sqrt {0.0083} \\\left[ {N{H_3}} \right] &= 0.091\,{\rm{M}}\end{aligned}\)