Calculate the number of moles of \(HI\) that are at equilibrium with \(1.25mol\)of \({H_2}\)and \(1.25\,mol\)of \({I_2}\) in a \(5.00\,L\)flask at \(44{8^0}C\).\({H_2} + {I_2} \rightleftharpoons 2HI\)

\({K_c} = 50.2\,at\,44{8^o}C\)

\({H_2} + {I_2} \rightleftharpoons 2HI\)

1. Volume of flask \(V = 5.00L\)
2. Value of equilibrium constant at \(44{8^0}C\)is \({K_c} = 50.2\)
3. The Number of moles of \({I_2}\) at equilibrium is \(1.25mol\)
4. The Number of moles of \({H_2}\) at equilibrium is \(1.25mol\)

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

\(\begin{array}{}\left[ {{H_2}} \right] = \frac{{1.25{\rm{mol}}}}{{5.00{\rm{L}}}}\\ = 0.25\,M\end{array}\)

\(\begin{array}{c}\left[ {{I_2}} \right] = \frac{{1.25{\rm{mol}}}}{{5.00{\rm{L}}}}\\ = 0.25\,M\end{array}\)

\(\begin{aligned}{}{K_C} &= \frac{{{{[HI]}^2}}}{{\left[ {{H_2}} \right] \times \left[ {{I_2}} \right]}}\\{[HI]^2} &= {K_c} \times \left[ {{H_2}} \right] \times \left[ {{I_2}} \right]\\{[HI]^2} &= 50.2 \times 0.25 \times 0.25\\{[HI]^2} &= 3.14\end{aligned}\)

\(\begin{array}{l}[HI] = \sqrt {3.14} \\(HI] = 1.77\,M\end{array}\)

\(\begin{aligned}{}\left[ {HI} \right]& = \frac{{{n_{HI}}}}{V}\\{n_{HI}} &= \left[ {HI} \right] \times V\\ &= 1.77\,{\rm{M}} \times 5.00\,L\\ &= 8.85\,mol\end{aligned}\)