91Ó°ÊÓ

Given information is

-The equilibrium constant is \({K_p} = 1.33 \times {10^{20}}\)

-0.50 atm of \({{\rm{H}}_2}\) and 0.500 atm of \({{\rm{O}}_2}\) are allowed to come to equilibrium

We have to find the partial pressures of the components.

We will assume that all \({{\rm{H}}_2}\)reacts with \({{\rm{O}}_2}\).

Two moles of \({{\rm{H}}_2}\)reacts with 1 mole of \({{\rm{O}}_2}\), to produce 2 moles of \({{\rm{H}}_2}{\rm{O}}\), so after the reaction the partial pressure of \({{\rm{H}}_2}\)will be 0 atm (0.500 atm -0.500 atm) of \({{\rm{O}}_2}\) will be 0.250 atm(0.500 atm -0.250 atm, and \({{\rm{H}}_2}{\rm{O}}\) will be 0.500 atm.

Since all of \({{\rm{H}}_2}\) reacted and the partial pressure is 0 , the equilibrium will move to the left

\({K_p} = \frac{{{{\left( {{H_2}{\rm{O}}} \right)}^2}}}{{{{\left( {{H_2}} \right)}^2} \times \left( {{O_2}} \right)}}\)

\(1.33 \cdot {10^{20}} = \frac{{{{(0.500 - x)}^2}}}{{{{(2x)}^2} \times (0.250 + x)}}\)

Since \({{\rm{K}}_ - }p\) is larger than \({10^4}\)

we will assume that 0.500 \( - {\bf{X}} \approx \)0.500

and that 0.250\( + x \approx 0.250\)

\(1.33 \times {10^{20}} = \frac{{{{(0.500)}^2}}}{{{{(2x)}^2} \times (0.250)}}\)

\(4{x^2} = \frac{{0.230}}{{1.33 \times {{10}^{20}} \times 0.250}}\)

\({x^2} = 1.88 \times {10^{ - 21}}\)

\(x = 4.34 \times {10^{ - 11}}{\rm{M}}\)

Therefore, the partial pressures of the components, are

\(\begin{array}{*{20}{c}}{{P_{{H_2}}} = 2x = 8.67 \times {{10}^{ - 11}}{\rm{M}}}\\{{P_{{O_2}}} = 0.250 + x = 0.250{\rm{M}}}\\{{P_{{H_2}O}} = 0.500 - x = 0.500{\rm{M}}}\end{array}\)