Answer:
[tex]2.55*10^{11[/tex]
Explanation:
Equation for the heterogeneous system is given as:
[tex]2A_{(aq)} + 3 B_{(g)} + C_{(l)}[/tex] ⇄ [tex]2D_{(s)}[/tex] [tex]+[/tex] [tex]3E_{(g)}[/tex]
The concentrations and pressures at equilibrium are:
[tex][A] = 9.68*10^{-2}M[/tex]
[tex]P_B = 9.54*10^3Pa[/tex]
[tex][C]=14.64M[/tex]
[tex][D]=10.11M[/tex]
[tex]P_E=9.56*10^4torr[/tex]
If we convert both pressure into bar; we have:
[tex]P_B = 9.54*10^3Pa[/tex]
[tex]P_B = (9.54*10^3)*\frac{1}{10^5} bar[/tex]
[tex]P_B=9.54*10^{-2}bar[/tex]
[tex]P_E=9.56*10^4torr[/tex]
1 torr = 0.001333 bar
[tex]9.54*10^4 *0.001333 = 127.5 bar[/tex]
[tex]K=\frac{[P_E]^3}{[A]^2[P_B]^3}[/tex]
[tex]K=\frac{(127.5)^3}{(9.68*10^{-2})^2(9.54*10^{-2})^3}[/tex]
[tex]K=2.55*10^{11[/tex]
