Answer:
[tex]6.25\times 10^{-6}[/tex] is the value of the equilibrium constant at this temperature.
Explanation:
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as [tex]K_{p}[/tex]
[tex]2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)[/tex]
Partial pressures at equilibrium:
[tex]p^o_{H_2O}=0.070 atm[/tex]
[tex]p^o_{H_2}=0.0035 atm[/tex]
[tex]p^o_{O_2}=0.0025 atm[/tex]
The equilibrium constant in terms of pressures is given as:
[tex]K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}[/tex]
[tex]K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}[/tex]
[tex]6.25\times 10^{-6}[/tex] is the value of the equilibrium constant at this temperature.
