Question

The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2OH2O , H2H2 , and O2O2 reach 0.070 atm,0.070 atm, 0.0035 atm,0.0035 atm, and 0.0025 atm,0.0025 atm, respectively at equilibrium. What is the value of the equilibrium constant at this temperature?

Answer 1

Answer:

$6.25\times 10^{-6}$ 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 $K_{p}$

$2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)$

Partial pressures at equilibrium:

$p^o_{H_2O}=0.070 atm$

$p^o_{H_2}=0.0035 atm$

$p^o_{O_2}=0.0025 atm$

The equilibrium constant in terms of pressures is given as:

$K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}$

$K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}$

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.