Question

The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2O, H2, and O2 reach 0.0750 atm, 0.00700 atm, and 0.00200 atm, respectively. What is the value of the equilibrium constant at this temperature?

Answer 1

Answer:

$1.742\times 10^{-5}$ is the value of the equilibrium constant at this temperature.

Explanation:

$2H_2O\rightleftharpoons 2H_2+O_2$

We are given:

Partial pressure of $H_2O=p^o_{H_2O}=0.0750 atm$

Partial pressure of $H_2=p^o_{H_2}=0.00700 atm$

Partial pressure of $O_2=p^o_{O_2}=0.00200 atm$

The expression of $K_p$ for the given chemical equation is:

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

Putting values in above equation, we get:

$K_p=\frac{(0.00700 atm)^2\times 0.00200 atm}{(0.0750 atm)^2}\\\\K_p=1.742\times 10^{-5}$

$1.742\times 10^{-5}$ is the value of the equilibrium constant at this temperature.