Answer:
The value of equilibrium constant is :
[tex]K_p=6.04\times 10^{-4}[/tex]
Explanation:
Partial pressure of nitrogen at equilibrium, [tex]p_1= 1.3241 atm[/tex]
Partial pressure of oxygen at equilibrium, [tex]p_2= 1.0844 atm[/tex]
Partial pressure of dinitrogen monoxide at equilibrium, [tex]p_3= 0.0339 atm[/tex]
[tex]2N_2(g)+O_2(g)\rightleftharpoons 2N_2O(g)[/tex]
The expression of [tex]K_p[/tex] will be given as:
[tex]K_p=\frac{p_3^{2}}{p_1^{2}\times p_2}[/tex]
[tex]K_p=\frac{(0.0339 atm)^2}{ (1.3241 atm)^2\times 1.0844 atm}=6.04\times 10^{-4}[/tex]
The value of equilibrium constant is :
[tex]K_p=6.04\times 10^{-4}[/tex]
