Answer:
[tex]K_{p}=4.35\times 10^{-4}[/tex]
Explanation:
We know, [tex]K_{p}=K_{c}(RT)^{\Delta n}[/tex]
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and [tex]\Delta n[/tex] is difference in sum of stoichiometric coefficient of products and reactants
Here [tex]\Delta n=(2)-(2+1)=-1[/tex] and T = 311 K
So, [tex]K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}[/tex]
Hence value of equilibrium constant in terms of partial pressure [tex](K_{p})[/tex] is [tex]4.35\times 10^{-4}[/tex]
