Question

Calculating `K_p`: In an equilibrium mixture at `500^@C`, we find that `P_(NH_3)=0.076` atm, `P_(N_2)=3.00` atm, and `P_(H_2)=1.85` atm. Calculate `K_p` at `500^@C` for the following reaction
`N_2(g)+3H_2(g)hArr 2NH_3(g)`

Answer 1

Strategy: Write the expression for `K_p` in terms of equilibrium partial pressures of all reacting substances and substitute partial pressures in atmospheres into it.
Solution: Equilibrium constant `K_p` is given by
`K_p=(P_(NH_3)^2)/(P_(N_2)P_(H_2)^3)`
Substituting the pressure, we find that
`K_p=((0.076)^2)/((0.300)(1.85)^3)=3.04xx10^-4`