In the equilibrium mixture of N2(g) + 3H2(g) 2NH3(g) at a total pressure of 10 atm, the mole fractions are 0.20 (N2), 0.50 (H2) and 0.30 (NH3). K_p in atm^-2 is

JEE Main levelSingle correct optionDifficulty 2 of 3ce073

In the equilibrium mixture of $N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)$ at a total pressure of 10 atm, the mole fractions are $0.20$ ( $N X_{2}$ ), $0.50$ ( $H X_{2}$ ) and $0.30$ ( $NH X_{3}$ ). $K_{p}$ in $at m^{- 2}$ is

K_{x} = x_{NH X_{3}}^{2} / (x_{N X_{2}} x_{H X_{2}}^{3}) = 0.09/ (0.20 \times 0.125) = 3.6

. Since

p_{i} = x_{i} P

K_{p} = K_{x} P^{Δ n} = K_{x} P^{- 2} = 3.6/100 = 0.036 at m^{- 2}