1.0 mol of NH3 decomposes in a closed vessel at constant total pressure of 5.0 atm: 2NH3(g) N2(g) + 3H2(g). At equilibrium, 30% of the NH3 has decomposed. Calculate K_p in atm^2, to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3ce039

1.0 mol of $NH X_{3}$ decomposes in a closed vessel at constant total pressure of 5.0 atm: $2 NH X_{3} (g) N X_{2} (g) + 3 H X_{2} (g)$ . At equilibrium, 30% of the $NH X_{3}$ has decomposed. Calculate $K_{p}$ in $at m^{2}$ , to two decimal places.

Numerical answer type, answer in atm^2, tolerance 0.005.

Moles:

NH X_{3}

= 0.70,

N X_{2}

= 0.15,

H X_{2}

= 0.45; total = 1.30 mol. Partial pressures: 2.692, 0.577, 1.731 atm.

K_{p} = p_{N X_{2}} p_{H X_{2}}^{3} / p_{NH X_{3}}^{2} = (0.577) (1.731)^{3} / (2.692)^{2} = 0.413 at m^{2}