Answer to Question #152119 in Chemistry for Alexis

Question

Combining Equilibrium Expression. (Note: this is like Hess’s Law and equilibrium together. If you have to reverse and equation make sure to take the reciprocal of the Kp value. Also if you have to double or triple an equation that means you should be squaring the equilibrium constant and not just multiplying by 2.

3) Given that, at 700K, Kp = 54.0 H2(g) + I2(g) ↔ 2HI(g)

Kp = 1.04 x 10-4 N2(g) + 3H2(g) ↔ 2NH3(g)

Find the value of Kp for 2NH3(g) + 3I2(g) ↔ 6HI(g) + N2(g)

Answer 1

Let's write the expressions for the equilibrium constants.

H₂(g) + I₂(g) ↔ 2HI(g), Kp1 = 54.0

$K_{p1} = \frac{p_{HI}^2}{p_{H_2}p_{I_2}}$

N₂(g) + 3H₂(g) ↔ 2NH₃(g), Kp2 = 1.04 x 10^-4

$K_{p2} = \frac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3}$

For 2NH₃(g) + 3I₂(g) ↔ 6HI(g) + N₂(g):

$K_{p3} = \frac{p_{HI}^6p_{N_2}}{p_{NH_3}^2p_{I_2}^3} = \left(\frac{p_{HI}^2}{p_{H_2}p_{I_2}}\right)^3·\left( \frac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3}\right)^{-1}$

$K_{p3} = (K_{p1})^3·(K_{p2})^{-1} = \frac{54.0^3}{1.04·10^{-4}} = 1.51·10^9$

Answer: the value of Kp for 2NH₃(g) + 3I₂(g) ↔ 6HI(g) + N₂(g) is 1.51·10⁹.