Solid $NH X_{4} I$ decomposes as $NH X_{4} I (s) NH X_{3} (g) + HI (g)$ with $K_{p 1} = 0.72 at m^{2}$ , and the $HI$ formed also dissociates as $2 HI (g) H X_{2} (g) + I X_{2} (g)$ with $K_{p 2} = 0.0144$ (dimensionless). Both equilibria are present simultaneously. Calculate the total pressure in atm at equilibrium, correct to two decimal places.

Numerical answer type, answer in atm, tolerance 0.01.

Answer: 1.89 atm

Let

p_{HI} = y

p_{H X_{2}} = p_{I X_{2}} = z

. From

K_{p 2} = z^{2} / y^{2}

z = 0.0144 y = 0.12 y

. Each

H X_{2}

uses two

HI

, and

NH X_{3}

equals the total

HI

formed, so

p_{NH X_{3}} = y + 2 z = 1.24 y

. Then

K_{p 1} = 1.24 y^{2} = 0.72

gives

y = 0.7620

atm. Total

= p_{NH X_{3}} + y + 2 z = 2.48 y = 1.890

atm.