Question

1-Calculate the partial pressure (in atm) of Cl₂ at equilibrium when 4.38 atm of COCl₂ dissociates at 350 K according to the following chemical equilbrium:

COCl2(g) ⇌ CO(g) + Cl2(g)

Kp = 8.77×10-10

***If the 5% approximation is valid, use the assumption to compute the partial pressure. Report your answer to three significant figures in scientific notation.

2-Calculate the partial pressure (in atm) of CH₃OH at equilibrium when 4.84 atm of CH₃I and 4.84 atm of H₂O react at 3000 K according to the following chemical equation:

CH3I (g) + H2O (g) ⇌ CH3OH (g) + HI (g)

Kp = 7.47×101

****Report your answer to three significant figures in scientific notation.

Answer 1

1. let x be the change in partial pressure of COCL2 at equilibrium

Kp = [CO][Cl2]/[COCl2]

8.77 x 10^-10 = x^2/4.38

Thus partial pressure of Cl2 at equilibrium (x) = 6.198 x 10^-5 atm

2. let x be the change in partial pressures of CH3I and H2O at equilibrium

Kp = [CH3OH][HI]/[CH3I][H2O]

7.47 x 10^1 = x^2/(4.84 - x)(4.84 - x)

1749.8923 + 7.47 x 10^1x^2 - 723.096x = x^2

73.7x^2 - 723.096x + 1749.8923 = 0

x = 0.0676 atm

Thus, partial pressure of CH3OH at equilibrium = 6.766 x 10^-2 atm