Question

At a temperature below room temperature but with all of the substances still in the gas phase, the equilibrium constant Kp for the decomposition of iodine monochloride (ICl) into I2 and Cl2 is 1.40 × 10–5. If a sealed vessel initially contains 1.70 atm of ICI but no I2 or Cl2. what are the partial pressures of all substances involved in the reaction when it comes to equilibrium?

Answer 1

Answer:

Partial pressure of ICl at equilibrium is 1.69 atm

Partial pressure of chloride gas at equilibrium is 0.00631 atm.

Partial pressure of iodine gas at equilibrium is 0.00631 atm.

Explanation:

The partial pressure of ICl initially = 1.70 atm

The equilibrium constant of the reaction = $K_p=1.40\times 10^{-5}$

$2ICl(g)\rightequilibrium Cl_2(g)+I_2(g)$

Initially

1.70 atm        0        0

At equilibrium

(1.70-2p)          p   p

The expression of equilibrium constant is given by :

$K_p=\frac{p_{Cl_2}\times p_{I_2}}{(p_{ICl})^2}$

$1.40\times 10^{-5}=\frac{p\times p}{(1.70-2p)^2}$

Solving for p :

p = 0.00631 atm

Partial pressure of ICl at equilibrium = 1.70 atm- 2 0.00631 atm = 1.69 atm

Partial pressure of chloride gas at equilibrium = 0.00631 atm.

Partial pressure of iodine gas at equilibrium = 0.00631 atm.