Question

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride:

I2(g)+Cl2(g)?2ICl(g)Kp=81.9 (at 298 K)
A reaction mixture at 298 K initially contains PI2=0.35atm and PCl2=0.35atm .What is the partial pressure of iodine monochloride when the reaction reaches equilibrium?

Also, Consider the reaction of A to form B:
2A(g)?B(g)Kc=2.1

Answer 1

The partial pressure of iodine monochloride when the reaction reaches equilibrium is 0.5733 atm

Step-by-step explanation:

Step 1: Data given

Kp = 81.9

PI2 = 0.35 atm

PCl2 = 0.35 atm

Step 2: The balanced equation

I2(g)+Cl2(g)⇌2ICl(g)

Step 3: The initial pressure

I2 = 0.35 atm

PCl2 = 0.35 atm

PICl = 0 atm

Since the mol ratio for I2: Cl2 is 1:1 there will react on both I2 and Cl2 1X

For every mol of I2 we get 2 mol of ICl

This means there will react +2X

Step 4: pressure at the equilibrium

I2 = (0.35 - X) atm

PCl2 = (0.35 -X) atm

PICl = 2X

Step 5: Calculate partial pressure

Kp = 81.9 = (pICl)² / (PI2*PCl2)

81.9 = (2X)² / ((0.35 - X)*(0.35 - X))

4X² = 81.9 * (0.35 - X)²

4X² = 81.9 *(0.1225 - 0.7X + X²)

0 = 77.9X² - 57.33X + 10.03275

X = 0.28665

I2 = (0.35 - X) = 0.06335 atm

PCl2 = (0.35 -X) = 0.06335 atm

PICl = 0.5733 atm

The partial pressure of iodine monochloride when the reaction reaches equilibrium is 0.5733 atm