Question

The equilibrium constant at 35 0 K for the reaction Br2(g) I2(g) 2IBr(g) has a value of 322. Bromine at an initial pressure of 0. 0500 atm is mixed with iodine at an initial partial pressure of 0. 0400 atm and held at 35 0 K until equilibrium is reached. Calculate the equilibrium pressure of each of the gases.

Answer 1

Answer:

Partial pressure of Br2 = 0.01158 atm

Partial pressure of I2 =  0.00158 atm

Partial pressure of IBr =0.07684 atm

Explanation:

Step 1: Data given

Temperature = 350 K

K = 322

Initial partial pressure of bromine = 0.0500 atm

Initial partial pressure of iodine = 0.0400 atm

Step 2: The balanced equation

Br2(g) + I2 ⇆ 2IBr(g)

Step 3: The initial pressures

pBr2 = 0.0500 atm

pI2 = 0.0400 atm

pIBr = 0 atm

Step 3: The pressure ate the equilibrium

pBr2 = 0.0500 - X atm

pI2 = 0.0400 -X atm

pIBr = 2X atm

Step 4: Define K

K = (pIBr)² / (pBr2)*(pI2)

Step 5: Calculate X

322 = (2X)² / (0.0500-X)(0.0400-X)

X = 0.03842

pBr2 = 0.0500 - 0.03842 = 0.01158 atm

pI2 = 0.0400 -0.03842 = 0.00158 atm

pIBr = 2*0.03842 = 0.07684 atm