Question

In the reaction below, 3.46 atm each of H2 and Cl2 were placed into a 1.00 L flask and allowed to react: H2(g) + Cl2(g) <=> 2 HCl(g) Given that Kc = 50.1, calculate the equilibrium pressure of HCl. Enter your answer to 2 decimal places.

Answer 1

The equilibrium pressure of HCl is 5.40 atm

Step-by-step explanation:

Kc = [HCl]^2/[H2][Cl2]

Initial pressure of H2 = 3.46 atm and Cl2 = 3.46 atm

Let the equilibrium pressure of HCl be y

From the equation of reaction, mole ratio of H2 to HCl is 1:2, equilibrium pressure of H2 is (3.46 - 0.5y)

Also, from the y of reaction, mole ratio of Cl2 to HCl is 1:2, equilibrium pressure of Cl2 is (3.46 - 0.5y)

Kc = 50.1

50.1 = y^2/(3.46 -y)(3.46 - y)

50.1(11.9716 - 3.46y + 0.25y^2) = y^2

12.525y^2 - 173.346y + 599.77716 = y^2

11.525y^2 - 173.346y + 599.77716 = 0

y^2 - 15.04y + 52.04 = 0

The value of y is obtained using the quadratic formula

y = [15.04 - (sqrt (-15.04^2 - 4×1×52.04))] ÷ 2(1) = (15.04 - 4.25) ÷ 2 = 10.79 ÷ 2 = 5.40 atm (to 2 decimal places)