2 HI(g) = H2(g) + 1,(9) K, - Party met = 0.0016 The decomposition of HI(g) at 298 K is represented by the equilibrium equation above. When 100. torr of HI(g) is added to a previously evacuated, rigid container and allowed to reach equilibrium, the partial pressure of 2(g) is approximately 3.7 torr. If the initial pressure of HI(g) is increased to 200. torr and the process is repeated at the same temperature, which of the following correctly predicts the equilibrium partial pressure of 1 (g), and why? a. Plz 14 torr, because it is directly proportional to the square of the initial pressure of HI. b. P =0.073 torr, because it is inversely proportional to the square of the initial pressure of HI. C. P = 7.4 torr, because it is directly proportional to the initial pressure of HI. d. P 1.9 torr, because it is inversely proportional to the initial pressure of HI.

(C) P = 7.4 torr, because it is directly proportional to the initial pressure of HI.

Explanation:

From the equation of reaction,

Kc = [H2][I2]/[HI]^2

Kc = 0.0016

Initial pressure of HI = 200 torr

Let the equilibrium pressure of I2 be y

Mole ratio of H2 to I2 from the equation of reaction is 1:1, equilibrium pressure of H2 is also y

Mole ratio of HI to I2 is 2:1, equilibrium pressure of HI is (200 - 2y)

0.0016 = y×y/(200 - 2y)^2

y^2 = 0.0016(40,000 - 800y + 4y^2)

y^2 = 64 - 1.28y + 0.0064y^2

y^2 - 0.0064y^2 + 1.28y - 64 = 0

0.9936y^2 + 1.28y - 64 = 0

y^2 + 1.29y - 64.41 = 0

The value of y must be positive and is obtained by the use of the quadratic formula.

y = [-1.29 + sqrt(1.29^2 - 4×1×-64.41)] ÷ 2(1) = [-1.29 + 16.10] ÷ 2 = 14.81 ÷ 2 = 7.4

Equilibrium pressure of I2 is 7.4 torr because the relationship between I2 and HI is direct in which increase in one quantity (initial pressure of HI) would result to a corresponding increase in the other quantity (equilibrium pressure of I2)

1 is the answer to this problem