Answer:
(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)
