HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C, what is the equilibrium concentration of H2 (g)? decomposes to and by the following equation: at If 1.0 is placed into a closed container and the reaction is allowed to reach equilibrium at 25, what is the equilibrium concentration of ()? a. 0.076 M b. 0.038 M c. 0.924 M d. 0.0017 M

(B) 0.038 M

Explanation:

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

Let the equilibrium concentration of H2 be y M

From the equation of reaction, mole ratio of H2 to I2 formed is 1:1, therefore equilibrium concentration of I2 is also y M

Also, from the equation of reaction, mole ratio of HI consumed to H2 formed is 2:1, therefore equilibrium concentration of HI is (1 - 2y) M

1.6×10^-3 = y×y/(1 - 2y)^2

y^2/1-4y+4y^2 = 0.0016

y^2 = 0.0016(1-4y+4y^2)

y^2 = 0.0016 - 0.0064y + 0.0064y^2

y^2-0.0064y^2+0.0064y-0.0016 = 0

0.9936y^2 + 0.0064y - 0.0016 = 0

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

y = [-0.0064 + sqrt(0.0064^2 - 4×0.9936×-0.0016)] ÷ 2(0.9936) = 0.0736 ÷ 1.9872 = 0.038 M

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >