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

what is the question here. i think you messed up.