Question

a.)If 1.00 mol of H2 is allowed to react with 1.00 mol of I2 in a 10.L reaction vessel at 700 K, what are the concentrations of H2, I2, and Hl at equilibrium given that Kc is 57.0?

b.)What are their concentrations in mol/L?

c.) If 0.100 M of H2 is allowed to react with 0.200 M of I2 in a reaction vessel at 700 K, what are the concentrations of H2, I2, and Hl at equilibrium given that Kc is 57.0?

Answer 1

Answer on Question #52490 – Chemistry – Inorganic Chemistry

Question:

a) If 1.00 mol of H₂ is allowed to react with 1.00 mol of I₂ in a 10.L reaction vessel at 700 K, what are the concentrations of H₂, I₂, and HI at equilibrium given that Kc is 57.0?

b) What are their concentrations in mol/L?

c) If 0.100 M of H₂ is allowed to react with 0.200 M of I₂ in a reaction vessel at 700 K, what are the concentrations of H₂, I₂, and HI at equilibrium given that Kc is 57.0?

Answer:

The reaction occurs: H₂ + I₂ → 2HI

The equilibrium equation for the reaction: K_c = μ₃² / μ₁ × μ₂, where μ₁ – equilibrium concentration of H₂, μ₂ – equilibrium concentration of I₂, μ₃ – equilibrium concentration of HI.

If y mol/l of H₂ reacts then μ₁ = 0.1 M – y, μ₂ = 0.2 M – y, μ₃ = 2y.

Thus, K_c = 57 = 4y² / ((0.1 - y)(0.2 - y)),

57 × (0.02 - 0.1y - 0.2y + y²) = 4y²

1.14 - 17.1y + 57y² = 4y²

53y² - 17.1y + 1.14 = 0

y = 0.093 ; 0.228. Only the first value has a physical sense. Therefore this is used for the further calculations:

H₂: μ₁ = 0.1 - 0.093 M = 0.007 M = 0.007 mol/l

I₂: μ₂ = 0.2 - 0.093 M = 0.107 M = 0.107 mol/l

HI: μ₃ = 0.186 M = 0.186 mol/l

Comment: 1 mol in 10 L equals 0.100 M. So answers a) and c) are identical.

www.AssignmentExpert.com