Question

Calculate the equilibrium concentrations of 12.00 mol iodine gas and 8.00 mol hydrogen gas for the production of hydrogen iodide gas in a 4.00 Liter vessel. The K = 86.5

Please use ICE method

Answer 1

we know that

concentration = moles / volume (L)

so

initially

[H2] = 8 / 4 = 2 M

[I2] = 12 / 4 = 3 M

now the reaction is

H2 + I2 --> 2HI

using ICE table

initial conc of H2 , I2 , HI are 2 , 3 , 0

change in conc of H2 , I2 , HI are -x , -x , +2x

equilibrium conc of H2 , I2 , HI are 2-x ,3-x , 2x

now

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

86.5 = [2x]^2 / [2-x] [3-x]

86.5 ( 2-x) (3-x) = 4x2

86.5 ( 6 - 5x + x2) = 4x2

86.5x2 - 432.5x + 519 = 4x2

82.5x2 - 432.5x + 519 = 0

x = 1.86

so

at equilibrium

[H2] = 2 - x = 2- 1.86 = 0.14

[I2] = 3 - x = 3 - 1.86 = 1.14

[HI] = 2x = 2 * 1.86 = 3.72