Question

Hydrogen iodide gas decomposes into hydrogen gas and iodine gas at 453°C. If a 2.00 L flask is filled with 0.200 mol of hydrogen iodide gas, 0.156 mol hydrogen iodide remains at equilibrium. What is the equilibrium constant, K_c, for the reaction at this temperature?

2 HI (g) ⇌ H₂ (g) + I₂ (g)

Answer 1

callculate equilibrium concentration of all species

initial conc of HI_(g) = 0.200 mol /2 L = 0.100 M

initial conc of H_2(g)  = 0 M

initial conc of I_2(g)  = 0 M

At equilibrium change in concentration

equilibrium concentration of HI_(g) = 0.156/2 = 0.078 M

equilibrium concentration of H_2(g) = 1 mole H₂ formed when there is 2 mole HI reacted total reacted HI = 0.200 - 0.156 = 0.044 mole thus 0.044 HI produce 0.022 mole H₂

hence equilibrium concentration of H₂ = 0.022 / 2 = 0.011 M

equilibrium concentration of I_2(g) = 1 mole I₂ formed when there is 2 mole HI reacted total reacted HI = 0.200 - 0.156 = 0.044 mole thus 0.044 HI produce 0.022 mole I₂

hence equilibrium concentration of I₂ = 0.022 / 2 = 0.011 M

K_c = [H₂][I₂] / [HI]²

substitute value at equilibrium concentration

K_c  = (0.011) (0.011) / (0.078)²  = 0.019888 M = 1.9888 X 10^-2

K_c   = 1.9888 X 10^-2