Answer to Question #74376 in General Chemistry for leul

Question

0.260 mol of HI gas is placed in a 2.00 L container at 450 degree Celsius and allowed to equilibrate by the reaction 2HI(g)= h2(g) + i2 (g), at equilibrium the concentration of i2 gas is 0.0150M, what is the value of k for this reaction at this temperature?

Answer 1

Concentration of I2: 0.0150 mol/L
Concentration of H2: 0.0150 mol/L (the same, according to the reaction equation)
Amount of I2: 0.0150 ∙ 2 = 0.0300 mol
Reacted amount of HI: 0.0300 ∙ 2 = 0.060 mol
Amount of HI remaining: 0.260 - 0.060 = 0.200 mol
Concentration of HI at equilibrium: 0.200 / 2 = 0.100 mol/L
So:
K = ([H_2 ][I_2])/〖[HI]〗^2 = (0.0150 ∙ 0.0150)/〖0.100〗^2 = 0.0225