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Kc for the following reaction is 0.35. H2 (g) + I2 (g) 2 HI (g). If...

Kc for the following reaction is 0.35. H2 (g) + I2 (g) 2 HI (g). If 1.0 mol of I2 and 1.0 mol of H2 are placed in a 1.0 L vessel, the equilibrium concentration of HI is :

The answer is 0.46 M, please explain how this value was reached. Thanks!

Expert Solution

Find out the initial concentration of the reactants as below.

[H₂] = (mol H₂)/(volume of container) = (1.0 mol)/(1.0 L) = 1.0 M

[I₂] = (mol I₂)/(volume of container) = (1.0 mol)/(1.0 L) = 1.0 M

Set up the ICE chart for the reaction as below.

H₂ (g) + I₂ (g) <=====> 2 HI (g)

initial 1.0 1.0 0.0

change -x -x +2x

equilibrium (1.0 – x) (1.0 – x) 2x

The equilibrium constant is given as

K_c = [HI]²/[H₂][I₂] = (2x)²/(1 – x).(1- x)

====> 0.35 = 4x²/(1 – x)²

====> 4x² = 0.35*(1 – x)²

Take the positive square root.

2x = 0.5916*(1 – x)

===> 2x = 0.5916 – 0.5916x

===> 2.5916x = 0.5916

===> x = (0.5916)/(2.5916) = 0.2283

The equilibrium concentration of HI is given as 2x = 2*(0.2283) M = 0.4566 M ≈ 0.46 M (ans).

queen_honey_blossom answered 1 year ago

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