At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3. H2(g)+I2(g)−⇀2HI(g)Kc=53.3 At this temperature, 0.800mol H2 and 0.800 mol I2 were placed in a 1.00 L container to

Question

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g)+I2(g)−⇀2HI(g)Kc=53.3

At this temperature, 0.800mol H2 and 0.800 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?​Therefore, a new amount of substances will be

Answer 1

H₂(g) + I₂(g) <==> 2HI(g) ... Kc = 53.3

(1). Write the equilibrium constant expression:

Kc = [HI]² / [H₂][I₂]

(2). Set up an ICE table:

H₂(g) + I₂(g) <==> 2HI(g)

0.800....0.800..........0............Initial

-x..........-x...............+2x.........Change

0.800-x...0.800-x.........2x..........Equilibrium

(3). Use the values in the ICE table in the Kc expression and solve for x:

53.3 = [2x]² / [0.8-x][0.8-x]

53.3 = (2x)² / (0.8-x)²

Taking square root of both sides, we have...

7.30 = 2x / 0.8-x

9.30x = 5.84

x = 0.628 M

(4). [HI] @ equilibrium (see ICE table) is 2x, therefore [HI] = 2 x 0.628 M = 1.26 M