Question

At a certain temperature, the equilibrium constant, Kc for this reaction is 53.3. H2(g)+I2(g) = 2HI(g) At this temperature, 0.300 mol of H2 and 0.300 mol of I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium? View comments (1)

Answer #1

since volume = 1 L no of moles = concentration

construct the ICE table

H2 (g) + I2 (g) <---------> 2HI (g)

I 0.3 0.2 0

C -x -x 2x

E 0.3-x 0.2-x 2x

Kc = [Hi]^{2} / [H2][I2]

53.3 = [2x]^{2} / [0.3-x][0.2-x]

53.3 = 4x^{2} / 0.06 -0.5x +x^{2}

49.3x^{2} -26.65x +3.198 = 0

solve the quadratic equation

there are two values for x

one is 0.36 and another is 0.18

if you consider 0.36 some of the equilibrium concentration terms will come -ve which is not possible so ignore this

x = 0.18

equilibrium concentration of HI = 2x

[Hi] = 2x = 2 x 0.18 = 0.36 M

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