Question

A quantity of HI was sealed in a tube, heated to 425°C, and held at this...

A quantity of HI was sealed in a tube, heated to 425°C, and held at this temperature until equilibrium was reached. The concentration of HI in the tube at equilibrium was found to be 0.0706 moles/liter. Calculate the equilibrium concentration of H2 (and I2). (Hint: H2 and I2 are equal).

For the reaction:                                 H2(g) + I2(g) ßà 2 HI(g)                  Kc = 54.6 at 425°C.

a. 4.78 × 10-3M

b. 9.55 × 10-3M

c. 2.34 × 10-3M

d. 1.17 × 10-3M

e. 1.85 × 10-4M

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Answer #1

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

Kc for the reaction

Kc = [HI]2 / [H2] [I2]

but he has given hint (Hint: H2 and I2 are equal)

so [H2] = [I2]

now the above expression will be

Kc = [HI]2 / [H2] [H2]

he has given Kc = 54.6 and

concentration of [HI] at equilibrium is 0.0706 moles/liter

put these values in above equation

54.6 = [0.0706]2 / [H2] [H2]

[H2]2 = 0.00498 / 54.6

[H2]2 = 9.128 x 10-5

[H2] = squareroot of 9.128 x 10-5

[H2] = 0.00955

[H2] = 9.55 × 10-3M

but at equilibrium concentration of [H2] = [I2]

so concentration of [I2] = 9.55 × 10-3M

answer is b. 9.55 × 10-3M

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