A student ran the following reaction in the laboratory at 720 K: H2(g) + I2(g) 2HI(g)...

A student ran the following reaction in the laboratory at 720 K: H2(g) + I2(g) 2HI(g) When she introduced 0.189 moles of H2(g) and 0.218 moles of I2(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 6.05×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =

Expert Solution

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

I 0.189 0.218 0

C -x -x 2x

E 0.189-x 0.218 -x 2x

but he has given equilibrium concentration I2 = 0.0605

0.218 -x = 0.0605

x = 0.218 - 0.0605 = 0.1575

[H2] = 0.189-x = 0.189 - 0.1575 = 0.0315 M

[HI] = 2x = 2 x 0.1575 = 0.315

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

Kc = [0.315]2 / [0.0315] [ 0.0605]

Kc = 0.099225 / 0.00190 = 52.066

queen_honey_blossom answered 2 years ago

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