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A student ran the following reaction in the laboratory at 686 K: H2(g) + I2(g) 2HI(g) When she introduced 0.200 moles of...

A student ran the following reaction in the laboratory at 686 K:

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

When she introduced 0.200 moles of H2(g) and 0.230 moles of I2(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 6.18×10-2 M.

Calculate the equilibrium constant, Kc, she obtained for this reaction.

Kc=?

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Answer #1

initial [H₂] = 0.g nod-0.9M [I₂] = 0.23 mol = 0.23 M TL How at equilibrium [[2] = 0.0618 M H₂(g) + 12 (g) → 2HI 0.2 0.23 - -x

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