At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) â‡” 2HF(g) is...

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At a particular temperature the equilibrium constant for thereaction: H2(g) + F2(g) â‡” 2HF(g) is K = 25.0. A reaction mixture ina 10.00-L flask contains 0.33 moles each of hydrogen and fluorinegases plus 0.30 moles of HF. What will be the concentration of H2when this mixture reaches equilibrium?

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3.7 Ratings (537 Votes)

Â Â H2(g)Â Â + F2(g) Â Â <===> 2HF(g)

initialÂ Â Â Â Â Â Â Â Â 0.33/10 = 0.033 MÂ Â Â Â Â 0.033 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.03 M

changeÂ Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2x

equilÂ Â Â Â Â Â Â Â Â Â Â Â 0.033-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.033 - xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.03+2x

Reaction quotient(Q) = [HF]^2/[H2][F2]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.03^2/(0.033^2)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.826

Â Â K = 25

at equilibrium

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â K = [HF]^2/[H2][F2]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 25 = (0.03+2x)^2/(0.033-x)^2

Â Â Â Â x = 0.02

concentration of H2 at equil = 0.033 - x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.033- 0.02 = 0.013 M

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At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) â‡” 2HF(g) is...

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Chemistry

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