For the reaction HI(g) <----> 1/2H2(g) + 1/2I2(g) at 458 C Kc=0145. If a 1.00L vessel...

For the reaction HI(g) <----> 1/2H2(g) + 1/2I2(g) at 458 C Kc=0145. If a 1.00L vessel contains 0.355 mole H2(g), 0.355 mole I2(g), and 1.25 mole HI(g) is this reaction at equilibrium, and if not how will it shift Calculate Kc for H2(g) + I2(g) <----> 2HI(g)

Expert Solution

HI(g) <----> 1/2H2(g) + 1/2I2(g) kc = 0.145

Qc = [H2]^1/2*[I2]^1/2/[HI]

= (0.355)^1/2*(0.3550^1/2/(1.25)

= 0.355/1.25 = 0.284

Qc>Kc the equilibrium shift to left side.

HI(g) <----> 1/2H2(g) + 1/2I2(g) kc = 0.145

Kc = [H2]^1/2*[I2]^1/2/[HI]

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

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

1/0.145 = [HI]^2/[H2][I2]

[HI]^2/[H2][I2] = 6.89 = Kc >>>>answer

H2(g) + I2(g) <----> 2HI(g) Kc = 6.89 >>>>answer

queen_honey_blossom answered 2 years ago

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