For the following reaction at a certain temperature H2 (g) + F2 (g) <--> 2HF (g)

Question

For the following reaction at a certain temperature
H2 (g) + F2 (g) <--> 2HF (g)
it is found that the equilibrium concentrations in a 5.00-Lrigid container are [H2] = 0.0500M, [F2] = 0.0100 M, and [HF] =0.400M. If 0.200 mol F2 is added to this equilibrium mixture,calculate the concentrations of all gases once equilibrium isreestablished.
H2 (g) + F2 (g) <--> 2HF (g)
it is found that the equilibrium concentrations in a 5.00-Lrigid container are [H2] = 0.0500M, [F2] = 0.0100 M, and [HF] =0.400M. If 0.200 mol F2 is added to this equilibrium mixture,calculate the concentrations of all gases once equilibrium isreestablished.

Explanation / Answer

First, wite the K expression

K = [HF]^2 /([H2][F2])

K = (0.4^2)/(0.05*0.01)

K = 320

then

initially

[HF] = 0.40

[H2] = 0.05

[F2] = 0.01 + 0.2/5 = 0.05

after equilibrium

[HF] = 0.40 +2x

[H2] = 0.05 - x

[F2] = 0.05 -x

substitute in K;

K = [HF]^2 /([H2][F2])

320 = ( 0.40 +2x)^2 / ( 0.05 - x)^2

sqrt(320) = ( 0.40 +2x) / ( 0.05 - x)

17.889*0.05 - 17.889x = 0.40 + 2x

19.889x = 17.889*0.05-0.40

x = (17.889*0.05-0.40) / (19.889)

x = 0.02486

[HF] = 0.40 +2*0.02486 = 0.44972 M

[H2] = 0.05 - 0.02486 = 0.02514 M

[F2] = 0.05 -0.02486 = 0.02514 M

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