Question

Susan performed the following experiment:

2.50 L flask (530 degrees celsius)

49.98 grams HF gas present along with 0.50 grams of H2 gas, 14.24 grams of F2 gas

Kc for this reaction at this temp (530.0 degrees celsius) = 1.00 x 10^-2

Balanced chemical reaction: 2HF (g) <--> H2 (g) + F2 (g)

1) Is the reaction at equilibirum? (show work)

If no, then what direction does it need to shift to be at equilibirum?

2) Determine the equilibirum concentration for each component in the reaction

3) What is the kp value of this reaction?

4) Is the reaction reactant/product favored? Why?

Answer 1

V = 2.5 L

MW of HF = 20.01

MW of F2 = 37.996

MW of H2 = 2

then

M = mol/L

M of Hf = (49.98/20.02)/2.5 = 0.99860

M of H2 = (0.5/2)/2.5 = 0.1

M of F2 = (14.24/37.996) /2.5 = 0.1499105

now

K = 10^-2

solve for Q

Q = [H2][F2] / [HF]^2

Q = (0.1)(0.1499105) / (0.99860^2) = 0.0150331

since

Q >> K; this is not in equilibrium

this is shifterd toward products

b)

find concnetration of each

[HF] = 0.99860 - 2x

[H2] = 0.1 - 2x

[F2] = 0.1499105 + x

in equilibrium

K = 10^-2 = 0.01

substituein k

0.01 = (0.99860 - 2x)(0.1 - 2x) / (0.1499105 + x)

solve for x

0.99860 *0.1 + (-2*0.99860 -2*0.1)x + (4x^2) = 0.01(0.1499105 ) + 0.01x

0.099860 - 2.1972x + 4x^2 = 0.01499105 + 0.01x

4x^2 -2.2072x + 0.08486895 = 0

x = 0.0415 and 0.510; the later can be since it will give negativ concentrations

then

[HF] = 0.99860 - 2*0.0415 = 0.9156

[H2] = 0.1 - 2*0.0415 = 0.017

[F2] = 0.1499105 + 0.0415 = 0.1914105

3) What is the kp value of this reaction?

Kp = Kc*(RT)^dn

dn = change in mol of products-reactants

dn = 2-2 = 0

then

Kp = Kc*(RT)^0

Kp = Kc = 10^-2 (in this specfic case ate 530)

4)

the reaction is favoured toward reactants, since K <<< 1

that is, REactants > Products (ratio)