Rate information for 25.0 degrees celsius is given in the table above A. Use the

Question

Rate information for 25.0 degrees celsius is given in the table above

A. Use the data above to write the rate law.

B. Determine the value of the rate constant at this temperature

C. Determine the rate of the 5th experiment

D. The energy of activation for this reaction is 134kj/mole. At what temperature will the rate double for this reaction?

Exp. (H+) (CH3COCH3) (Br2) Rate(M/S) 1 .30 .050 .100 5.7 x 10^-5 2 .30 .10 .100 5.7 x 10^-5 3 .30 .050 .200 1.1 x 10^-4 4 .90 .100 .400 2.0 x 10^-3 5 .025 .100 .250 ?

Explanation / Answer

let the rate law be

rate = k [H+]^a [CH3COCH3]^b [Br2]^c

consider exp1 and exp2

[H+] and [Br2] are constant

doubling the conc of CH3COCH3 has no affect of the rate

so

b = 0

now

consider exp1 and exp3

[H+] and [CH3COCH3] are constant

doubling the conc of Br2 doubles the rate

so

c =1

now

consider exp3 and exp4

2 x 10-3 / 1.1 x 10-4 = [0.9 / 0.3]^a [0.1 / 0.5]^0 [0.4/0.2]^1

a = 2

so

the rate law is

rate = k [H+]^2 [Br2]

B)

consider exp1

5.7 x 10-5 = k [0.3]^2 [0.1]

k = 6.333 x 10-3

C)

now

rate = 6.333 x 10-3 x [0.025]^2 [0.25]

rate = 9.9 x 10-7

D)

we know that

ln ( k2/k1) = (Ea / R) (1/T1 - 1/T2)

so

ln2 = ( 134 x 1000 / 8.314 ) ( 1/ 298 - 1/T2)

T2 = 301.87

so

T2 = 301.87 - 273 = 28.87

so

the rate will double at 28.87 C

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