a,The activation energy of a certain reaction is 37.1 kJ/mol . At 30 ∘C , the...

a,The activation energy of a certain reaction is 37.1 kJ/mol . At 30 ∘C , the rate constant is 0.0170s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

b,Given that the initial rate constant is 0.0170s−1 at an initial temperature of 30 ∘C , what would the rate constant be at a temperature of 170. ∘C for the same reaction described in Part A?

Expert Solution

a)
T1 = 30 oC
= (273+30) k
= 303 k
use,
2.303log(K2/K1) = (E/R)*[1/T1 - 1/T2]
K2 = 2*K1
2.303*log(2) = (37100/8.3)*[1/303 - 1/T2]
0.69 = 4469.8*[1/303 - 1/T2]
1.6*10^-4 = [1/303 - 1/T2]
0.16*10^-3 = 3.3*10^-3 -1/T2
1/T2 = 3.3*10^-3 - 0.16*10^-3
1/T2 = 3.14*10^-3
T2 = 318.5 k
= (318.5-273) oC
= 45.5 oC
b)
T1 = 30 oC
= (273+30) k
= 303 k
T2 = 170 oC
= (273+170) k
= 443 k
use,
2.303log(K2/K1) = (E/R)*[1/T1 - 1/T2]
2.303*log(K2/0.017) = (37100/8.3)*[1/303 - 1/443]
2.303*log(K2/0.017) = 4469.8*(3.3*10^-3 - 2.3*10^-3)
2.303*log(K2/0.017) = 4469.8*10^-3
2.303*log(K2/0.017) = 4.4698
log(K2/0.017) = 1.9
K2/0.017 = 79.4
K2 = 1.35 S^-1

queen_honey_blossom answered 2 years ago

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