Question

In: Chemistry

a) The activation energy of a certain reaction is 47.0 kJ/mol . At 24 ∘C ,...

a) The activation energy of a certain reaction is 47.0 kJ/mol . At 24 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

b)Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24 ∘C , what would the rate constant be at a temperature of 100. ∘C for the same reaction described in Part A?

Expert Solution

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

k2 / k1 = 2 .

Ea = 47 x 10³ joules / mol

T1 = 273 +42 = 315 K

T2 = ?

ln 2 = -47 x 10³ / 8.314 [1/T2 - 1/315]

0.6931 = -5.65 x 10³ [1/T2 - 1/315]

0.6931 = -5.65 x 10³ [1/T2 - 0.00317]

[1/T2 - 0.00317] = -0.6931 / 5.65 x 10³

[1/T2 - 0.00317] =- 0.1226 x10^-3

1/T2 = 0.00305

T2 = 327

At 327 K or 54 degrees Celsius would this reaction go twice as fast

Question b

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

ln k2 / 0.0190 = -47 x 10³ / 8.314 [1/373 - 1/297]

ln k2 / 0.0190 =3.878

K2 = 0.9184s−1

K2 = 0.9184 s−1 at a temperature of 100∘C

queen_honey_blossom answered 2 years ago

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