Question

In: Chemistry

Part A: The activation energy of a certain reaction is 34.4 kJ/mol . At 24 ∘C...

Part A:

The activation energy of a certain reaction is 34.4 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? Express your answer with the appropriate units.

Part 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 130. ∘C for the same reaction described in Part A? Express your answer with the appropriate units.

Solutions

Expert Solution

Part A

By using Arrhenius equation:

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

Ea, activation energy = 34.4 kJ/mol or 34.4 *10³ J/mol

Temperature, T1 = 24^oC = (24+273) K,

T2 = ?

Rate constant k2 = 2k1

Gas constant, R = 8.314 J/mol-K

Substituting all these values in the above equation:
ln(2k1/k1) = (34.4*10³ / 8.314)(1/(24 + 273) - 1/T2)
0.693147 = 4137.59923 (1/297 - 1/T2)

1.68 *10^-4 = 1/297 - 1/T2

1/T2 = -0.0032

T2 = 312.5 K

T2 = 312.5 – 273 K = 39.5 ^oC

Part B

By using Arrhenius equation:

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

Ea, activation energy = 34.4 kJ/mol or 34.4 *10³ J/mol

Temperature, T1 = 24^oC = (24+273) K,

T2 = 130 oC = (130+273)K

Rate constant k1 = 0.0190 /s

k2 = ?

Gas constant, R = 8.314 J/mol-K

Substituting all these values in the above equation:
ln(k2/0.0190) = (34.4*10³ / 8.314)[1/(24 + 273) - 1/(130+273)]
ln(k2/0.0190) = 4137.59923 (1/297 - 1/403)

ln(k2/0.0190) = 4137.59923 * 0.000885614

ln(k2/0.0190) = 3.664314931

(k2/0.0190) = e^3.664314931

k2 = 39.02938918 * 0.019

rate constant k2 = 0.74156 s^-1

queen_honey_blossom answered 1 year ago

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