Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temperature of 150. ∘C for the same reaction described in Part A?

Part A said: The activation energy of a certain reaction is 45.5 kJ/mol . At 21 ∘C , the rate constant is 0.0110s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?
T2= 32 degrees

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2022-07-25T18:47:24+02:00

The rate of reaction at 150 degrees is 1.878.

We have the formula as;

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

ln(2( 0.0110)/0.0110) = 45.5 * 10^3/8.314(1/294 - 1/T2)

0.693 = 5473(1/294 - 1/T2)

0.693/ 5473 = (1/294 - 1/T2)

1.27 * 10^-4 = 1/294

1/T2 = 1/294 - 1.27 * 10^-4

1/T2 = 340 * 10^-4 - 1.27 * 10^-4

T2 = 30 degrees

Again;

ln(k2/0.0110) = 45.5 * 10^3/8.314(1/303 - 1/423)

ln(k2/0.0110) = 5473(3.3 * 10^3 - 2.36 * 10^3)

ln(k2/0.0110) = 5.14

k2/0.0110 = e^5.14

k2 = 0.0110 * e^5.14

k2 = 1.878

The temperature that the reaction will go twice as fast is 30 degrees while the rate construction at 150 degrees is 1.878.

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