a. When the temperature is changed from 25°C to 56.8°C, the reaction rate triple

Question

a. When the temperature is changed from 25°C to 56.8°C, the reaction rate triples. What is the Activation Energy (Ea) in kJ/mol for this reaction?

b. A first order reaction has a half life of 73.0 minutes at 30.0°C. What is the half life (in minutes) when the reaction is heated to 69.0°C? The activation energy for this reaction is 115.0 kJ/mol.

c. 2 A  ? 3 B + C

Experiment has shown that the half life of this reaction was 30.20 minutes and that this did not depend on the initial concentration of A. If you start with an initial concentration of [A] = 0.0400 M, how many minutes until the concentration of [B] = 0.0194 M?

Explanation / Answer

(a)

Arrhenius equation,

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

ln(3/1) = ( Ea / 0.008314 ) * [ ( 1 / 298.15 ) - ( 1 / 329.95 ) ]

Ea = Activation energy = 28.26 kJ

(b)

k1 = 0.693 / t1/2 = 0.693 / 73.0 = 0.00949 min-1

Arrhenius equation,

ln(k2 / 0.00949) = ( 115.0 / 0.008314 ) * [ ( 1 / 303.15 ) - ( 1 / 342.15 ) ]

k2 / 0.00949 = e5.20

k2 = 181.4 * 0.00949

k2 = 1.72 min-1

Therefore,

Half life time = 0.693 / k2 = 0.693 / 1.72 = 0.402 min-1

(c)

Since the half life is not dependent on initial concentration the reaction follows first order kinetics.

k = 0.693 / t1/2 = 0.693 / 30.20 = 0.0229 min-1

From the balanced equation.

3 mol of B is formed from 2 mol of A

then, 0.0194 mol of B is formed from 2 * 0.0194 / 3 = 0.0129 mol of A

Therefore, remaining concentration of A = 0.0400 - 0.0129 = 0.0271 M

First order integrated rate constant equation is,

k = ( 1 / t ) * ln[A]0/[A]t

0.0229 = ( 1 / t ) * ln(0.0400 / 0.0271)

t = 17.0 min

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