Question

4. A slightly bruised apple will rot extensively in about 4.00 days at room temperature 25.0°C. If the same apple is kept in the refrigerator at 5.0°C, the same extent of rotting takes about 16.0 days. What is the activation energy for the rotting reaction?

Answer 1

Given data:

Rotting reaction 1: Temperature (T1) = 5oC = 278K; Duration = 16.0 days.

Rotting reaction 2: Temperature (T2) = 25oC = 298K; Duration = 4.00 days.

Solution:

Arrhenius equation for a reaction is

k = Ae-(Ea/RT) --------> 1

Where k is rate constant; A is frequency factor or pre-exponential factor; Ea is activation energy; R is gas constant, 8.314 J/mol.K; T is kelvin temperature.

Different form for equation 1 created by a mathematical operation on the standard one is

lnk = lnA - (Ea/RT) ------> 2

Equation 2 can be written for rotten reaction 1 and 2 as below.

lnk1 = lnA - (Ea/RT1) ------> 3

lnk2 = lnA - (Ea/RT2) ------> 4

Equation 4 / equation 3;

lnk2 / lnk1 = [lnA - (Ea/RT2)] / [lnA - (Ea/RT1)] ----> 5

Since there is small temperature change, the frequency factor (A) in the equation 5 is approximately constant for both the rotting reaction. Hence it can be cancelled.

Solving and simplifying equation 5

ln( k2 / k1) = (Ea/R)[(T2 - T1) / ((T1 x T2)] -------> 6

At 278K, rotting reaction takes place for 16 days. That means k1 = 1/12, 1/12 portion rots per day. Similarly,  at 298K, rotting reaction takes place for 4 days. That means k2 = 1/4, 1/4 portion rots per day.

Substitute the values of k1, k2, R,  T1 & T2 values in equation 6.

ln[(1/4) / (1/12)] = (Ea/8.314J mol-1K-1)[(298K - 278K) / ((278K x 298K)]

ln(3) = (Ea/8.314J mol-1K-1)[(20K) / ((82844K2)]

ln(3) = Ea [(2.414 x 10-4) / 8.314J mol-1]

1.0986 =  Ea(2.9035 x 10-5J-1 mol)

Ea = 1.0986 / (2.9035 x 10-5J mol-1) = 37837J mol-1

Hence the activation energy for the rotting reaction is 37837J mol-1.