Suppose the activation energy for some reaction is 33 kJ/mol, and further, suppose you measure the reaction rate at 300 K

To keep the concentrations constant but increase the
rate by a factor of 2 by increasing the temperature, what temperature would you select?
Answer in units of K

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The activation energy is the minimum energy required for a reaction to proceed. In other words, it is the energy that must be overcome in order for the chemical reaction to happen. It can be derived in many ways and one of which is the use of the reaction rate constants and changes in temperature. The equation can be written as:

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


k1 and k2 are reaction rate constants

R = gas constant 8.314 J/mol-K

T1 and T2 are absolute temperature (K)

Ea = energy of activation = 33kJ/mol = 33000j/mol

The problem is stating that the reaction rate increase by a factor of 2. This means that if k1 is 1 then k2 would be 1x2 or 2. Therefore k2/k1 = 2.

Now we can solve the problem since it is asking for the change in temperature when the k1 is increased by a factor of 2.

We try to transform the equation first and then substitute the given values.

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

`ln [(2)/(1)] = (33000)/(8.314) ((1)/(300) - (1)/(T2))`

`0.6931 = 3969.21 ((1)/(300) - (1)/(T2))`

` ` `(0.6931)/(3969.21) =((1)/(300) - (1)/(T2))`

`0.0001746 =((1)/(300) - (1)/(T2))`

`(1)/(T2) = (1)/(300) - 0.0001746`

`(1)/(T2) = 0.0033333 - 0.0001746`

`(1)/(T2) = 0.0031587`

`T2 = (1)/(0.0031587)`

T2 = 316.59 K


See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial