We’re being asked to **determine the activation energy** of the reaction when the rate increases by 7.5x from 25°C to 30°C

This means we need to use the ** two-point form of the Arrhenius Equation**:

$\overline{){\mathbf{ln}}\left(\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}\right){\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}\frac{\mathbf{Ea}}{\mathbf{R}}\left[\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}\right]}$

where:

**k _{1}** = rate constant at T

**k _{2}** = rate constant at T

**E _{a}** = activation energy (in J/mol)

**R** = gas constant (8.314 J/mol•K)

**T _{1} and T_{2}** = temperature (in K).

Increasing the reaction temperature from 25°C to 30°C raises the reaction rate 7.5- fold. What is the activation energy?

A. 5.5 kJ

B. 303 kJ

C. 43.1 kJ

D. 16.2 kJ

E. 110. kJ

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Arrhenius Equation concept. You can view video lessons to learn Arrhenius Equation. Or if you need more Arrhenius Equation practice, you can also practice Arrhenius Equation practice problems.

What is the difficulty of this problem?

Our tutors rated the difficulty of*Increasing the reaction temperature from 25°C to 30°C raises...*as medium difficulty.