Question

If the rate of reaction is doubled by increasing the temperature from 298 K to 308K, then calculate the energy of activation of the reaction.Calculate the given example.

Answer 1

Activation Energy Problem

Step 1: Convert temperatures from degrees Celsius to Kelvin. T = degrees Celsius + 273.15. T1 = 3 + 273.15. ...

Step 2 - Find Ea ln(k2/k1) = Ea/R x (1/T1 - 1/T2) ...

Answer: The activation energy for this reaction is 4.59 x 104 J/mol or 45.9 kJ/mol.

Answer 2

Answer:

$E_{a}=52897.77 Joules.$

Explanation:

Since we know the relation between the rate of reaction and temperature of the reaction.

Assume $K_{1}$ and $K_{2}$ are the rate constant and $T_{1}$ and $T_{2}$ are the temperature then,

$log\dfrac{K_{2} }{K_{1}} =\dfrac{E_{a}(T_{2}-T_{1}) }{2.303RT_{1}T_{2}}$...............(1)

We have

$\dfrac{K_{2} }{K_{1}} = 2$ and $T_{1}=298 k$ and $T_{2}=308k$

Using equation (1) we get,

$E_{a} = \dfrac{log2\times2.303\times 8.314\times 298\times 308}{10}$

After solving we get,

$E_{a}=52897.77 Joules.$