Question

A reaction has a rate constant of 0.0117 s-1 at 400.0 k and 0.689 s-1 at 450.0 k. calculate the activation energy in kilojoules for this reaction.

Answer 1

The activation energy for the given reaction can be calculated using the Arrhenius equation. By plugging in the appropriate values and solving for Ea, we find that the activation energy is approximately 63.4 kJ/mol.

Step-by-step explanation:

The activation energy (Ea) can be calculated using the Arrhenius equation, which relates the rate constant (k), temperature (T), and frequency factor (A):

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

Using the given data:

k1 = 0.0117 s-1 at 400.0 K

k2 = 0.689 s-1 at 450.0 K

T1 = 400.0 K

T2 = 450.0 K

R = 8.314 J/mol K

Plugging in these values:

ln(0.689/0.0117) = ((Ea/8.314)((1/400.0) - (1/450.0)))

Solving for Ea:

Ea ≈ 63.4 kJ/mol

Therefore, the activation energy for this reaction is approximately 63.4 kilojoules per mole.

Answer 2

Answer: The activation energy of the reaction is 122.007 kJ.

Step-by-step explanation:

Rate constant at
`T_1,K_1=0.0117 s^(-1)`

`T_1=400 K`

Rate constant at
`T_2,K_2=0.689 s^(-1)`

`T_2=450 K`

Activation of the energy is given by formula:

`\log(K_2)/(K_1)=(E_a)/(2.303* R)* (T_2-T_1)/(T_1T_2)`

`\log(0.689 s^(-1))/(0.0117 s^(-1))=(E_a)/(2.303* 8.314 J/K mol)}* (450 K-400 K)/(450 K* 400 K)`

On solving for
`E_a`

`E_a=122,007.88 Joules=122.007 kilo-Joules (1000 J = 1kJ)`

The activation energy of the reaction is 122.007 kJ.