Question

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy of the reaction? enter your answer in scientific notation.

Answer 1

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :  

The relation between the rate constant the activation energy is,  

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Answer 2

Answer;

139.8 kJ/mol

Explanation;

The activation energy is given by the formula;

Eₐ = R•ln[k₂/k₁]•T₁T₂/(T₂ – T₁)  

After converting ˚C to K:

T1 = 531 K

T2 = 468 k (after converting ˚C to K):  

R = 8.314 J/mol•K

Eₐ = (8.314 J/mol•K) × ln[(3.2 × 10⁻³)/(4.5 × 10⁻⁵)] × (531 K)(468 K)/(531 – 468 K)  

Eₐ = 139.8 kJ/mol