Rate constant of a first order reaction follows following equation $\log k = 8.0-\frac{5000}{T}$. The $E_a$ for the reaction is: (Given $R=8.3 JK^{-1}mol^{-1}$)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $95.57\, kJ\, mol^{-1}$
To find the activation energy \( E_a \) of a first-order reaction given the equation for the rate constant \( k \):
\(\log k = 8.0 - \frac{5000}{T}\) ------(i)
we can relate this to the Arrhenius equation, which is given by:
\(k = A e^{-\frac{E_a}{RT}}\)
Taking the logarithm of both sides, we get:
\(\log k = \log A - \frac{E_a}{2.303RT}\) ------(ii)
Comparing equation (i) and (ii), we get:
\(\frac{E_a}{2.303R} = 5000\)
Now substituting the value of the gas constant \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \):
\(E_a = 5000 \times 2.303 \times 8.314\)
\(E_a = 95579.5 J/mol\)
\(E_a = 95.57 kJ/mol\)
Conclusion
Thus, the activation energy \( E_a \) for the reaction is approximately: 95.57 kJ/mol