A certain reaction rate increases 1000 folds in the presence of a catalyst at 27^° C . The activation energy of the original pathway is 98 ~kJ / mol . What is the activation energy of the new pathway?

Correct answer: (A) 80.77 kJ. log k_2/k_1=log 1000 = (E_1-E_2)/(R T) =(98000-E_2)/(8.314 × 300) E_2 (catalyst) =80.77 ~kJ

Q54 / 200 · Manipal (MET) 2013 · Chemistry

A certain reaction rate increases 1000 folds in the presence of a catalyst at $27^{\circ} \mathrm{C}$. The activation energy of the original pathway is $98 \mathrm{~kJ} / \mathrm{mol}$. What is the activation energy of the new pathway?

Correct answer
A
80.77 kJ
B
56.38 kJ
C
24.67 kJ
D
90.43 kJ

Solution

$\log \frac{k_2}{k_1}=\log 1000$ $\begin{aligned}= & \frac{E_1-E_2}{R T} \\ & =\frac{98000-E_2}{8.314 \times 300} \\ E_2 & \text { (catalyst) }=80.77 \mathrm{~kJ}\end{aligned}$

Hence, the correct answer is option: 'A'.

A certain reaction rate increases 1000 folds in the presence of a catalyst at $27^{\circ} \mathrm{C}$. The activation energy of the original pathway is $98 \mathrm{~kJ} / \mathrm{mol}$. What is the activation energy of the new pathway?

Solution

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