Question

A reaction rate increases by a factor of 500. in the presence of a catalyst at 37oC. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factor being equal

Answer 1

Answer:

$E_2=999984KJ/mole$

Explanation:

From the question we are told that:

Factor $dK=500$

Temperature $T=37 C=310k$

Activation energy $E=10^6kJ/mol$

Generally the Arhenius equation is mathematically given by

$ln \frac{K_2}{K_1}=\frac{ E_1-E_2}{RT}$

Where

$\frac{K_2}{K_1}=500$

$ln 500=\frac{ 10^6-10^3-E_2}{8.314*310}$

$E_2=999984KJ/mole$

Answer 2

The activation energy of the new reaction is 105.99 kJ/mol.

Using the Arrhenius equation;

ln(k2/k1) = -Ea2/RT2 +  Ea1/RT1

Now, from the information in the question;

k2/k1 = 500

Ea = ?

R = 8.314 JKmol-1

T2 = 37oC + 273 = 310 K

T1 = 37oC + 273 = 310 K

Substituting values;

ln (500) =- Ea2 + Ea1

6.2 = -Ea2 + 106 × 10^3 J

Ea = 106 × 10^3 J - 6.2

Ea = 105.99 × 10^3 J  or 105.99 kJ/mol

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