Question

The rate constant for a reaction at 40.0°C is exactly 4 times that at 20 0°C Calculate the energy of activation for the reaction.

Answer 1

Explanation:

Data provided:

T2 = 40.0 °C

(absolute temperature = T2 = 40.0 °C + 273 = 313 K)

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T1 (absolute) = 20.0 °C + 273 = 293 K

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The rate constant for a reaction at 40.0 °C is exactly 4 times that at 20.0 °C, mathematically:

k2/k1 = 4

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Here is used the Arrhenius expression as follows:

$ln\text{ }\frac{k2}{k1}=\text{ }\frac{Ea}{R}x\lbrack\frac{1}{T1}-\frac{1}{T2}\rbrack$

R = universal gas constant = 8.314 J/mol K

Ea = activation energy

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Procedure:

$\begin{gathered} ln\text{ 4 = }\frac{Ea}{8.314\text{ J/mol K}}x\lbrack\frac{1}{293\text{ K}}-\frac{1}{313\text{ K}}\rbrack \\ 1.386\text{ = }\frac{Ea}{8.314}x(2.18x10^{-4}) \\ Ea\text{ = 52858.73 J/mol} \\ \end{gathered}$

Answer: Ea = 52858.73 J/mol