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

Step-by-step 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{ }(k2)/(k1)=\text{ }(Ea)/(R)x\lbrack(1)/(T1)-(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{ = }(Ea)/(8.314)x(2.18x10^(-4)) \\ Ea\text{ = 52858.73 J/mol} \\ \end{gathered}`

Answer: Ea = 52858.73 J/mol