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A. a first order reaction has rate constants of 4.6×10-2rnsrn-1rnat 0° c and 20° c, respectively. rnwhat is the value of the activation energy?
Accepted Answer
The Arrhenius equation can be used to determine the activation energy of a reaction:
$text{k} = ext{Ae}^{-frac{text{E}_ ext{a}}{ ext{RT}}}$
where:
$text{k}$ is the rate constant
$text{A}$ is the pre-exponential factor
$text{E}_ ext{a}$ is the activation energy
$text{R}$ is the gas constant
$text{T}$ is the temperature in Kelvin
We can take the natural logarithm of both sides of the equation to get:
$ln(k) = ln(A) - frac{E_a}{RT}$
If we plot $ln(k)$ versus $frac{1}{T}$, the slope of the line will be $-frac{E_a}{R}$ and the intercept will be $ln(A)$.
Using the given rate constants and temperatures, we can calculate the activation energy:
$ln(4.6 times 10^{-2}) - ln(20) = -frac{E_a}{(8.314 J/mol K)(273 K)}$
$rightarrow E_a = 52.3 text{kJ/mol}$
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