calculate the activation energy of a first order reaction whose temperature coefficient is 2 obtained by studying the reaction between 25 degrees centigrade and 35 degrees centigrade
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The activation energy can be calculated using the Arrhenius Equation which is:
Ln(k2/k1) = - (Ea/R) * ((1/t2) – (1/t1)) (Equation 1)
where k2 = rate constant at temperature t2
k1 = rate constant at temperature t1
Ea = activation energy in J/mol
R = gas constant = 8.314 J/mol-K
t1 = reaction temperature 1 in Kelvin
t2 = reaction temperature 2 in Kelvin
Now, temperature coefficient = 2 (given)
We know that temperature coefficient = ratio of rate of the reaction
Therefore, k2/k1 = 2
Also, T1 = 25C = 298.15 K (given)
and T2 = 35C = 308.15 K (given)
Now substituting the values of k2/k1, R, t1 and t2 in equation 1, we get
ln(2) = - (Ea/8.314) * ((1/308.15) - (1/298.15))
Rearranging and solving the above equation, we get Ea = 5.29 * 10^4 J/mol (Answer)
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