Respuesta :
Answer
The variation in the rate k (k1/k2) is 0.5707 s^-1
Step-by-step explanation:
Given the following parameters
Activation energy = 43.5Kj.mol
T1 = 300K
T2 = 310K
R = 8.314 J.Kmol
[tex]K\text{ =A}e^{-\frac{Ea}{RT}}^{}^{}[/tex][tex]\begin{gathered} \text{where K = rate constant},\text{ Ea = Activation energy, R= gas constant and T = temperature} \\ \ln (k)\text{ = }\ln A\text{ - }\frac{Ea}{RT} \\ \text{ A is a constant} \\ \text{ if the reaction occurs at two temperature T1 and T2} \\ \ln k1\text{ = }\ln A\text{ - }\frac{Ea}{RT1}--------\text{ equation 1} \\ \ln k2\text{ = }\ln A\text{ - }\frac{Ea}{RT2}\text{ --------- EQUATION 2} \\ \text{ substracting equation 1 from 2},\text{ we have} \\ \ln (\frac{k1}{k2})\text{ = }-\frac{Ea}{RT1}\text{ + }\frac{Ea}{RT2} \\ \ln (\frac{k1}{k2})\text{ = }\frac{Ea}{R}(\frac{1}{T2}\text{ - }\frac{1}{T1}) \\ \\ \ln (\frac{k1}{k2})\text{ = }\frac{43500}{8.314}(\frac{1}{310}\text{ - }\frac{1}{300}) \\ \ln (\frac{k1}{k2})\text{ = 5232.139 }(0.00322\text{ - 0.003333)} \\ \ln (\frac{k1}{k2})\text{ = 5232.139 (}-\text{ 0.0001071)} \\ \ln (\frac{k1}{k2})\text{ = -0.560851} \\ \text{Take the exponential of both sides} \\ \frac{k1}{k2}\text{ = }e^{-0.560851} \\ \frac{k1}{k2\text{ }}=0.5707s^{-1} \end{gathered}[/tex]