1. Tardigrade
  2. Question
  3. Chemistry
  4. The slope of the line for the graph of log k versus(1/T) for the reaction, N2O5→2NO2+(1/2)O2 is - 5000. Calculate the energy of activation of the reaction(kJ K-1mol-1)

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Q. The slope of the line for the graph of log $k$ versus$\frac{1}{T}$ for the reaction, $N_{2}O_{5}\to2NO_{2}+\frac{1}{2}O_{2}$ is $- 5000.$ Calculate the energy of activation of the reaction$\left(kJ K^{-1}mol^{-1}\right)$

Chemical Kinetics

Solution:

$K=Ae^{-E_{a/RT}}$
In $K=In \,A-\frac{E_{a}}{RT}$
$2.303,log \,K=-\frac{E_{a}}{RT}+2.303 \,log A$
$log\, K=-\frac{E_{a}}{2.303RT}+log \,A$
Slope $\frac{E_{a}}{2.303R}=5000$
$E_{a}=5000\times8.31\times2.303$
$=95.7kJ k^{-1}mol^{-1}$