The slope of the line for the graph of log k versus 1T for the reaction, N2O5→2NO2+12O2 is -5000. Calculate the energy of activation of the reaction (kJ K−1 mol−1)

K=Ae−Ea/RTln⁡ K=ln⁡A−EaRT2.303log⁡K=−EaRT+2.303log⁡Alog⁡K=−Ea2.303RT+log⁡ASlope =-Ea2.303R=-5000Ea=5000×8.31×2.303×10-3=95.7 kJ k−1 mol−1

The slope of the line for the graph of log k versus 1T for the reaction, N2O5→2NO2+12O2 is -5000. Calculate the energy of activation of the reaction (kJ K−1 mol−1)

K=Ae−Ea/RTln⁡ K=ln⁡A−EaRT2.303log⁡K=−EaRT+2.303log⁡Alog⁡K=−Ea2.303RT+log⁡ASlope =-Ea2.303R=-5000Ea=5000×8.31×2.303×10-3=95.7 kJ k−1 mol−1

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The slope of the line for the graph of log k versus $\frac{1}{T}$ for the reaction, $N_{2} O_{5} \to 2 {NO}_{2} + \frac{1}{2} O_{2}$ is -5000. Calculate the energy of activation of the reaction $({kJ K}^{- 1} {mol}^{- 1})$

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Detailed Solution

$K = {Ae}^{- E_{a} / RT}$

$\ln K = \ln A - \frac{E_{a}}{RT}$

$\begin{array}{l} 2 .303 \log K = - \frac{E_{a}}{RT} + 2 .303 \log A \\ \log K = - \frac{E_{a}}{2 .303 RT} + \log A \end{array}$

Slope $= - \frac{E_{a}}{2 .303 R} = - 5000$

$E_{a} = 5000 \times 8 .31 \times 2 .303 \times 10^{- 3}$

$= 95 .7 kJ k^{- 1} {mol}^{- 1}$

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