Question

The rate constant for the gas-phase decomposition of N2O5.

$N_2{O}_5 \to 2N{O}_2 + \frac{1}{2} O_2$

It has the following temperature dependence:

Make the appropriate graph using this data, and determine

the activation energy for this reaction.

Short answer

The graph between the Ink vs 1/T is given below. The activation energy for this reaction is$1.0359⨯{10}^{2}kJ/molK.$

Step-by-step solution

Drawing the appropriate graph between T(K) and k(s-1)

The given reaction is unimolecular.

Describing the activation energy for this reaction

Data

$$\begin{gathered} {\text{T}}_1 = 338 \text{K} and {\text{k}}_1 = 4.9 \times {10}^{-3} \left({\text{s}}^{-1}\right) \\ {\text{T}}_{\text{2}} = 298 \text{K} and {\text{k}}_2= 3.5\times {10}^{-5} \left(s^{-1}\right) \end{gathered}$$

The Arrhenius equation is

$\text{Ink} = \text{InA} - \frac{\text{E}}{\text{RT}}$

role="math" localid="1663754648845" $For{T}_1=338Kandk1=4.9⨯{10}^{-3}$

${\text{Ink}}_1 = \text{InA} - \frac{\text{E}}{{\text{RT}}_{\text{1}}} \dots \left(1\right)$

$For{T}_2=298Kand{k}_2=3.5⨯{10}^{-5}$

${\text{Ink}}_2 = \text{InA} - \frac{\text{E}}{{\text{RT}}_{\text{2}}} \dots \left(2\right)$

Subtracting (2) from (1), we get

$$\begin{gathered} \text{In}\frac{{\text{k}}_1}{{\text{k}}_2} = \left[\frac{\text{E}}{{\text{RT}}_2} - \frac{\text{E}}{{\text{RT}}_1} \right] \\ \text{In}\left(\frac{{\text{k}}_1}{{\text{k}}_2}\right) = \frac{\text{E}}{\text{R}} \left[\frac{1}{{\text{T}}_2} - \frac{1}{{\text{T}}_1} \right] \\ \text{In}\left(\frac{4.9 \times {10}^{-3}}{3.5\times {10}^{-5}}\right) = \frac{\text{E}}{8.314} \left[\frac{1}{298} - \frac{1}{338} \right] \end{gathered}$$

$$\begin{gathered} \text{In}\left(1.4 \times {10}^2\right) = \frac{\text{E}}{8.314} \left[\frac{40}{298 \times 338} \right] \\ \text{ln}\left(1.4 \times {10}^2\right) = \frac{\text{E}}{8.314} \left(0.0003971 \right) \\ \text{In}\left(1.4 \times {10}^2\right) = \text{E} \left(0.0000477\right) \\ 9416 = \text{E}( 4.77 \times {10}^{-5}) \\ \text{E} = 1.0359 \times {10}^5 \frac{\text{J}}{\text{mol} \text{K}} \\ \text{E} = 1.0359 \times {10}^2 \frac{\text{KJ}}{\text{mol} \text{K}} \end{gathered}$$

Therefore, the activation energy for the reaction (E) is$1.0359⨯102kJ/molK.$