Question

Experimental values for the temperature dependence of the rate constant for the gas-phase reaction are as follows:

$N{O}_{\left(g\right)} + {O_3}_{\left(g\right)} \to N{O}_2{ }_{\left(g\right)} + {O_2}_{\left(g\right)}$

Make the appropriate graph using these data, and determine the activation energy for this reaction.

Make the appropriate graph using these data, and determine the activation energy for this reaction.

Short answer

The graph between Ink vs T is given as below. The activation energy of the reaction is$4.0828kJ/mol.$

Step-by-step solution

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

Describing the activation of the reaction

Data

$$\begin{gathered} T_1=195K,k_1=1.08⨯109s{}^{-1} \\ {T}_2=369K,k_2=35.5⨯109s^{-1} \end{gathered}$$

The Arrhenius reaction is

$\text{In}{k}_{} = \text{In}A - \frac{\text{E}}{\text{RT}}$

For (1)

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

For (2)

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

Subtracting (2) from (1)

role="math" localid="1663755412677" $$\begin{gathered} In\left(\frac{k_1}{k_2}\right)= \frac{E}{R{T}_2} - \frac{E}{R{T}_1} \\ \text{In}\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}\right) = \frac{\text{E}}{\text{R}} \left(\frac{1}{{\text{T}}_{\text{2}}} - \frac{1}{{\text{T}}_{\text{1}}}\right) \\ \text{In}\left(\frac{1.08 \times {10}^9}{35.5 \times {10}^9}\right)= \frac{\text{E}}{8.314} \left(\frac{1}{369} - \frac{1}{195}\right) \\ \text{In}\left(0.304\right)= \frac{\text{E}}{8.314} \left(\frac{1}{369} - \frac{1}{195}\right) \\ - 1.18744= \frac{E}{8.314} \left(\frac{-174}{71,955}\right) \\ 1.18744= \frac{E}{8.314} \left(0.002418\right) \\ 9.87 = E \left(0.002418\right) \\ E = 4082.8685 \\ E = 4.0828Kj/mol \end{gathered}$$

The activation energy of the reaction is$4.0828kJ/mol.$