# The rate constant of the reaction O(g) + Na_2(g) -> NO(g) + N(g), occurring in the stratosphere, is 9.7 x 10^10 L mol^-1s^-1 at 800 C. The activation energy of the reaction is 315 kJ/mol. How do you determine the rate constant at 700 C?

Jul 20, 2016

You can do it like this:

#### Explanation:

The Arrhenius Equation gives:

$\textsf{k = A {e}^{- {E}_{a} / \left(R T\right)}}$

$\textsf{k}$ is the rate constant

$\textsf{A}$ is the frequency factor, which is constant for a particular reaction

$\textsf{R}$ is the gas constant, $\textsf{8.31 \text{J/K/mol}}$

$\textsf{T}$ is the absolute temperature

Taking natural logs:

$\textsf{\ln k = \ln A - {E}_{a} / \left(R T\right)}$

If $\textsf{{T}_{1} = 800 \textcolor{w h i t e}{x} K}$ and $\textsf{{T}_{2} = 700 \textcolor{w h i t e}{x} K}$ then:

$\textsf{\ln {k}_{1} = \ln A - {E}_{a} / \left(R {T}_{1}\right) \text{ } \textcolor{red}{\left(1\right)}}$

and

sf(lnk_2=lnA-E_a/(RT_2)" "color(red)((2))

Subtracting both sides of $\textsf{\textcolor{red}{\left(1\right)}}$ from sf(color(red)((2))rArr

$\textsf{\ln \left[{k}_{1} / {k}_{2}\right] = - {E}_{a} / \left(R {T}_{1}\right) - \left(- {E}_{a} / \left(R {T}_{2}\right)\right)}$

$\therefore$$\textsf{\ln \left[{k}_{1} / {k}_{2}\right] = {E}_{a} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]}$

Putting in the numbers:

$\textsf{\ln \left[{k}_{1} / {k}_{2}\right] = \frac{315 \times {10}^{3}}{8.31} \left[\frac{1}{700} - \frac{1}{800}\right]}$

$\textsf{\ln \left[{k}_{1} / {k}_{2}\right] = 6.747}$

$\therefore$$\textsf{{k}_{1} / {k}_{2} = 851.728}$

$\therefore$$\textsf{{k}_{2} = \frac{9.7 \times {10}^{10}}{851.728} = 1.1 \times {10}^{8} \textcolor{w h i t e}{x} l . m o {l}^{- 1} . {s}^{- 1}}$

This is less than $\textsf{{k}_{1}}$ which is what you would expect since the reaction occurs at a lower temperature.