Question

Oxidation of NO2 and SO2 in the gas-phase produces nitric acid and sulfuric acid, respectively, and proceeds predominantly through reaction with the hydroxyl radical.

OH + NO2 → HNO3 k1 = 6E-11 cm3 molecule-1 s-1

For this problem, you can assume the initial mixing ratio of NO2 is 4.0 ppbv (1.08E11 mlc cm-3) and a steady-state [OH] = 3E6 mlc cm-3. We will assume there are no additional sources of NO2 for this system, and there is no initial nitric acid present at t0.

i. Write the rate equation for NO2 loss

ii. Integrate your rate equation to obtain an expression for [NO2] as a function of t

iii. Make a plot of [NO2] vs. time for t = 0-3 hours (use enough data points to create a smooth curve)

Answer 1

Hi, since you have the following rate law:

, we could start by rearranging the equation as:

take integral on both sides

we are assuming that the concentration of [OH] remains constant otherwise we would have to treat it as a variable and that would make the integration process a lot more complex!

We can apply the upper and lower limits

as we know that , we can write

, since

and

so here is the integrated equation, if you like you can compare it with y = mx + c where

x = t

I hope it helps. If you need information, please do not hesitate to ask.