# Problem: The reaction2NO (g) + O2 (g) → 2NO2 (g)was studied, and the following data were obtained whereWhat would be the initial rate for an experiment where [NO] 0 = 6.21x 1018 molecules/cm3 and [O2]0 =  7.36 x 1018 molecules/cm3?

🤓 Based on our data, we think this question is relevant for Professor Giles' class at VCU.

###### FREE Expert Solution

Recall that the rate law only focuses on the reactant concentrations and has a general form of:

k = rate constant
A & B = reactants
x & y = reactant orders

Step 1. Calculate order with respect to NO:

x = order

$\frac{\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{×}{\mathbf{10}}^{\mathbf{17}}}{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{16}}}\mathbf{=}{\left[\frac{3.00×{10}^{18}}{1.00×{10}^{18}}\right]}^{\mathbf{x}}\phantom{\rule{0ex}{0ex}}\mathbf{9}\mathbf{=}{\left[3\right]}^{\mathbf{x}}$

3 raised to the 2nd power = 9

x = 2 → 2nd order with respect to NO

Step 2. Calculate order with respect to O2:

y = order

since the numerator and the denominator is raised to the same power, you can simplify the equation to:

0.4 raised to the power of 1 = 0.4

y = 1 → 1st order with respect to O2

Step 3Calculate the value of the rate constant

Go back to the given data. You can use any experimental data to calculate for k, but let’s use the 1st data for this calculation:

[NO] = 1.00 x 1018

[O2] = 1.00 x 1018

rate = 2.00 x 1016

k = 2.0 x 10-38

###### Problem Details

The reaction

2NO (g) + O(g) → 2NO(g)

was studied, and the following data were obtained where

What would be the initial rate for an experiment where [NO] 0 = 6.21x 1018 molecules/cm3 and [O2]0 =  7.36 x 1018 molecules/cm3?