Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the following chemical equation. 2 NO₂(g) → 2 NO(g) + O₂(g) .
A sample of NO₂(g) is initially placed in a 2.50-L reaction vessel at 300°C.If the half-life and the rate constant at 300°C are 11 seconds and 0.54 M^-1 s^-1,respectively,how many moles of NO₂ were in the original sample?

Question

Quizplus · Accepted Answer

Since the reaction is second-order, the rate law is given by:

rate = k[N02]^2

Using the half-life equation for a second-order reaction:

t1/2 = 1 / (k [N02]0)

Where [N02]0 is the initial concentration of nitrogen dioxide.

Substituting the given values:

11 s = 1 / (0.54 x 10^-3 M^-1 s^-1 [N02]0)

[N02]0 = 0.42 M

To calculate the number of moles, we need to multiply the concentration by the volume of the reaction vessel:

moles of N02 = 0.42 mol/L x 2.50 L = 1.05 mol

Therefore, the correct answer is choice B.

Nitrogen Dioxide Decomposes at 300°C Via a Second-Order Process to Produce