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 L mol^-1 s^-1,respectively,how many moles of NO₂ were in the original sample?

Question

Quizplus · Accepted Answer

The second-order rate law can be written as follows:

Rate = k[N2O4]^2

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

t1/2 = 1 / k[N2O4]

Plugging in the given values:

11 s = 1 / (0.54 L mol^-1 s^-1)[N2O4]

[N2O4] = 0.42 mol L^-1

To find the total moles of N2O4 in the original sample:

moles = concentration × volume = 0.42 mol L^-1 × 2.50 L = 1.05 mol

Therefore, the answer is B) 0.42 mol.

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