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 44 seconds and 0.54 M^-1 s^-1,respectively,how many moles of NO₂ were in the original sample?

Question

Quizplus · Accepted Answer

We can use the second-order integrated rate law for a reactant to solve this problem:

1/[NOS]t - 1/[NOS]0 = kt

where [NOS]t is the concentration of NOS at time t, [NOS]0 is the initial concentration of NOS, k is the rate constant, and t is the time.

We are given that the half-life (t1/2) of the reaction is 44 seconds, which means that [NOS]t = 0.5[NOS]0 when t = t1/2. Substituting these values into the integrated rate law, we can solve for [NOS]0:

1/(0.5[NOS]0) - 1/[NOS]0 = k(44 s)
1/[NOS]0 = 2/(k(44 s)) + 1/(0.5[NOS]0)
1/[NOS]0 = 4/(k(44 s)) (since 0.5[NOS]0 cancels out)

Now we just need to plug in the given values for k and solve for [NOS]0:

1/[NOS]0 = 4/(0.54 MS-1)
[NOS]0 = 0.11 mol/L

Finally, we can calculate the total number of moles of NOS in the 2.50-L reaction vessel:

moles of NOS = concentration x volume = 0.11 mol/L x 2.50 L = 0.275 mol

Therefore, the best choice is B) 0.11 mol, which is the initial number of moles of NOS in the sample.

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