Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen 14) 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?
0.11
To determine the number of moles of NO2 in the original sample, we need to use the half-life and rate constant given. Here's how you can calculate it step-by-step:
Step 1: Write down the given information.
- Half-life (t1/2) = 11 seconds
- Rate constant (k) = 0.54 M-1 s-1
- Initial concentration of NO2 (C0) (unknown)
- Final concentration of NO2 (Ct) = ?
Step 2: Understand the second-order reaction.
In a second-order reaction, the rate of reaction is proportional to the square of the concentration of the reactant(s). The rate law equation for a second-order reaction is:
rate = k * [NO2]^2
Step 3: Determine the relationship between the initial concentration and the final concentration of NO2 using the half-life.
The half-life (t1/2) can be related to the rate constant (k) and the initial concentration ([NO2]0) using the equation:
t1/2 = 1 / (k * [NO2]0)
Rearranging the equation, we get:
[NO2]0 = 1 / (k * t1/2)
Step 4: Substitute the values into the equation to calculate the number of moles of NO2 in the original sample.
[NO2]0 = 1 / (0.54 M-1 s-1 * 11 s)
Calculate [NO2]0 using the equation:
[NO2]0 = 1 / (5.94 M-1 s-1)
Now, you have the concentration of NO2 in moles per liter ([NO2]0). To determine the number of moles in the original sample, you need to multiply the concentration by the volume.
We know that the initial volume (V) is 2.50 L, so to calculate the number of moles (n), we can use the equation:
n = [NO2]0 * V
Substitute the values:
n = (1 / 5.94 M-1 s-1) * 2.50 L
Finally, calculate the number of moles (n):
n = 0.42 moles
Therefore, there were 0.42 moles of NO2 in the original sample.