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?
A) 0.17 mol
B) 0.42 mol
C) 5.9 mol
D) 15 mol
The correct answer and explanation is :
To solve this problem, we need to use the integrated rate law for a second-order reaction. The general equation for a second-order reaction is:
$$
\frac{1}{[A]} = \frac{1}{[A]_0} + kt
$$
where:
- $[A]$ is the concentration of the reactant at time $t$,
- $[A]_0$ is the initial concentration of the reactant,
- $k$ is the rate constant,
- $t$ is the time.
For a second-order reaction, the half-life is given by the formula:
$$
t_{1/2} = \frac{1}{k[A]_0}
$$
Step 1: Rearranging to find the initial concentration $[A]_0$
We are given the half-life ($t_{1/2} = 11$ seconds) and the rate constant ($k = 0.54$ M$^{-1}$ s$^{-1}$). We can rearrange the half-life equation to solve for the initial concentration $[A]_0$:
$$
[A]0 = \frac{1}{k \cdot t{1/2}}
$$
Substituting the known values:
$$
[A]_0 = \frac{1}{(0.54 \, \text{M}^{-1} \text{s}^{-1}) \cdot (11 \, \text{s})}
$$
$$
[A]_0 = \frac{1}{5.94 \, \text{M}^{-1}} = 0.168 \, \text{M}
$$
Thus, the initial concentration of NO2 is 0.168 M.
Step 2: Using the volume of the reaction vessel to find the number of moles
Now that we know the initial concentration, we can use the volume of the reaction vessel (2.50 L) to find the number of moles of NO2. The number of moles is given by:
$$
\text{moles} = \text{concentration} \times \text{volume}
$$
Substituting the values:
$$
\text{moles} = 0.168 \, \text{M} \times 2.50 \, \text{L} = 0.42 \, \text{mol}
$$
Final Answer:
The correct answer is B) 0.42 mol.
Explanation:
The process involves using the second-order rate law and the given half-life to first determine the initial concentration of NO2. Once we have the concentration, we apply the volume of the reaction vessel to calculate the moles of NO2 initially present. The result of this calculation is 0.42 mol, making option B the correct answer.