For the first-order reaction,2 N₂O(g)→ 2 N₂(g)+ O₂(g),what is the concentration of N₂O after 3 half-lives if mol of N₂O is initially placed into a 1.00-L reaction vessel?

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Quizplus · Accepted Answer

Since the reaction is first-order, we can use the equation for the concentration of a reactant after n half-lives: \begin{equation} [ ext{N₂O}]_n = \frac{1}{2^n}[ ext{N₂O}]_0 \end{equation} where $[ ext{N₂O}]_n$ is the concentration of N₂O after n half-lives and $[ ext{N₂O}]_0$ is the initial concentration of N₂O. We are given that the initial concentration of N₂O is 11ea6923_183a_cd75_80d0_b50f1eedbb9c_TB6105_11 mol/L, and we want to know the concentration after 3 half-lives. Three half-lives correspond to $n=3$, so we can substitute these values into the equation above: \begin{equation} [ ext{N₂O}]_3 = \frac{1}{2^3}[ ext{N₂O}]_0 = \frac{1}{8}(11ea6923_183a_cd75_80d0_b50f1eedbb9c_TB6105_11\; ext{mol/L}) \approx 3.1 imes 10^{-12}\; ext{mol/L} \end{equation} We can also use stoichiometry to find the concentration of O₂ produced after 3 half-lives. From the balanced equation, we see that for every 2 mol of N₂O that react, 1 mol of O₂ is produced: \begin{equation} 2 ext{ N₂O} ightarrow ext{ O₂} + ext{ 2 N₂} \end{equation} Therefore, the concentration of O₂ after 3 half-lives is: \begin{equation} [ ext{O₂}]_3 = \frac{1}{2^3}(\frac{1}{2}[ ext{N₂O}]_0) = \frac{1}{16}(11ea6923_183a_cd75_80d0_b50f1eedbb9c_TB6105_11\; ext{mol/L}) \approx 1.6 imes 10^{-12}\; ext{mol/L} \end{equation} Therefore, the closest answer choice is B: 3.1 × 10^-² M.

For the First-Order Reaction,2 N2O(g)→ 2 N2(g)+ O2(g),what Is the Concentration

For the First-Order Reaction,2 N₂O(g)→ 2 N₂(g)+ O₂(g),what Is the Concentration