Question

For \(2N_2O_5 \rightarrow 4NO_2 + O_2\), if rate is \(1.02 \times 10^{-4}\) and \(k = 3.4 \times 10^{-5}\,s^{-1}\), find the concentration of \(N_2O_5\).

• A. \(1.0\,\text{mol/L}\)
• B. \(2.0\,\text{mol/L}\)
• C. \(3.0\,\text{mol/L}\)
• D. \(4.0\,\text{mol/L}\) \bigskip

Hint:

In first–order reactions, concentration can be directly found using \(Rate = k[\text{Reactant}]\).

Answer 1

The Correct Option is C

Concept: For a first–order reaction, the rate law is given by: :contentReference[oaicite:0]{index=0} where \begin{itemize} \item Rate = reaction rate \item \(k\) = rate constant \item \([N_2O_5]\) = concentration of the reactant \end{itemize} Step 1: {\color{red}Substitute the given values.} \[ Rate = 1.02 \times 10^{-4} \] \[ k = 3.4 \times 10^{-5} \] \[ 1.02 \times 10^{-4} = (3.4 \times 10^{-5}) [N_2O_5] \] Step 2: {\color{red}Solve for the concentration.} \[ [N_2O_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} \] Step 3: {\color{red}Simplify the calculation.} \[ [N_2O_5] = 3 \] \[ [N_2O_5] = 3.0\,\text{mol/L} \] Thus, the concentration of \(N_2O_5\) is: \[ \boxed{3.0\,\text{mol/L}} \] \bigskip