Question:
The rate constant of the reaction, \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \) at 300 K is \( 3 \times 10^{-5} \) s\(^{-1}\). If the rate of the reaction is \( 2.4 \times 10^{-5} \) mol dm\(^{-3}\) s\(^{-1}\), then the molar concentration of \( N_2O_5 \) is
The rate constant of the reaction, \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \) at 300 K is \( 3 \times 10^{-5} \) s\(^{-1}\). If the rate of the reaction is \( 2.4 \times 10^{-5} \) mol dm\(^{-3}\) s\(^{-1}\), then the molar concentration of \( N_2O_5 \) is
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Unit of rate constant tells the order directly.
Updated On: May 2, 2026
- 0.4 M
- 0.8 M
- 0.04 M
- 0.08 M
- 0.6 M
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The Correct Option is B
Solution and Explanation
Concept: First order reaction
Since unit of $k$ = s$^{-1}$ ⇒ first order: \[ \text{Rate} = k[N_2O_5] \] ---
Step 1: Substitute values
\[ 2.4 \times 10^{-5} = 3 \times 10^{-5} \times [N_2O_5] \] ---
Step 2: Solve
\[ [N_2O_5] = \frac{2.4}{3} = 0.8 \] --- Final Answer: \[ \boxed{0.8 \, M} \]
Since unit of $k$ = s$^{-1}$ ⇒ first order: \[ \text{Rate} = k[N_2O_5] \] ---
Step 1: Substitute values
\[ 2.4 \times 10^{-5} = 3 \times 10^{-5} \times [N_2O_5] \] ---
Step 2: Solve
\[ [N_2O_5] = \frac{2.4}{3} = 0.8 \] --- Final Answer: \[ \boxed{0.8 \, M} \]
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