Question:
Observe the following reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$. At T(K), the concentration of $N_2O_5(g)$ changed from 2 mol $L^{-1}$ to 1.5 mol $L^{-1}$ in 100 min. What is the average rate (in mol $L^{-1} min^{-1}$) of this reaction?
Observe the following reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$. At T(K), the concentration of $N_2O_5(g)$ changed from 2 mol $L^{-1}$ to 1.5 mol $L^{-1}$ in 100 min. What is the average rate (in mol $L^{-1} min^{-1}$) of this reaction?
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Always divide the rate of disappearance of the reactant by its stoichiometric coefficient.
Updated On: Jun 10, 2026
- $5 \times 10^{-3}$
- $2.5 \times 10^{-3}$
- $2.5 \times 10^3$
- $1.25 \times 10^{-3}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Reaction rate = $-\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t}$.
Step 2: Analysis
$\Delta [N_2O_5] = 1.5 - 2.0 = -0.5$ mol $L^{-1}$. Rate $= -\frac{1}{2} \times (-0.5 / 100) = 0.5 / 200 = 0.0025 = 2.5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Step 3: Conclusion
The average rate is $2.5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Final Answer: (B)
Reaction rate = $-\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t}$.
Step 2: Analysis
$\Delta [N_2O_5] = 1.5 - 2.0 = -0.5$ mol $L^{-1}$. Rate $= -\frac{1}{2} \times (-0.5 / 100) = 0.5 / 200 = 0.0025 = 2.5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Step 3: Conclusion
The average rate is $2.5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Final Answer: (B)
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