Question:
In a reaction 2A $\rightarrow$ product, the concentration of A decreases from 0.5 M to 0.4 M in 10 minutes. The rate of the reaction (in mol $L^{-1} min^{-1}$) is:
In a reaction 2A $\rightarrow$ product, the concentration of A decreases from 0.5 M to 0.4 M in 10 minutes. The rate of the reaction (in mol $L^{-1} min^{-1}$) is:
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Don't forget the stoichiometric coefficient (1/2)!
Updated On: Jun 6, 2026
- $5 \times 10^{-1}$
- $5 \times 10^{-2}$
- $5 \times 10^{-3}$
- $1 \times 10^{-2}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Rate of reaction $= -\frac{1}{2} \frac{d[A]}{dt}$.
Step 2: Meaning
Concentration change $= 0.5 - 0.4 = 0.1$ M. Time $= 10$ min.
Step 3: Analysis
Rate $= -\frac{1}{2} \frac{0.1}{10} = \frac{0.1}{20} = 0.005 = 5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Step 4: Conclusion
The rate is $5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Final Answer: (C)
Rate of reaction $= -\frac{1}{2} \frac{d[A]}{dt}$.
Step 2: Meaning
Concentration change $= 0.5 - 0.4 = 0.1$ M. Time $= 10$ min.
Step 3: Analysis
Rate $= -\frac{1}{2} \frac{0.1}{10} = \frac{0.1}{20} = 0.005 = 5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Step 4: Conclusion
The rate is $5 \times 10^{-3}$ mol $L^{-1} min^{-1}$.
Final Answer: (C)
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