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:

Question

For a reaction $ A + B \to \text{products} $, the rate of the reaction was doubled when the concentration of $ A $ was doubled. When the concentrations of $ A $ and $ B $ were doubled, the rate was again doubled. The order of the reaction with respect to $ A $ and $ B $ are:

Answer 1

Step 1: Know the difference between rate of reaction and rate of change of A.
For $2A \rightarrow$ product, two molecules of A are used up for the reaction to happen once. So the rate of reaction is the rate at which A falls, divided by 2.

Step 2: Write the formula.
\[ \text{Rate} = -\frac{1}{2}\frac{\Delta[A]}{\Delta t} \]

Step 3: Find how much A dropped.
A went from 0.5 M down to 0.4 M, a fall of $0.1$ M over 10 minutes.

Step 4: Put the values in.
Rate $= \dfrac{1}{2} \times \dfrac{0.1}{10} = \dfrac{0.1}{20} = 0.005$ mol $L^{-1}$ min$^{-1}$.

Step 5: Write in scientific form.
$0.005 = 5 \times 10^{-3}$ mol $L^{-1}$ min$^{-1}$.

Step 6: Conclusion.
So the answer is $5 \times 10^{-3}$ mol $L^{-1}$ min$^{-1}$. \[ \boxed{5 \times 10^{-3}} \]

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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The Correct Option is C

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