Question

Consider a first order reaction: \[ A \rightarrow \text{products} \] 3 different solutions are taken and the rate of reaction of: Solution 1: 100 mL 10M \( A \) \( \rightarrow r_1 \) Solution 2: 200 mL 10M \( A \) \( \rightarrow r_2 \) Solution 3: 100 mL 10M \( A \) + 100 mL water \( \rightarrow r_3 \) The correct order of the rates of reactions is,

• A. \( r_1 = r_2 = r_3 \)
• B. \( r_1 = r_2<r_3 \)
• C. \( r_1 = r_2>r_3 \)
• D. \( r_1<r_2 = r_3 \)

Hint:

For a first-order reaction, the rate depends linearly on the concentration of the reactant.

Answer 1

The Correct Option is C

To determine the correct order of the rates of the first-order reactions, we need to consider the formula for the rate of a first-order reaction:

\[\text{Rate} = k[A]\]

where \( k \) is the rate constant and \([A]\) is the concentration of reactant \( A \).

1. Solution 1: Consists of 100 mL of 10 M \( A \). The concentration of \( A \) remains 10 M as there is no dilution. Therefore, the rate \( r_1 \) is proportional to 10 M. 

\[r_1 = k \times 10\]

1. Solution 2: Consists of 200 mL of 10 M \( A \). The concentration of \( A \) is still 10 M because the concentration doesn't change with volume as long as no additional solvent is added. Therefore, the rate \( r_2 \) is also proportional to 10 M. 

\[r_2 = k \times 10\]

1. Solution 3: Consists of 100 mL of 10 M \( A \) and 100 mL of water, effectively diluting the solution to 200 mL. The concentration of \( A \) becomes 5 M because the volume doubled while the amount of substance stays the same (\(\frac{100}{200} \times 10 = 5\; \text{M}\)). Therefore, the rate \( r_3 \) is proportional to 5 M. 

\[r_3 = k \times 5\]

From the above reasoning, we have:

• \( r_1 = k \times 10 \)
• \( r_2 = k \times 10 \)
• \( r_3 = k \times 5 \)

Thus, the correct order of the rates is:

\[r_1 = r_2 > r_3\]

Therefore, the correct answer is \((r_1 = r_2 > r_3)\).