Question

\(A \rightarrow \text{Product}\) (First order reaction). Three sets of experiments were performed for a reaction under similar experimental conditions: \[ \text{Run 1} \Rightarrow 100\ \text{mL of } 10\ \text{M solution of reactant } A \] \[ \text{Run 2} \Rightarrow 200\ \text{mL of } 10\ \text{M solution of reactant } A \] \[ \text{Run 3} \Rightarrow 100\ \text{mL of } 10\ \text{M solution of reactant } A + 100\ \text{mL of } H_2O \] The correct variation of rate of reaction is

• A. Run 3 \(<\) Run 1 \(=\) Run 2
• B. Run 1 \(=\) Run 2 \(=\) Run 3
• C. Run 1 \(<\) Run 2 \(<\) Run 3
• D. Run 3 \(<\) Run 1 \(<\) Run 2

Hint:

For first order reactions, rate depends only on concentration, not on volume directly.

Answer 1

The Correct Option is D

The given problem involves a first-order reaction where reactant \(A\) converts to a product. We have three different experimental setups to analyze:

1. Run 1: 100 mL of 10 M solution of reactant \(A\).
2. Run 2: 200 mL of 10 M solution of reactant \(A\).
3. Run 3: 100 mL of 10 M solution of reactant \(A\) + 100 mL of water.

Since this is a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The rate law expression for a first-order reaction can be written as:

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

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

Let's analyze each run in terms of concentration and volume:

• Run 1: Concentration = 10 M. Rate of reaction depends on \([A] = 10 \, \text{M}\).
• Run 2: Since the amount of reactant is doubled but the concentration remains the same, the concentration is still 10 M. However, the total amount of reactant is higher.
• Run 3: The solution is diluted with 100 mL of water. Therefore, the new concentration is:

\(\text{New concentration} = \frac{10 \times 100}{200} = 5 \, \text{M}\)

Based on the calculated concentrations, the rate of reactions can be compared as:

• Run 1: Rate is proportional to 10 M.
• Run 2: Rate is also proportional to 10 M, but the total amount of reactant is higher, hence higher rate.
• Run 3: Rate is proportional to 5 M due to dilution.

Thus, the rate of reaction should vary in the order:

Run 3 < Run 1 < Run 2

Hence, the correct option is: Run 3 < Run 1 < Run 2.