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

Since the reaction is first order, the rate depends directly on the concentration of reactant \(A\): \[ \text{Rate} = k[A]. \] Step 1: Analyze Run 1.
In Run 1, the concentration of \(A\) is \(10\ \text{M}\). So, \[ \text{Rate}_1 \propto 10. \] Step 2: Analyze Run 2.
In Run 2, although the volume is doubled, the concentration remains \(10\ \text{M}\). However, the total number of moles of \(A\) is doubled, leading to a higher overall reaction rate. Thus, \[ \text{Rate}_2>\text{Rate}_1. \] Step 3: Analyze Run 3.
In Run 3, dilution occurs: \[ \text{New concentration} = \frac{100 \times 10}{200} = 5\ \text{M}. \] Hence, \[ \text{Rate}_3 \propto 5. \] Step 4: Compare all rates.
\[ \text{Rate}_3<\text{Rate}_1<\text{Rate}_2. \] Final Answer: \[ \boxed{\text{Run 3 }<\ \text{Run 1 }<\ \text{Run 2}} \]