A$$\rightarrow$$ product (First order reaction).
Three sets of experiment were performed for a reaction under similar experimental conditions:

Run 1 $$\Rightarrow$$ 100 mL of 10 M solution of reactant A

Run 2 $$\Rightarrow$$ 200 mL of 10 M solution of reactant A

Run 3 $$\Rightarrow$$ 100 mL of 10 M solution of reactant A + 100 mL of $$H_{2}O$$ added.

The correct variation of rate of reaction is

The reaction is first order: A → products. The rate law is given by:

$$\text{rate} = k [A]$$

where $$k$$ is the rate constant and $$[A]$$ is the concentration of A. The rate depends only on the concentration of A.

Now, analyze each run:

Run 1: 100 mL of 10 M solution of reactant A.

Volume = 100 mL = 0.1 L

Moles of A = concentration × volume = 10 mol/L × 0.1 L = 1 mol

Concentration $$[A]_1 = \frac{\text{moles}}{\text{volume}} = \frac{1 \text{ mol}}{0.1 \text{ L}} = 10 \text{ M}$$

Rate for Run 1: $$\text{rate}_1 = k \times 10$$

Run 2: 200 mL of 10 M solution of reactant A.

Volume = 200 mL = 0.2 L

Moles of A = 10 mol/L × 0.2 L = 2 mol

Concentration $$[A]_2 = \frac{2 \text{ mol}}{0.2 \text{ L}} = 10 \text{ M}$$

Rate for Run 2: $$\text{rate}_2 = k \times 10$$

Run 3: 100 mL of 10 M solution of reactant A + 100 mL of H₂O added.

Moles of A = 10 mol/L × 0.1 L = 1 mol (from 100 mL solution)

Total volume = 100 mL + 100 mL = 200 mL = 0.2 L

Concentration $$[A]_3 = \frac{1 \text{ mol}}{0.2 \text{ L}} = 5 \text{ M}$$

Rate for Run 3: $$\text{rate}_3 = k \times 5$$

Comparing the rates:

$$\text{rate}_1 = 10k$$, $$\text{rate}_2 = 10k$$, $$\text{rate}_3 = 5k$$

Thus, $$\text{rate}_3 < \text{rate}_1 = \text{rate}_2$$

Evaluating the options:

A. Run 1 = Run 2 = Run 3 → Incorrect, because rates are not equal.

B. Run 1 < Run 2 < Run 3 → Incorrect, because Run 3 has the smallest rate.

C. Run 3 < Run 1 < Run 2 → Incorrect, because Run 1 and Run 2 have equal rates.

D. Run 3 < Run 1 = Run 2 → Correct, as $$\text{rate}_3 < \text{rate}_1 = \text{rate}_2$$.

The correct answer is option D.