Step-by-step Solution:

1. Understand the Stoichiometry and Initial Conditions

The reaction is given as $2X + Y \xrightarrow{k} P$. Initial conditions:

• Moles of X = 2.0 mol
• Moles of Y = 1.0 mol
• Volume of solution = 1.0 L

Therefore, the initial concentrations are:

• $[X]_0 = \frac{2.0 \text{ mol}}{1.0 \text{ L}} = 2.0$ M
• $[Y]_0 = \frac{1.0 \text{ mol}}{1.0 \text{ L}} = 1.0$ M

2. Analyze the Reaction Progress at t = 50 s

At t = 50 s, the moles of Y left are 0.5 mol.

• Concentration of Y at 50 s, $[Y]_{50} = \frac{0.5 \text{ mol}}{1.0 \text{ L}} = 0.5$ M.

The change in concentration of Y is $\Delta[Y] = [Y]_0 - [Y]_{50} = 1.0 - 0.5 = 0.5$ M. According to the stoichiometry ($2X + Y \to P$), for every mole of Y that reacts, 2 moles of X react. Therefore, the change in concentration of X is $\Delta[X] = 2 \times \Delta[Y] = 2 \times 0.5 = 1.0$ M.

The concentration of X at 50 s is: $[X]_{50} = [X]_0 - \Delta[X] = 2.0 - 1.0 = 1.0$ M.

3. Analyze the Rate Law and Evaluate Option B

The rate of reaction is given by the rate law: Rate $= \frac{d[P]}{dt} = k[X]$. From the stoichiometry, the rate can also be expressed in terms of reactants: Rate $= -\frac{1}{2}\frac{d[X]}{dt} = -\frac{d[Y]}{dt} = \frac{d[P]}{dt}$.

Combining these, we get: $-\frac{1}{2}\frac{d[X]}{dt} = k[X]$, which implies $\frac{d[X]}{dt} = -2k[X]$. This shows that the concentration of X decreases following first-order kinetics with an effective rate constant of $2k$.

At $t=0$ s, $[X]_0 = 2.0$ M. At $t=50$ s, $[X]_{50} = 1.0$ M. Since the concentration of X becomes half of its initial value in 50 s, the half-life of X ($t_{1/2,X}$) is 50 s. Therefore, option B is correct.

4. Calculate the Rate Constant, k, and Evaluate Option A

For a first-order decay of X with an effective rate constant of $2k$, the half-life is given by: $t_{1/2, X} = \frac{\ln 2}{2k}$ We know $t_{1/2, X} = 50$ s. $50 = \frac{\ln 2}{2k}$ $k = \frac{\ln 2}{2 \times 50} = \frac{0.693}{100} = 6.93 \times 10^{-3} \text{ s}^{-1}$.

Option A states that $k = 13.86 \times 10^{-4} \text{ s}^{-1} = 1.386 \times 10^{-3} \text{ s}^{-1}$. Since our calculated value $6.93 \times 10^{-3} \text{ s}^{-1}$ does not match the value in option A, option A is incorrect.

5. Evaluate Option C

We need to find the rate of disappearance of X, $-\frac{d[X]}{dt}$, at $t = 50$ s. We have the relation: $-\frac{d[X]}{dt} = 2k[X]$. At $t=50$ s, $[X]_{50} = 1.0$ M. $-\frac{d[X]}{dt} |_{t=50s} = 2 \times (6.93 \times 10^{-3} \text{ s}^{-1}) \times (1.0 \text{ mol L}^{-1})$ $-\frac{d[X]}{dt} |_{t=50s} = 13.86 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$. This matches the value in option C. Therefore, option C is correct.

6. Evaluate Option D

We need to find the rate of disappearance of Y, $-\frac{d[Y]}{dt}$, at $t = 100$ s. From the rate expressions, Rate $= -\frac{d[Y]}{dt} = k[X]$. First, we need to find the concentration of X at $t = 100$ s. Since the half-life of X is 50 s:

• At $t=0$ s, $[X] = 2.0$ M.
• After one half-life ($t=50$ s), $[X] = \frac{2.0}{2} = 1.0$ M.
• After two half-lives ($t=100$ s), $[X] = \frac{1.0}{2} = 0.5$ M. So, $[X]_{100} = 0.5$ M.

Now, calculate the rate: $-\frac{d[Y]}{dt} |_{t=100s} = k [X]_{100}$ $-\frac{d[Y]}{dt} |_{t=100s} = (6.93 \times 10^{-3} \text{ s}^{-1}) \times (0.5 \text{ mol L}^{-1})$ $-\frac{d[Y]}{dt} |_{t=100s} = 3.465 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$. This value, when rounded, matches the value in option D ($3.46 \times 10^{-3}$). Therefore, option D is correct.

Conclusion: The correct statements are B, C, and D.