Question

For the following reaction;
\(\begin{array}{l}2X + Y\overset{k}{\rightarrow}P\end{array}\)

The rate of reaction is
\(\begin{array}{l}\frac{d[P]}{dt} = k[X]\end{array}\)
Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are). (Use: ln 2 = 0.693)

• A. The rate constant, k, of the reaction is 13.86 × 10^-4 s^-1.
• B. Half-life of X is 50 s.
• C. At 50 s, – d[X] / dt = 13.86 × 10^-3 mol L^-1 s^-1.
• D. At 100 s, d[Y] / dt = 3.46 × 10^-3 mol L^-1 s^-1.

Answer 1

The Correct Option is B, C, D

Given reaction: \( 2X + Y \xrightarrow{k} P \)

Rate law: \( \dfrac{d[P]}{dt} = k[X] \)

This shows a pseudo first-order reaction with respect to X.

Initial concentrations:

• \([X]_0 = \dfrac{2\, \text{mol}}{1.0\, \text{L}} = 2\, \text{mol/L}\)
• \([Y]_0 = \dfrac{1\, \text{mol}}{1.0\, \text{L}} = 1\, \text{mol/L}\)

Let at time t = 50 s, the concentration of Y is 0.5 mol/L, so 0.5 mol of Y is consumed.

From the stoichiometry, 2 mol of X reacts with 1 mol of Y. Therefore, X consumed = \(2 \times 0.5 = 1\) mol ⇒ final \([X] = 2 - 1 = 1 \, \text{mol/L}\)

---

Checking Option B:

From rate law: \( \text{Rate} = k[X] \), a first-order reaction in X.

For a first-order reaction:

\[ [X] = [X]_0 e^{-kt} \]

We know: \( [X]_0 = 2, [X] = 1 \) at \( t = 50\, \text{s} \), so: \[ 1 = 2e^{-k \cdot 50} \Rightarrow \frac{1}{2} = e^{-50k} \Rightarrow \ln \frac{1}{2} = -50k \Rightarrow -\ln 2 = -50k \Rightarrow k = \frac{\ln 2}{50} = \frac{0.693}{50} \]

Therefore, half-life \( t_{1/2} = \frac{\ln 2}{k} = 50\, \text{s} \)

✓ Correct: Option B

---

Checking Option C:

At 50 s, \([X] = 1 \, \text{mol/L}\), and from earlier: \[ k = \frac{0.693}{50} = 0.01386\, \text{s}^{-1} \] So, rate: \[ -\frac{d[X]}{dt} = k[X] = 0.01386 \times 1 = 0.01386 \, \text{mol L}^{-1} \text{s}^{-1} \]

✓ Correct: Option C

---

Checking Option D:

At 100 s, since half-life is 50 s, the concentration of X would halve again:

\([X] = 1 \rightarrow 0.5\, \text{mol/L}\)

Rate at 100 s: \[ \text{Rate} = k[X] = 0.01386 \times 0.5 = 0.00693 \, \text{mol L}^{-1} \text{s}^{-1} \]

Now, from stoichiometry: \[ \text{Rate of disappearance of Y} = \frac{1}{2} \times \text{Rate of disappearance of X} = \frac{1}{2} \times 0.00693 = 0.003465 \approx 3.46 \times 10^{-3} \]

✓ Correct: Option D

---

Correct Answers: Options B, C, and D