Question

\(X \rightarrow 2Y\) is a first order reaction where \(1.0 \, \text{mol/L}\) of the reactant yields \(0.4 \, \text{mol/L}\) of \(Y\) in 200 minutes. Calculate the half-life period of the reaction in minutes.

• A. 151.24
• B. 203.69
• C. 620.96
• D. 271.34

Hint:

In reactions like \(X \rightarrow 2Y\), always use stoichiometry first to find how much reactant is consumed before applying the first order rate equation.

Answer 1

The Correct Option is C

Step 1: Understanding the reaction stoichiometry.
The given reaction is:
\[ X \rightarrow 2Y \] This means 1 mole of \(X\) gives 2 moles of \(Y\).

Step 2: Initial concentration of reactant.
Initial concentration of \(X\) is given as:
\[ [X]_0 = 1.0 \, \text{mol/L} \] The concentration of \(Y\) formed after 200 minutes is:
\[ [Y] = 0.4 \, \text{mol/L} \]

Step 3: Calculating concentration of \(X\) consumed.
Since 1 mole of \(X\) gives 2 moles of \(Y\), the amount of \(X\) consumed is half of the amount of \(Y\) formed.
\[ \text{Concentration of } X \text{ consumed} = \frac{0.4}{2} \] \[ = 0.2 \, \text{mol/L} \]

Step 4: Calculating remaining concentration of \(X\).
\[ [X]_t = [X]_0 - \text{concentration consumed} \] \[ [X]_t = 1.0 - 0.2 \] \[ [X]_t = 0.8 \, \text{mol/L} \]

Step 5: Applying first order rate equation.
For a first order reaction:
\[ k = \frac{2.303}{t} \log \frac{[X]_0}{[X]_t} \] Substitute the values:
\[ k = \frac{2.303}{200} \log \frac{1.0}{0.8} \] \[ k = \frac{2.303}{200} \log(1.25) \]

Step 6: Calculating the rate constant.
\[ \log(1.25) = 0.09691 \] \[ k = \frac{2.303 \times 0.09691}{200} \] \[ k = \frac{0.2232}{200} \] \[ k = 0.001116 \, \text{min}^{-1} \]

Step 7: Calculating half-life period.
For a first order reaction:
\[ t_{1/2} = \frac{0.693}{k} \] \[ t_{1/2} = \frac{0.693}{0.001116} \] \[ t_{1/2} = 620.96 \, \text{minutes} \] Final Answer:
\[ \boxed{620.96 \, \text{minutes}} \] Hence, the correct answer is option (C).