Question

For a chemical reaction, half-life period \(t_{1/2}\) is 10 minutes. How much reactant will be left after 20 minutes if one starts with 100 moles of reactant and the order of the reaction is:

• A. zero, (ii) one, and (iii) two?
• B. 0, 25, 33.33
• C. 25, 0, 33.33
• D. 33.33, 25, 0

Hint:

Tip for reaction order problems In reaction order problems, remember the different rate laws for zero, first, and second-order reactions. Zero-order reactions have linear decay, first-order reactions decay exponentially, and second-order reactions decay with the square of time. Use the appropriate formula for each case.

Answer 1

The Correct Option is A

Given: - \( t_{1/2} = 10 \) minutes (half-life) - Initial reactant: 100 moles - Reaction time: 20 minutes Calculate the remaining reactant after 20 minutes for different reaction orders. (i) Zero Order: Rate is constant; concentration decreases linearly. Equation: \[ [A] = [A]_0 - kt \] Where: - \( [A]_0 \) is initial concentration (100 moles) - \( k \) is the rate constant - \( t \) is time From \( t_{1/2} = 10 \) minutes, \( k \) is: \[ k = \frac{[A]_0}{t_{1/2}} = \frac{100}{10} = 10 \, \text{moles/min} \] After 20 minutes: \[ [A] = 100 - (10 \times 20) = 100 - 200 = 0 \]
For zero-order, 0 moles remain. (ii) First Order: Concentration decreases exponentially. Equation: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] Where \( [A]_0 \) is initial concentration and \( k \) is the rate constant. Half-life is: \[ t_{1/2} = \frac{0.693}{k} \] Given \( t_{1/2} = 10 \) minutes, solve for \( k \): \[ k = \frac{0.693}{10} = 0.0693 \, \text{min}^{-1} \] Remaining reactant after 20 minutes: \[ \ln \left( \frac{100}{[A]} \right) = 0.0693 \times 20 \] \[ \ln \left( \frac{100}{[A]} \right) = 1.386 \] \[ \frac{100}{[A]} = e^{1.386} \approx 4.0 \] \[ [A] \approx \frac{100}{4.0} = 25 \, \text{moles} \]
For first-order, 25 moles remain. (iii) Second Order: Concentration decreases with the square of time. Equation: \[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \] Half-life is: \[ t_{1/2} = \frac{1}{k[A]_0} \] Given \( t_{1/2} = 10 \) minutes, solve for \( k \): \[ k = \frac{1}{10 \times 100} = 0.001 \, \text{min}^{-1} \] Remaining reactant after 20 minutes: \[ \frac{1}{[A]} - \frac{1}{100} = 0.001 \times 20 \] \[ \frac{1}{[A]} = \frac{1}{100} + 0.02 = 0.03 \] \[ [A] = \frac{1}{0.03} = 33.33 \, \text{moles} \]
For second-order, 33.33 moles remain.