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Q3.19:
A first order reaction takes 40 min for 30% decomposition. Calculate (t_{1/2}).

Solution :

Step 1: Establishing the First-Order Kinetic Equation

For a first-order reaction, the relationship between the rate constant ($k$), initial concentration ($[R]_0$), and concentration at time $t$ ($[R]_t$) is given by the integrated rate law:

$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$ [Per the First-Order Integrated Rate Law]

Step 2: Defining Given Variables and Concentration Values

  • Time elapsed ($t$) = $40 \text{ min}$
  • Percentage decomposition = $30\%$
  • Initial concentration ($[R]_0$) = $100$
  • Remaining concentration ($[R]_t$) = $100 - 30 = 70$

Step 3: Calculation of the Rate Constant ($k$)

Substituting the values into the integrated rate law:

$k = \frac{2.303}{40} \log \frac{100}{70}$

$k = \frac{2.303}{40} (\log 10 - \log 7)$

Given $\log 10 = 1$ and $\log 7 \approx 0.8451$:

$k = \frac{2.303}{40} (1 - 0.8451)$

$k = \frac{2.303 \times 0.1549}{40}$

$k \approx \frac{0.3567}{40} \approx 0.008918 \text{ min}^{-1}$

Step 4: Calculation of Half-Life ($t_{1/2}$)

For a first-order reaction, the half-life period is independent of the initial concentration and is derived as follows:

$t_{1/2} = \frac{0.693}{k}$ [Per the derived half-life formula for first-order kinetics]

$t_{1/2} = \frac{0.693}{0.008918}$

$t_{1/2} \approx 77.7 \text{ min}$

Visual Representation of Kinetic Decay

[R] t t=40min ([R]=70)

Final Solution: The half-life ($t_{1/2}$) of the reaction is approximately 77.7 minutes.


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