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Q3.16:
The rate constant for a first order reaction is (60 s^{-1}). How much time will it take to reduce the initial concentration of the reactant to its (\frac{1}{16}^{th}) value?

Solution :

Given Variables & Initial Setup

For a first-order chemical reaction, the rate constant ($k$) and the concentration of the reactant at any given time ($t$) are governed by the integrated rate law:

$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$

Where:

  • $k = 60 \, \text{s}^{-1}$ [Given rate constant]
  • $[R]_0 = \text{Initial concentration}$
  • $[R]_t = \frac{1}{16} [R]_0$ [Target concentration]
  • $t = \text{Time required (to be determined)}$

Step 1: Substitution of Values into the Rate Equation

Substitute the ratio of concentrations into the integrated rate law equation [Per the First-Order Kinetics derivation]:

$60 = \frac{2.303}{t} \log \frac{[R]_0}{\frac{1}{16}[R]_0}$

$60 = \frac{2.303}{t} \log(16)$

Step 2: Logarithmic Calculation

To evaluate $\log(16)$, we express the value as a power of 2 [By the property of logarithms $\log(a^n) = n \log a$]:

$16 = 2^4$

$\log(16) = \log(2^4) = 4 \times \log(2)$

Given $\log(2) \approx 0.3010$:

$\log(16) = 4 \times 0.3010 = 1.2040$

Step 3: Algebraic Solving for Time ($t$)

Rearrange the equation to isolate $t$:

$t = \frac{2.303 \times 1.2040}{60}$

$t = \frac{2.7728}{60}$

$t = 0.04621 \, \text{s}$

Visual Representation of Concentration Decay

[R] t [R]â‚€ [R]â‚€/16

Final Verification

The calculation is based on the standard exponential decay model $N(t) = N_0 e^{-kt}$. Using the conversion $t = \frac{\ln(16)}{k}$, we confirm the consistency of the logarithmic base conversion [Logarithmic Identity: $\log_{10}(x) = \frac{\ln(x)}{2.303}$].

Final Solution: The time taken for the reaction to reach $\frac{1}{16}$ of its initial concentration is approximately $4.62 \times 10^{-2} \, \text{s}$.


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