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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 :

Step 1: Establishing the Rate Law for First-Order Reactions

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

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

[Per the Integrated Rate Law for First-Order Kinetics]

Step 2: Identifying Given Variables

  • Rate constant ($k$) = $60 \, s^{-1}$
  • Initial concentration = $[R]_0$
  • Final concentration ($[R]_t$) = $\frac{1}{16} [R]_0$

Step 3: Calculating the Time Required ($t$)

Substitute the given values into the integrated rate equation:

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

[By algebraic substitution]

Simplify the concentration ratio:

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

Express $16$ as a power of $2$ to utilize log properties:

$16 = 2^4$

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

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

$\log(16) \approx 4 \times 0.3010 = 1.204$

Step 4: Final Algebraic Manipulation

$t = \frac{2.303 \times 1.204}{60}$

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

$t \approx 0.0462 \, s$

Graphical Representation of First-Order Decay

Time (s) [R] [R]â‚€ [R]â‚€/16

Final Solution: The time required to reduce the initial concentration to its $\frac{1}{16}$th value is $4.62 \times 10^{-2} \, s$.


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