MP Board · Class 12 · Chemistry · Chemical KineticsA first-order reaction has a rate constant of $1.15 \times 10^{-3} \text{ s}^{-1}$. How long will 5 g of this reactant take to reduce to 3 g? Also calculate the half-life period ($t{1/2}$) of the reaction.

Step-by-Step Numerical Solution

Given Data:

• Order of reaction = First order
• Rate constant ($k$) = $1.15 \times 10^{-3} \text{ s}^{-1}$
• Initial concentration ($[A]_0$) = 5 g
• Final concentration ($[A]$) = 3 g

Part 1: Calculation of Time ($t$)

\nThe integrated rate equation for a first-order reaction is: $$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$$ \nSubstitute the given values into the formula: $$t = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3}$| \nCalculate the logarithmic term: $$\log(5/3) = \log(1.6667) = 0.2218$$ \nNow substitute this value back: $$t = \frac{2.303 \times 0.2218}{1.15 \times 10^{-3}}$$ $$t = \frac{0.5108}{1.15 \times 10^{-3}}$$ $$t = 444.17 \text{ seconds}$$

Part 2: Calculation of Half-Life Period ($t_{1/2}$)

\nThe formula for the half-life of a first-order reaction is: $$t_{1/2} = \frac{0.693}{k}$| \nSubstitute the value of $k$: $$t_{1/2} = \frac{0.693}{1.15 \times 10^{-3}}$$ $$t_{1/2} = 602.61 \text{ seconds}$$

Answer:

• Time taken for 5 g to reduce to 3 g is 444.17 seconds.
• Half-life of the reaction is 602.61 seconds.