MP Board · Class 12 · Chemistry · Chemical KineticsThe thermal decomposition of a compound is a first-order reaction. If 50% of a sample of the compound decomposes in 120 minutes, calculate the rate constant ($k$) for the reaction. Also, calculate the time required for 90% completion of the reaction. (Given: $\log 2 = 0.3010$, $\log 10 = 1$)
Step-by-Step Numerical Solution:
Given Data:
- Type of reaction = First-order reaction
- Half-life ($t_{1/2}$) = 120 minutes
Part 1: Calculation of Rate Constant ($k$)\nFor a first-order reaction, the half-life period is related to the rate constant by the formula: $$t_{1/2} = \frac{0.693}{k}$| \nRearranging the formula to solve for $k$: $$k = \frac{0.693}{t_{1/2}}$$ \nSubstituting the given value of $t_{1/2}$ (120 minutes): $$k = \frac{0.693}{120 \text{ min}}$$ $$k = 0.005775 \text{ min}^{-1} \text{ or } 5.78 \times 10^{-3} \text{ min}^{-1}$$
Part 2: Calculation of Time Required for 90% Completion ($t_{90%}$)\nFor a first-order reaction, the integrated rate equation is: $$t = \frac{2.303}{k} \log \left( \frac{[R]0}{[R]} \right)$$ \nLet the initial concentration $[R]0 = 100$.\nSince 90% of the reaction is complete, the remaining concentration $[R]$ at time $t$ will be: $$[R] = 100 - 90 = 10$$ \nSubstituting the values of $[R]0$, $[R]$, and $k$ into the integrated rate equation: $$t{90%} = \frac{2.303}{0.005775} \log \left( \frac{100}{10} \right)$$ $$t{90%} = \frac{2.303}{0.005775} \log(10)$$ \nSince $\log(10) = 1$: $$t{90%} = \frac{2.303}{0.005775} \times 1$$ $$t_{90%} = 398.78 \text{ minutes}$|
Answer:
- Rate constant ($k$) = $5.78 \times 10^{-3} \text{ min}^{-1}$
- Time for 90% completion = 398.78 minutes