In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_1$ and $t_2$ (s), respectively. The ratio $t_1/t_2$ will be:
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Easy Solution
1
Understand First Order Kinetics
For a first order reaction, the amount remaining after $n$ half-lives is $[A] = [A]_0 \cdot (1/2)^n$. The total time $t = n \times t_{1/2}$.
2
Analyze t1 (1/4th concentration)
To reach $1/4$ of initial concentration: $[A] = [A]_0 / 4 = [A]_0 \cdot (1/2)^2$. This takes exactly 2 half-lives.
$$t_1 = 2 \cdot t_{1/2}$$
3
Analyze t2 (1/8th concentration)
To reach $1/8$ of initial concentration: $[A] = [A]_0 / 8 = [A]_0 \cdot (1/2)^3$. This takes exactly 3 half-lives.
$$t_2 = 3 \cdot t_{1/2}$$
4
Calculate the Ratio
Divide $t_1$ by $t_2$ to find the required ratio.
$$\frac{t_1}{t_2} = \frac{2 \cdot t_{1/2}}{3 \cdot t_{1/2}}$$
5
Simplify the Expression
The half-life terms cancel out as they are constant for a specific first order reaction.
$$\frac{t_1}{t_2} = \frac{2}{3}$$
6
Mathematical Confirmation
Using $t = \frac{1}{k} \ln \frac{[A]_0}{[A]}$: $t_1 = \frac{\ln(4)}{k} = \frac{2\ln2}{k}$ and $t_2 = \frac{\ln(8)}{k} = \frac{3\ln2}{k}$. Ratio = $2/3$.
7
Final Answer Match
The ratio $t_1/t_2$ is $2/3$, which corresponds to Option D in the provided question image.
Final Answer: 2/3
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Theory
1. First Order Rate Law
A first-order reaction is a chemical reaction in which the rate of the reaction is directly proportional to the concentration of only one reactant. The differential rate law is expressed as $Rate = k[A]$. Upon integration, we obtain the expression $k = (2.303/t) \log([A]_0/[A])$. This indicates that the time required for a specific percentage of completion is independent of the initial concentration. This property distinguishes first-order reactions from zero-order or second-order reactions. In such systems, a plot of $\ln[A]$ versus time yields a straight line with a slope equal to $-k$.
2. Concept of Half-life (t1/2)
The half-life of a reaction is the time required for the reactant concentration to decrease to half of its initial value. For a first-order reaction, the half-life is given by the formula $t_{1/2} = 0.693/k$. Remarkably, this value remains constant throughout the reaction regardless of how much reactant is present. This means it takes the same amount of time for the concentration to drop from 100% to 50% as it does to drop from 50% to 25%. This consistency allows us to treat total time as an integer multiple of half-lives for simple fractions like 1/4, 1/8, or 1/16.
3. Logarithmic Relationships in Kinetics
Chemical kinetics frequently utilizes logarithmic math because of the exponential nature of decay in first-order processes. When comparing two time intervals $t_1$ and $t_2$ for different levels of completion, we can use the ratio: $t_1/t_2 = \log([A]_0/[A]_1) / \log([A]_0/[A]_2)$. If the final concentrations are powers of the same base (like $1/2^2$ and $1/2^3$), the ratio simplifies to the ratio of the exponents. This shortcut is extremely effective for competitive exams like JEE Main where time management is crucial. It bypasses the need to calculate the actual rate constant $k$.
4. Fractional Life of Reactions
Fractional life refers to the time taken for a reactant to reduce to a fraction ‘1/f’ of its initial concentration. In first-order kinetics, $t_{1/f} = (1/k) \ln(f)$. Common examples include $t_{75\%}$ (where $f=4$) and $t_{87.5\%}$ (where $f=8$). Understanding these relationships helps in predicting the behavior of various chemical and physical processes, such as drug metabolism in the body or radioactive dating in geology. The ratio of any two fractional lives is a constant numerical value characteristic of the order of the reaction.
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FAQs
1
Why is the ratio constant?
For first-order reactions, the rate constant ‘k’ and half-life are constants. Thus, the time for any fractional change is always a constant multiple of half-life.
2
What if it was a zero order reaction?
In zero order, $t \propto ([A]_0 – [A])$. Time to 1/4 (3/4th completion) would be $0.75[A]_0/k$ and to 1/8 would be $0.875[A]_0/k$. The ratio would be $0.75/0.875 = 6/7$.
3
How to handle non-integer half-lives?
Use the formula $t = (1/k) \ln([A]_0/[A])$. For example, if it was to 1/5th concentration, $t = \ln(5)/k$.
4
What does ‘decomposition to one fourth’ mean?
It means the final concentration $[A]$ is $1/4$ of $[A]_0$. It is NOT 1/4th decomposed (which would mean 3/4 remaining).
5
Is t1/t2 same as t25%/t12.5%?
Yes, if the percentages refer to the remaining concentration. If they refer to decomposition, the ratios change.
6
Can I use $t = 1/k \ln(2)$ directly?
That is the formula for $t_{1/2}$. For other times, use $\ln(\text{initial}/\text{final})$.
7
Does temperature affect this ratio?
No, temperature changes ‘k’, but ‘k’ cancels out in the ratio $t_1/t_2$. The ratio remains constant.
8
What is the significance of the graph of log[A] vs t?
It’s a straight line. The ratio of distances on the time axis for two points on the y-axis (log scale) gives the time ratio.
9
Is there any reaction that is 100% completed?
Theoretically, first-order reactions take infinite time to reach 0 concentration ($100\%$ completion).
10
What if I confuse 1/4 with 1/2?
Always remember the power of 2. $4 = 2^2$ (2 half-lives), $8 = 2^3$ (3 half-lives).
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