t99.9% with respect to t90% for a first-order reaction is:

Question

Prepp Team · Accepted Answer

Understanding First-Order Reaction Kinetics

Let's analyze the relationship between the time required for 99.9% completion and the time required for 90% completion for a first-order reaction. For a first-order reaction, the rate of the reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction is given by:

\( \ln[A]_t - \ln[A]_0 = -kt \)

Where:

\( [A]_t \) is the concentration of the reactant at time \( t \)
\( [A]_0 \) is the initial concentration of the reactant
\( k \) is the rate constant
\( t \) is the time

This equation can be rearranged to find the time \( t \):

\( t = \frac{1}{k} (\ln[A]_0 - \ln[A]_t) = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right) \)

We can express the amount of reactant remaining after a certain percentage of reaction is completed. If \( x \% \) of the reaction is completed, then the amount of reactant remaining is \( (100-x)\% \) of the initial concentration. So, \( [A]_t = [A]_0 \times \frac{100-x}{100} \).

Substituting this into the time equation:

\( t_{x\%} = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_0 \times \frac{100-x}{100}}\right) = \frac{1}{k} \ln\left(\frac{100}{100-x}\right) \)

Calculating Time for 90% Completion (t90%)

For 90% completion, \( x = 90 \). Using the formula \( t_{x\%} = \frac{1}{k} \ln\left(\frac{100}{100-x}\right) \):

\( t_{90\%} = \frac{1}{k} \ln\left(\frac{100}{100-90}\right) = \frac{1}{k} \ln\left(\frac{100}{10}\right) = \frac{1}{k} \ln(10) \)

Calculating Time for 99.9% Completion (t99.9%)

For 99.9% completion, \( x = 99.9 \). Using the same formula:

\( t_{99.9\%} = \frac{1}{k} \ln\left(\frac{100}{100-99.9}\right) = \frac{1}{k} \ln\left(\frac{100}{0.1}\right) = \frac{1}{k} \ln(1000) \)

Finding the Ratio t99.9% / t90%

Now, we need to find the ratio of \( t_{99.9\%} \) to \( t_{90\%} \):

\( \frac{t_{99.9\%}}{t_{90\%}} = \frac{\frac{1}{k} \ln(1000)}{\frac{1}{k} \ln(10)} \)

The term \( \frac{1}{k} \) cancels out:

\( \frac{t_{99.9\%}}{t_{90\%}} = \frac{\ln(1000)}{\ln(10)} \)

We know that \( 1000 = 10^3 \). Using the logarithm property \( \ln(a^b) = b \ln(a) \):

\( \ln(1000) = \ln(10^3) = 3 \ln(10) \)

Substitute this back into the ratio expression:

\( \frac{t_{99.9\%}}{t_{90\%}} = \frac{3 \ln(10)}{\ln(10)} \)

The term \( \ln(10) \) cancels out:

\( \frac{t_{99.9\%}}{t_{90\%}} = 3 \)

Thus, \( t_{99.9\%} \) is three times \( t_{90\%} \) for a first-order reaction.

Summary of Calculation Steps

Recall the integrated rate law for a first-order reaction.
Derive the general formula for the time taken for \( x \% \) completion: \( t_{x\%} = \frac{1}{k} \ln\left(\frac{100}{100-x}\right) \).
Calculate \( t_{90\%} \) using the formula.
Calculate \( t_{99.9\%} \) using the formula.
Find the ratio \( \frac{t_{99.9\%}}{t_{90\%}} \) and simplify using logarithm properties.

Comparison with Options

Let's compare our calculated ratio with the given options:

Option	Value	Matches Calculation?
1	Two	No
2	One	No
3	Three	Yes
4	Four	No

Our calculated ratio is 3, which matches Option 3.

Revision Table: First-Order Reaction Kinetics Key Times

Time	Percentage Completion (x%)	Formula (\( t_{x\%} \))	Simplified Expression
t90%	90%	\( \frac{1}{k} \ln\left(\frac{100}{10}\right) \)	\( \frac{1}{k} \ln(10) \)
t99%	99%	\( \frac{1}{k} \ln\left(\frac{100}{1}\right) \)	\( \frac{1}{k} \ln(100) = \frac{2}{k} \ln(10) \)
t99.9%	99.9%	\( \frac{1}{k} \ln\left(\frac{100}{0.1}\right) \)	\( \frac{1}{k} \ln(1000) = \frac{3}{k} \ln(10) \)
t\(_{1/2}\) (Half-life)	50%	\( \frac{1}{k} \ln\left(\frac{100}{50}\right) \)	\( \frac{1}{k} \ln(2) \)

From the table, we can clearly see that \( t_{99.9\%} = \frac{3}{k} \ln(10) \) and \( t_{90\%} = \frac{1}{k} \ln(10) \). Their ratio is indeed 3.

Additional Information: First-Order Reaction Concepts

First-order reactions are fundamental in chemical kinetics. Here are some related concepts:

Half-life (\( t_{1/2} \)): For a first-order reaction, the half-life is independent of the initial concentration. \( t_{1/2} = \frac{\ln(2)}{k} \). This means it takes the same amount of time for the concentration to halve, regardless of how much reactant is present.
Rate Constant (\( k \)): The rate constant \( k \) is characteristic of the reaction at a given temperature. Its units for a first-order reaction are typically s\(^{-1}\) or time\(^{-1}\).
Integrated Rate Law: The integrated rate law relates concentration to time, allowing us to calculate concentrations at different times or the time required to reach a certain concentration.
Examples: Radioactive decay is a classic example of a first-order process. Many chemical reactions also follow first-order kinetics under specific conditions.

Understanding the relationship between time and completion percentage for first-order reactions is crucial for solving kinetics problems.