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Magazine Chapter 18: Problem 36
An organic compound undergoes first-order decomposition. The time taken for its decomposition to \(1 / 8\) and \(1 / 10\) of its initial concentration are \(t_{1 / 8}\) and \(t_{1 / 10}\) respectively. What is the value of \(\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10\) ? \(\left(\log _{10} 2=0.3\right)\)
Short Answer
Expert verified
The value is 9.
Step by step solution
01
Understanding First-Order Kinetics
The rate of a first-order reaction is directly proportional to the concentration of the reactant. The formula for first-order decay is \[k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right)\] where \(k\) is the rate constant, \(t\) is time, \([A]_0\) is the initial concentration, and \([A]\) is the concentration at time \(t\).
02
Apply for Concentration to 1/8
If the compound decomposes to \(\frac{1}{8}\) of its initial concentration, the formula becomes \[k = \frac{1}{t_{1/8}} \ln \left( 8 \right)\]. Since \(8 = 2^3\), the expression simplifies to \[k = \frac{1}{t_{1/8}} \cdot 3 \cdot \ln(2)\].
03
Apply for Concentration to 1/10
Similarly, for the compound to decompose to \(\frac{1}{10}\) of its initial concentration, the formula becomes \[k = \frac{1}{t_{1/10}} \ln \left( 10 \right)\].
04
Equating the Rate Constants
Since both expressions represent the same rate constant \(k\), equate the two: \[\frac{3 \cdot \ln(2)}{t_{1/8}} = \frac{\ln(10)}{t_{1/10}}\].
05
Solve for the Ratio
From the equation \(\frac{3 \cdot \ln(2)}{t_{1/8}} = \frac{\ln(10)}{t_{1/10}}\), rearrange to find the ratio \(\frac{t_{1/8}}{t_{1/10}}\): \[\frac{t_{1/8}}{t_{1/10}} = \frac{3 \cdot \ln(2)}{\ln(10)}\].
06
Calculate logs and the Final Answer
Given \(\log_{10} 2 = 0.3\) and knowing \(\log_{10} 10 = 1\), substitute to find \(\ln(2) = 0.3 \cdot \ln(10) = 0.3\ln(10)\) and thus \(\ln(10) = 1\). So, \[\frac{t_{1/8}}{t_{1/10}} = 3 \cdot 0.3\]. Calculating this gives \(0.9\). Now, \[\left( \frac{t_{1/8}}{t_{1/10}} \right) \times 10 = 0.9 \times 10 = 9\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rate Constant
In first-order reactions, the rate constant, denoted by the symbol \( k \), is an intrinsic value that describes the speed of a chemical reaction. It's crucial for determining how rapidly a reaction proceeds over time. The key thing to remember about the rate constant is that it remains constant for a given reaction under set conditions, such as temperature and pressure, regardless of the concentrations of reactants.
If you're working on problems involving first-order kinetics, you'll often encounter the formula:\[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]
This formula is useful for finding \( k \) when initial and current concentrations are known, allowing you to quantify the decay rate of the substance undergoing decomposition. Remember, the natural logarithm \( \ln \) is a key component in this equation, used to simplify calculations of non-linear decays; reflect this in your calculations when solving similar problems.
If you're working on problems involving first-order kinetics, you'll often encounter the formula:\[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]
- Where \( k \) is the rate constant,
- \( t \) represents time,
- \([A]_0\) is the starting concentration,
- and \([A]\) is the concentration at time \( t \).
This formula is useful for finding \( k \) when initial and current concentrations are known, allowing you to quantify the decay rate of the substance undergoing decomposition. Remember, the natural logarithm \( \ln \) is a key component in this equation, used to simplify calculations of non-linear decays; reflect this in your calculations when solving similar problems.
Reaction Rate
The reaction rate in a first-order kinetics scenario is directly proportional to the concentration of the reactant. Essentially, this means that as the concentration decreases, the rate of the reaction also diminuishes. This characteristic is distinctive in first-order reactions, allowing them to be mathematically modeled effectively. For example, consider a reaction where the concentration of a compound reduces over time; the rate can be signified by the simple equation:
\[ ext{Rate} = k[A] \]
By using this relationship, you can predict how fast a reaction is occurring at any given moment. For reactions that follow first-order kinetics, the initial concentration \([A]_0\) will decay at a rate influenced by \( k \), emphasizing its importance as a predictive tool in chemistry.
\[ ext{Rate} = k[A] \]
- Where \( k \) is the rate constant determined from experiments or calculations,
- and \([A]\) is the concentration of the reactant at a given time.
By using this relationship, you can predict how fast a reaction is occurring at any given moment. For reactions that follow first-order kinetics, the initial concentration \([A]_0\) will decay at a rate influenced by \( k \), emphasizing its importance as a predictive tool in chemistry.
Concentration Decay
Concentration decay refers to the decrease in the concentration of a reactant in a chemical reaction over time. In first-order reactions, this decay is exponential, meaning that the rate at which the concentration diminishes is proportional to the current amount present. This kind of decay pattern is intuitive鈥攍ess reactant leads to fewer collisions possible, slowing down the reaction.
The mathematical expression for first-order concentration decay can be written as:
\[ [A] = [A]_0 \, e^{-kt} \]
As the compound decomposes, its concentration decreases in a predictable manner that can be easily plotted on a graph as an exponentially decaying curve. An excellent example from the original exercise explains the decomposition to specific fractions (such as \(1/8\) or \(1/10\) of initial concentration), which showcases how concentration decay can be quantitively linked to time and the rate constant \( k \). Understanding this helps in predicting how long a reaction will take to reach a certain stage of completion.
The mathematical expression for first-order concentration decay can be written as:
\[ [A] = [A]_0 \, e^{-kt} \]
- Here, \([A]\) is the concentration at time \( t \),
- \([A]_0\) is the initial concentration,
- \( k \) is the rate constant,
- and \( e \) is the base of the natural logarithm.
As the compound decomposes, its concentration decreases in a predictable manner that can be easily plotted on a graph as an exponentially decaying curve. An excellent example from the original exercise explains the decomposition to specific fractions (such as \(1/8\) or \(1/10\) of initial concentration), which showcases how concentration decay can be quantitively linked to time and the rate constant \( k \). Understanding this helps in predicting how long a reaction will take to reach a certain stage of completion.
