Chapter 10: Problem 184
Match the following (a) \(\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}\) (p) Zero order (b) \(\mathrm{t}_{1 / 2}=\frac{\mathrm{a}}{2 \mathrm{k}}\) (q) First order (c) \(\mathrm{t}=\frac{1}{\mathrm{k}}\) (r) Average life (d) \(\mathrm{t}_{\frac{1}{2}}=\frac{1}{\mathrm{ak}}\) (s) Second order (t) Nuclear disintegration
Short Answer
Step by step solution
Understand the Terms
Match Equation (a)
Match Equation (b)
Match Equation (c)
Match Equation (d)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Life Equations
In a first-order reaction, the half-life (\( t_{1/2} \)) is independent of the initial concentration of the reactants. This results in a consistent half-life, regardless of how much reactant is present. The formula is given by:\[ t_{1/2} = \frac{0.693}{k} \]Where \( k \) is the rate constant.
- First-order reactions: Half-life does not change with concentration.
First-Order Reactions
The rate law for a first-order reaction is:\[ \text{Rate} = k[A] \]Here, \( [A] \) represents the concentration of the reactant A, and \( k \) is the rate constant. Manipulating this rate law gives the integrated form, which is a logarithmic equation that links reactant concentration over time:\[ \ln[A] = \ln[A]_0 - kt \]This demonstrates how the natural logarithm of the concentration decreases linearly as time goes on.
- Linear decrease: A straight line is seen when plotting \( \ln[A] \) against time.
- Constant half-life: Unique as the half-life stays the same no matter how much reactant you started with.
Zero-Order Reactions
The rate law for zero-order reactions is:\[ \text{Rate} = k \]Since the rate is independent of concentration, it suggests that the reaction progresses at a steady rate. Integrating the rate law gives the concentration of reactants over time:\[ [A] = [A]_0 - kt \]This indicates a linear decrease of \([A] \) as time progresses. Here are some key characteristics:
- Linear relationship: When you plot concentration versus time, youll get a straight line.
- Dependent half-life: Half-life increases with an increase in initial concentration \( a \).
Second-Order Reactions
The rate law for a simple second-order reaction takes the form:\[ \text{Rate} = k[A]^2 \]If the reaction is of the type \( 2A \rightarrow \) Products or \( A + B \rightarrow \) Products for different reactants contributing equally.The integrated rate law for second-order reactions can be expressed as:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]This suggests that a plot of inverse concentration vs. time will yield a straight line, with the slope equal to the rate constant \( k \).
- Non-linear kinetics: Unlike zero- and first-order, the relationship here is quadratic.
- Inverse half-life: Half-life tends to decrease with higher initial concentratons.