How does the half-life of a first-order reaction change as the initial concentration of the reactant increases?

The half-life of a first-order reaction remains constant. It is independent of the initial concentration, as shown by the formula $t_{1/2} = 0.693/k$.

When comparing a zero-order and a second-order linear plot, what is the most striking difference in their slopes?

A zero-order plot ($[A]$ vs $t$) has a negative slope ($-k$), while a second-order plot ($1/[A]$ vs $t$) has a positive slope ($+k$).

Which axis transformation is required to identify a first-order reaction graphically?

The natural logarithm of the concentration ($\ln[A]$) must be plotted on the y-axis against time on the x-axis. A straight line confirms first-order kinetics.

What is a common error when calculating the rate constant $k$ from a first-order $\ln[A]$ vs time graph?

Students often report $k$ as a negative value because the slope is negative. However, the rate constant $k$ is by definition a positive value ($k = -slope$).

What mistake is made if a student identifies a reaction as second-order based on a linear $\ln[A]$ vs time plot?

This is a conceptual error; a linear $\ln[A]$ plot identifies a first-order reaction. A second-order reaction would only be linear on a $1/[A]$ vs time plot.

Why is it incorrect to assume a reaction is zero-order just because the concentration decreases over time?

Concentration decreases over time for all reaction orders (except for products). Zero-order is specifically defined by a constant rate of decrease, resulting in a straight line for $[A]$ vs $t$.

Define the 'Integrated Rate Law'.

It is a mathematical equation derived from the rate law that expresses the concentration of a reactant as a function of time, allowing for the calculation of concentration at any specific moment.

What does the y-intercept represent in a $1/[A]$ vs time plot?

The y-intercept represents the reciprocal of the initial concentration, $\frac{1}{[A]_0}$. This is derived from the linear form of the second-order integrated rate law.

What is the relationship between the rate constant $k$ and the slope of a zero-order concentration-time graph?

The slope of the line is equal to $-k$. Therefore, the rate constant is the absolute value of the slope of the $[A]$ vs $t$ graph.

How can you use the half-life of a reaction to distinguish between zero and second order if the initial concentration is doubled?

In a zero-order reaction, doubling $[A]_0$ doubles the half-life. In a second-order reaction, doubling $[A]_0$ halves the half-life.

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Concentration-Time Graphs & Rate Constants

Summary

Graphical analysis of concentration-time data allows for the determination of reaction order and the calculation of the rate constant ( $k$ ). By transforming concentration data into linear forms—using raw concentration, natural logarithms, or reciprocals—chemists can identify whether a reaction follows zero, first, or second-order kinetics based on which plot yields a straight line.

1. Definition & Core Concepts

Integrated Rate Laws: These are mathematical expressions that relate the concentration of reactants to the elapsed time, derived from the differential rate laws using calculus.
Reaction Order: The power to which a reactant's concentration is raised in the rate law, which dictates the specific mathematical relationship between concentration and time.
Rate Constant ( $k$ ): A proportionality constant unique to a specific reaction at a given temperature; its value can be determined from the slope of the linearized concentration-time graph.
Linearization: The process of applying mathematical functions (like $\ln$ or $1/x$ ) to experimental data to create a straight-line plot for easier analysis.

Comparison of linear plots for zero, first, and second order reactions showing the specific axis transformations required for each.

2. Underlying Principles

3. Methods & Techniques

4. Key Distinctions

5. Exam Strategy & Tips

6. Common Pitfalls & Misconceptions

Confusing ln and log: Integrated rate laws specifically use the natural logarithm (base $e$ ), not the common logarithm (base 10).
Ignoring the Negative Sign: Forgetting that $k$ is always a positive value, even if the slope of the graph is negative.
Misidentifying Linearity: Students often mistake a slightly curved line for a straight one; always compare all three plots before making a final determination.