Ch.14 - Chemical Kinetics

Problem 39b

Chapter 14, Problem 39b

(b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

Verified step by step guidance

Identify the type of graph you have created in part (a). For a first-order reaction, you should have plotted the natural logarithm of the concentration of the reactant, \( \ln[A] \), versus time.

Understand that the slope of the line in this graph represents the negative of the rate constant, \( k \), for the reaction. This is derived from the integrated rate law for a first-order reaction: \( \ln[A] = -kt + \ln[A_0] \), where \( [A_0] \) is the initial concentration.

Determine the slope of the line from your graph. This can be done by selecting two points on the line and using the formula for the slope: \( \text{slope} = \frac{\Delta y}{\Delta x} \), where \( \Delta y \) is the change in \( \ln[A] \) and \( \Delta x \) is the change in time.

Recognize that the slope you calculated is equal to \( -k \). Therefore, the rate constant \( k \) is the negative of the slope value you obtained.

Ensure that the units of the rate constant \( k \) are consistent with the units of time used in your graph. For a first-order reaction, \( k \) typically has units of \( \text{time}^{-1} \), such as \( ext{s}^{-1} \) or \( ext{min}^{-1} \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Reactions

First-order reactions are chemical reactions where the rate is directly proportional to the concentration of one reactant. This means that if the concentration of the reactant doubles, the reaction rate also doubles. The rate law for a first-order reaction can be expressed as rate = k[A], where k is the rate constant and [A] is the concentration of the reactant.

Rate Constant (k)

The rate constant (k) is a proportionality factor in the rate law that is specific to a given reaction at a specific temperature. For first-order reactions, the rate constant can be determined from the slope of a plot of the natural logarithm of the concentration of the reactant versus time. The units of k for a first-order reaction are typically s⁻¹.

Graphical Analysis of Reaction Kinetics

Graphical analysis involves plotting data to visualize relationships between variables. For first-order reactions, plotting ln[A] versus time yields a straight line, where the slope of the line is equal to -k. This method allows for the determination of the rate constant from experimental data, making it a powerful tool in kinetics studies.

Recommended video:

Guided course

06:11

Dimensional Analysis

Related Practice

Textbook Question

Consider the reaction of peroxydisulfate ion (S₂O₈^2-) with iodide ion (I^-) in aqueous solution:

S₂O₈^2-(aq) + 3 I^-(aq) → 2 SO₄^2-(aq) + I₃^-(aq)

At a particular temperature, the initial rate of disappearance of S₂O₈^2- varies with reactant concentrations in the following manner:

Experiment [S₂O₈^2-] (M) [I^-] (M) Initial Rate (M/s)

1 0.018 0.036 2.6 × 10^-6

2 0.027 0.036 3.9 × 10^-6

3 0.036 0.054 7.8 × 10^-6

4 0.050 0.072 1.4 × 10^-5

(a) Determine the rate law for the reaction and state the units of the rate constant.

Textbook Question

(a) For the generic reaction A → B what quantity, when graphed versus time, will yield a straight line for a first-order reaction?

Textbook Question

The decomposition of sodium bicarbonate (baking soda), NaHCO₃(s), into Na₂CO₃(s), H₂O(l), and CO₂(g) at constant pressure requires the addition of 85 kJ of heat per two moles of NaHCO₃. (b) Draw an enthalpy diagram for the reaction.

Textbook Question

(a) The gas-phase decomposition of SO₂Cl₂, SO₂Cl₂(g) → SO₂(g) + Cl₂(g), is first order in SO₂Cl₂. At 600 K the half-life for this process is 2.3 × 10⁵ s. What is the rate constant at this temperature?

views

Textbook Question

(b) At 320°C the rate constant is 2.2 × 10^-5 s^-1. What is the half-life at this temperature?