Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Define half-life. Write the equation relating the half-life of a first-order reaction to the rate constant.

Short Answer

Expert verified
Half-life is the time for a reactant to reduce to half, given by \(\text{t}_{1/2} = \frac{0.693}{k}\) for first-order reactions.

Step by step solution

01

Define Half-life

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration. It is a measure of how quickly a reactant is consumed in a reaction.
02

Identify Reaction Order

For this exercise, we need to focus on a reaction that is first-order. In first-order reactions, the rate of reaction is directly proportional to the concentration of one reactant.
03

First-Order Half-Life Equation

The equation relating the half-life ( ext{t}_{1/2}) of a first-order reaction to the rate constant (k) is given by:\[\text{t}_{1/2} = \frac{0.693}{k}\]This means that the half-life is inversely proportional to the rate constant.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Reaction
A first-order reaction is a type of chemical reaction where the reaction rate depends linearly on the concentration of a single reactant. Imagine you're observing a substance breaking down over time. In a first-order reaction, if you increase the concentration of the reactant, the rate at which the reaction occurs will also increase proportionally. This is because the reaction rate formula for a first-order reaction is:
\[ ext{Rate} = k[A] \\]where \(k\) is the rate constant and \( [A] \) is the concentration of the reactant.
To remember this easily, think about how baking a cake might work. If you have more cake mix, the batter forms faster assuming your mixing efforts are consistent. Similarly, in first-order reactions, more reactant leads to a faster reaction, up to a linear point. Other reactions might depend on multiple reactants, but first-order reactions keep it simple with just one variety.
  • Directly proportional to reactant concentration
  • Simple relationship between rate and concentration
Rate Constant
The concept of a rate constant, denoted by \( k \), is fundamental in understanding the speed of a reaction. In reaction kinetics, the rate constant is what determines how quickly a reaction moves to completion once it has started.
For first-order reactions, \( k \) has a specific significance. The equation relating the half-life of a first-order reaction to its rate constant is given by:
\[\text{t}_{1/2} = \frac{0.693}{k}\]This formula tells us that the half-life and the rate constant are inversely related.
This means a larger \( k \) results in a shorter half-life, implying the reactant concentration decreases faster. Think of \( k \) as a "rate dial"; a higher setting speeds up the reaction.
  • Central to calculating reaction speed
  • Inversely related to half-life in first-order reactions
Reaction Kinetics
Reaction kinetics is the study of reaction rates and the steps that lead to the final products. It focuses not just on how quickly reactions occur, but also on the pathway and transformations reactants undergo on their way to becoming products.
In the case of first-order reactions, kinetics simplifies to examining the rate dependent on a single reactant. For such reactions, the study is straightforward, and the rate equation can help predict how concentrations will change over time. Kinetics allows chemists to:
  • Determine the reaction order
  • Predict the time needed for a reaction to reach a certain extent
  • Understand how different conditions affect reaction rates
By applying kinetics principles, one can visualize the pace at which reactions happen, similar to estimating travel time based on speed and distance in everyday life.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Distinguish between average rate and instantaneous rate. Which of the two rates gives us an unambiguous measurement of reaction rate? Why?

Explain why most metals used in catalysis are transition metals.

A protein molecule \(\mathrm{P}\) of molar mass \(\mathscr{M}\) dimerizes when it is allowed to stand in solution at room temperature. A plausible mechanism is that the protein molecule is first denatured (i.e., loses its activity due to a change in overall structure) before it dimerizes: \(\mathrm{P} \stackrel{k}{\longrightarrow} \mathrm{P}^{*}(\) denatured \() \quad\) (slow) $$ 2 \mathrm{P}^{*} \longrightarrow \mathrm{P}_{2} $$ (fast) where the asterisk denotes a denatured protein molecule. Derive an expression for the average molar mass (of \(\mathrm{P}\) and \(\left.\mathrm{P}_{2}\right), \bar{U},\) in terms of the initial protein concentration \([\mathrm{P}]_{0}\) and the concentration at time \(t,[\mathrm{P}]_{t},\) and \(\mathscr{M} .\) Describe how you would determine \(k\) from molar mass measurements.

The concentrations of enzymes in cells are usually quite small. What is the biological significance of this fact?

The following data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$ $$ \begin{array}{cccc} \text { Experiment } & {\left[\mathrm{H}_{2}\right](M)} & {[\mathrm{NO}](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume that the oxygen atom is the intermediate.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free