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Explain the important distinctions between each pair of terms: (a) first-order and second-order reactions; (b) rate law and integrated rate law; (c) activation energy and enthalpy of reaction; (d) elementary process and overall reaction; (e) enzyme and substrate.

Short Answer

Expert verified
First-order and second-order reactions differ in how the reaction rate responds to changes in reactant concentration. Rate law relates reactant concentrations to reaction rate, while the integrated rate law shows reactant concentrations over time. Activation energy is the energy required to reach the transition state, whereas enthalpy of reaction is the overall energy change. An elementary process is a single step in a reaction, while the overall reaction summarizes the complete process. Enzyme is a catalyst, while a substrate is the molecule upon which an enzyme acts.

Step by step solution

01

Comparison (a)

First-order reactions only depend on the concentration of one reactant. If you double the concentration of the reactant, the rate of reaction also doubles. In second-order reactions, the rate depends on either the concentrations of two first-order reactants, or the square of the concentration of a single reactant. If you double the concentration of the reactant(s), the reaction rate quadruples.
02

Comparison (b)

The rate law defines the relationship between the rate of a reaction and the concentrations of the reactants. This equation usually has to be determined experimentally. The integrated rate law, on the other hand, comes from integrating the rate law and shows the concentrations of the reactants as a function of time.
03

Comparison (c)

Activation energy is the minimum amount of energy required for a reaction to occur. This is the energy necessary to reach the transition state. The enthalpy of reaction, on the other hand, is the overall change in energy during a chemical reaction. This is the difference in energy between the reactants and the products.
04

Comparison (d)

An elementary process is a single step in a reaction mechanism. These steps are the simplest individual actions that describe the progression of a reaction. An overall reaction is the net effect of these elementary processes and summarizes what happens in the whole reaction.
05

Comparison (e)

An enzyme is a biological catalyst that speeds up chemical reactions by reducing the activation energy. A substrate is the molecule upon which an enzyme acts. The enzyme binds to the substrate, performs its catalytic function, and then releases the changed substrate (or the products formed from the substrate).

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Key Concepts

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

Reaction Orders
Understanding the order of a chemical reaction is important because it tells us how the concentration of reactants affects the reaction rate. In the simplest terms, a first-order reaction is one where the rate depends on the concentration of a single reactant. This means if you double the concentration of that reactant, the reaction rate doubles too.
An example of this would be radioactive decay. For a second-order reaction, the rate can depend on one of two scenarios:
  • It might depend on two different reactants, each to the power of one (for example, Rate = k[A][B]).
  • Or it could depend on the square of a single reactant's concentration (Rate = k[A]^2).
For second-order reactions, doubling the concentration leads to four times the rate of reaction. This distinction helps chemists predict how changing conditions will impact the speed of reactions.
Rate Laws
The rate law of a reaction is like the equation of a line—it provides a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. This is typically determined through experiments and helps us understand how fast a reaction proceeds.
Rate laws are usually expressed in the form:
  • Rate = k [A]^m [B]^n
where k is the rate constant, and m and n are the orders of the reaction with respect to reactants A and B, respectively.
Integrated rate laws, however, give us a relationship between the concentration of reactants and time. They provide insight into how concentrations change over time, allowing us to calculate how long it takes for a reaction to occur or reach a certain point.
Activation Energy
Activation energy is a core concept in understanding how reactions occur at a molecular level. It is the minimum energy required to initiate a chemical reaction. Imagine pushing a boulder over a hill—activation energy is the initial push needed to get it moving down the other side.
This energy helps molecules overcome the energy barrier that keeps them from reacting spontaneously.
The concept is related to the transition state, which is the highest energy point during a reaction where old bonds are breaking and new bonds are forming.
Oh, and it's important to note that activation energy is not the same as the enthalpy of reaction, which is the difference in energy between reactants and products, representing the overall energy change in the reaction.
Elementary Process
In chemistry, understanding how reactions proceed in steps is crucial. Each of these steps is known as an elementary process.
This is the smallest unit of a reaction mechanism, and it describes a simple interaction where one or two reactants form transition states to products.
The overall reaction you see in a chemical equation is actually the result of several elementary processes that occur in sequence.
These steps are vital because they give us insight into the kinetics and dynamics of the overall reaction, helping us understand and predict reaction behavior more accurately.
Enzymes and Substrates
Enzymes play a special role in the world of chemistry and biology. They are biological catalysts that significantly speed up the rate of reactions by lowering the activation energy needed.
This means reactions that might take hours or days to occur naturally can happen in milliseconds with the help of enzymes.
Enzymes work by binding to substrates, which are the specific molecules they act upon, at a region known as the active site.
When an enzyme binds its substrate, it forms an enzyme-substrate complex that facilitates the conversion of substrates into products much faster.
This interaction is crucial for various biological processes, including digestion, cellular metabolism, and DNA replication.

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Most popular questions from this chapter

A first-order reaction, \(\mathrm{A} \longrightarrow\) products, has a halflife of \(75 \mathrm{s},\) from which we can draw two conclusions. Which of the following are those two (a) the reaction goes to completion in 150 s; (b) the quantity of \(A\) remaining after 150 s is half of what remains after 75 s; (c) the same quantity of A is consumed for every 75 s of the reaction; (d) one- quarter of the original quantity of A is consumed in the first 37.5 s of the reaction; (e) twice as much A is consumed in 75 s when the initial amount of \(\mathrm{A}\) is doubled; (f) the amount of \(\mathrm{A}\) consumed in 150 s is twice as much as is consumed in 75 s.

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ Which of these sets of data corresponds to a (a) zero-order, (b) first-order, (c) second-order reaction?

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

The decomposition of dimethyl ether at \(504^{\circ} \mathrm{C}\) is $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$$ The following data are partial pressures of dimethyl ether (DME) as a function of time: \(t=0\) s, \(P_{\text {DME }}=\) \(312 \mathrm{mmHg} ; 390 \mathrm{s}, 264 \mathrm{mmHg} ; 777 \mathrm{s}, 224 \mathrm{mmHg} ; 1195 \mathrm{s},187 \mathrm{mmHg} ; 3155 \mathrm{s}, 78.5 \mathrm{mmHg}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is the total gas pressure at 390 s? (d) What is the total gas pressure when the reaction has gone to completion? (e) What is the total gas pressure at \(t=1000\) s?

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