For the reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}$$ the rate expression is rate \(=k[\mathrm{~A}][\mathrm{B}]\) (a) Given three test tubes, with different concentrations of \(\mathrm{A}\) and \(\mathrm{B}\), which test tube has the smallest rate? (1) \(0.10 M \mathrm{~A}_{\mathrm{i}} 0.10 \mathrm{M} \mathrm{B}\) (2) \(0.15 \mathrm{MA}_{;} 0.15 \mathrm{M} \mathrm{B}\) (3) \(0.06 M \mathrm{~A} ; 1.0 M \mathrm{~B}\) (b) If the temperature is increased, describe (using the words increases, decreases, or remains the same) what happens to the rate, the value of \(k\), and \(E_{x}\)

Short Answer

Expert verified
Answer: Test tube 1 has the smallest rate of reaction. An increase in temperature results in (a) an increased rate of reaction, (b) an increased value of k, and (c) no change in E_x as it is a property of the reaction.

Step by step solution

01

Calculate the rate for each test tube using the rate expression

We are given the rate expression as \(rate = k[\mathrm{A}][\mathrm{B}]\). Using the concentrations of A and B for each test tube, we can calculate the rates as follows: Test tube 1: \(rate_1 = k(0.10)(0.10)\) Test tube 2: \(rate_2 = k(0.15)(0.15)\) Test tube 3: \(rate_3 = k(0.06)(1.0)\)
02

Compare the rates to find the smallest rate

Now that we have the rate expressions for each test tube, let's compare them to find the smallest rate. Since we are only comparing the rates, we don't need to know the value of k. We can compare the products of concentrations instead: Test tube 1: \((0.10)(0.10) = 0.01\) Test tube 2: \((0.15)(0.15) = 0.0225\) Test tube 3: \((0.06)(1.0) = 0.06\) From the comparison, we can see that Test tube 1 has the smallest rate.
03

Describe the effect of increasing temperature on rate, value of k, and E_x

According to the Arrhenius equation, an increase in temperature typically results in an increase in the value of the rate constant, k. This happens because a higher temperature provides more energy for the reacting molecules, increasing the likelihood of successful collisions. (a) Rate: When the temperature is increased, the rate of the reaction increases. (b) Value of k: When the temperature is increased, the value of k also increases. (c) E_x: The activation energy, E_x, remains the same as it is a property of the reaction and is not directly affected by a change in temperature.

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.

Rate Expression
Understanding the rate expression is crucial when analyzing how fast a chemical reaction proceeds. The rate expression for a reaction, also known as the rate law, represents the relationship between the reaction rate and the concentration of reactants.

In the provided exercise, the rate expression given is \( rate = k[A][B] \), where \( k \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of the reactants A and B, respectively. The brackets denote concentration in molarity (M), and the rate is directly proportional to the product of these concentrations. This denotes that the reaction is first-order with respect to both A and B, implying that the rate of reaction will double if the concentration of either reactant is doubled, keeping everything else constant.

The simplicity of the rate expression allows us to predict how the reaction rate will change with varying concentrations of A and B without knowing the exact value of the rate constant, \( k \).
Concentration Effect on Reaction Rate
The impact of reactant concentration on the rate of a chemical reaction is straightforward yet profound. The rate expression encapsulates this relationship and informs us that the rate of reaction is contingent on the molar concentrations of the reactants involved.

Let's consider the example from the exercise. Comparing the given test tubes, we observe that even with different concentrations of A and B, it's the product of the concentrations that governs the rate. By comparing these products, we determined that Test tube 1, with lower concentrations of both reactants, has the smallest rate. This illustrates a general principle: as the concentration of reactants decreases, the rate of the reaction tends to diminish because there are fewer reactant molecules available to collide and react with each other.
Temperature Effect on Reaction Rate
Temperature is a pivotal factor in the rate of a chemical reaction. Increasing the temperature typically leads to a higher reaction rate. This is because molecules move more vigorously at higher temperatures, leading to more frequent and more energetic collisions among reactant molecules.

As per the provided exercise, when the temperature is increased, the rate of the reaction will increase as well. This is because an increase in temperature usually means that more molecules have enough kinetic energy to overcome the activation energy barrier of the reaction. However, it's crucial to note that the activation energy, \( E_x \), itself does not change with temperature; it is an inherent characteristic of the chemical reaction.
Arrhenius Equation
The Arrhenius equation is a formula that provides a quantitative basis for the effect of temperature on reaction rates. It is expressed as \( k = A e^{-\frac{E_x}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor (a constant), \( E_x \) represents the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

The equation clearly shows that the rate constant \( k \) increases exponentially with an increase in temperature. This exponential relationship underscores why even a slight rise in temperature can lead to a significant increase in the rate constant, and consequently, the reaction rate. It connects the macroscopic observation of increased reaction rate with temperature to the microscopic behavior of molecules, providing students with a deeper understanding of the underlying phenomena at play.

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

Express the rate of the reaction $$2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ in terms of (a) \(\Delta\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]\) (b) \(\Delta\left[\mathrm{CO}_{2}\right]\)

Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)

The uncoiling of deoxyribonucleic acid (DNA) is a first-order reaction. Its activation energy is \(420 \mathrm{~kJ}\). At \(37^{\circ} \mathrm{C}\), the rate constant is \(4.90 \times 10^{-4} \mathrm{~min}^{-1}\). (a) What is the half-life of the uncoiling at \(37^{\circ} \mathrm{C}\) (normal body temperature)? (b) What is the half-life of the uncoiling if the organism has a temperature of \(40^{\circ} \mathrm{C}\left(\approx 104^{\circ} \mathrm{F}\right)\) ? (c) By what factor does the rate of uncoiling increase (per \({ }^{\circ} \mathrm{C}\) ) over this temperature interval?

The chirping rate of a cricket \(\mathrm{X}\), in chirps per minute near room temperature, is $$\mathrm{X}=7.2 t-32$$ where \(t\) is the temperature in \({ }^{\circ} \mathrm{C}\). (a) Calculate the chirping rates at \(25^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). (b) Use your answers in (a) to estimate the activation energy for the chirping. (c) What is the percentage increase for a \(10^{\circ} \mathrm{C}\) rise in temperature?

Write the rate expression for each of the following elementary steps: (a) \(\mathrm{NO}+\mathrm{O}_{3} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\) (b) \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) (c) \(\mathrm{K}+\mathrm{HCl} \longrightarrow \mathrm{KCl}+\mathrm{H}\)

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