/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 The activation energy of an unca... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) At \(25^{\circ} \mathrm{C}\) (298.15 K), the catalyst increases the rate of the reaction by a factor of approximately 148.14. (b) At \(125^{\circ} \mathrm{C}\) (398.15 K), the catalyst increases the rate of the reaction by a factor of approximately 19.44.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we add 273.15 to the Celsius values. (a) For 25°C: \(T_1 = 25 + 273.15 = 298.15K\) (b) For 125°C: \(T_2 = 125 + 273.15 = 398.15K\)
02

Calculate the rate constant ratio

We will determine the ratio of the rate constants in the presence and absence of a catalyst, which can be represented as: \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{Ae^{-\frac{E_a^{catalyzed}}{RT}}}{Ae^{-\frac{E_a^{uncatalyzed}}{RT}}} \] Since the collision factor (A) remains the same, it cancels out: \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-\frac{E_a^{catalyzed}}{RT}}}{e^{-\frac{E_a^{uncatalyzed}}{RT}}} \] Also, we need to substitute the values of activation energies given in the problem: Uncatalyzed activation energy (\(E_a^{uncatalyzed}\)) = 95 kJ/mol = 95,000 J/mol Catalyzed activation energy (\(E_a^{catalyzed}\)) = 55 kJ/mol = 55,000 J/mol
03

Calculate the factor at each temperature

(a) For 25°C (298.15K) \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-\frac{55,000}{8.314 \times 298.15}}}{e^{-\frac{95,000}{8.314 \times 298.15}}} \] \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} \approx 148.14 \] (b) For 125°C (398.15K) \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-\frac{55,000}{8.314 \times 398.15}}}{e^{-\frac{95,000}{8.314 \times 398.15}}} \] \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} \approx 19.44 \]
04

Interpret the results

(a) At 25°C, the rate of the reaction will increase by a factor of approximately 148.14 with the addition of a catalyst. (b) At 125°C, the rate of the reaction will increase by a factor of approximately 19.44 with the addition of a catalyst.

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 91Ó°ÊÓ!

Key Concepts

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

Catalyst
A catalyst is a substance that can significantly speed up the rate of a chemical reaction without being consumed in the process. It works by lowering the activation energy required for the reaction to occur, which allows more reactant molecules to have enough energy to reach the transition state.

In the given exercise, we saw that introducing a catalyst to a reaction lowered the activation energy from 95 kJ/mol to 55 kJ/mol. This considerable reduction in activation energy indicates that the catalyst provides an alternative pathway for the reaction, one that requires less energy. As a result, even at the same temperature, more molecules have sufficient energy to surpass the lowered activation energy barrier, thus increasing the rate of the reaction.

It is critical to note that while catalysts accelerate reaction rates, they do not affect the thermodynamics of the reaction, such as the equilibrium position, which remains unchanged. By providing a more efficient reaction path, catalysts are invaluable in many industrial processes, allowing for faster production rates and lower energy costs.
Rate Constant
The rate constant, often denoted by 'k', is a proportionality constant in the rate equation of a chemical reaction that provides the relationship between the reactant concentrations and the reaction rate at a given temperature. It reflects the rate at which a reaction proceeds and is influenced by factors including temperature, activation energy, and the presence of a catalyst.

In the exercise, we calculated the rate constant ratio between catalyzed and uncatalyzed reactions at different temperatures. As shown in the solution, the catalyst significantly increased the rate constant, which in turn increases the reaction rate. Since the rate constant is higher in the presence of a catalyst, it means that for every successful collision between reactant molecules, the reaction is much more likely to proceed, leading to faster reaction times.

Understanding how the rate constant varies with conditions like temperature and the presence of a catalyst is crucial in controlling the rates of reactions, which is especially important when designing industrial processes or in lab experiments.
Arrhenius Equation
The Arrhenius equation mathematically expresses the relationship between the rate constant 'k', the activation energy of the reaction 'Ea', and the temperature 'T'. It is given by the formula: \[ k = Ae^{-\frac{E_a}{RT}} \]where 'A' is known as the frequency factor or pre-exponential factor, 'Ea' is the activation energy, 'R' is the gas constant, and 'T' is the absolute temperature in Kelvin.

The Arrhenius equation is pivotal because it shows that even small changes in activation energy or temperature can have massive impacts on the reaction rate. As seen in the exercise, by exponentiating the negative ratio of activation energy over the product of the gas constant and temperature (\(-\frac{E_a}{RT}\)), the resulting rate constant 'k' can differ greatly with or without a catalyst. Moreover, it explains why reactions speed up with increasing temperature – as 'T' rises, the value of \(e^{-\frac{E_a}{RT}}\) increases, leading to a higher 'k' and thus a faster reaction rate.

The equation also demonstrates the profound effect of catalysts on reaction rates by significantly reducing the activation energy, 'Ea', thereby increasing the rate constant, 'k'. This relation allows chemists to predict and control the speed of chemical reactions effectively.

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

The following mechanism has been proposed for the gas-phase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) Write rate laws for each elementary reaction in the mechanism. (d) If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

(a) Two reactions have identical values for \(E_{a}\). Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a higher rate constant than the other. Account for these observations.

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64{ }^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\)

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

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.