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Chapter 14: Problem 61

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol}\) ? (b) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\) What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ? (c) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

Short Answer

Expert verified
Using the Arrhenius equation, we find the rate constants at 60潞C for both reactions: (a) For the first reaction, with an activation energy of \(E_a = 75.5 \times 10^3 J\text{mol}^{-1}\), we calculate the rate constant at 60潞C as \(k_2 = 2.07 \times 10^{-1} \mathrm{s^{-1}}\). (b) For the second reaction, with an activation energy of \(E_a = 125 \times 10^3 J\text{mol}^{-1}\), we calculate the rate constant at 60潞C as \(k_2 = 6.81 \times 10^{-3} \mathrm{s^{-1}}\). Assumptions made in this calculation include: (1) the reactions follow the first-order kinetics, (2) the activation energy and pre-exponential factor are constant over the temperature range considered, and (3) the temperature change does not affect the reaction mechanism, and the rate-determining step remains the same.

Step by step solution

01

Get the k value at the initial temperature

Use the Arrhenius equation: \(k_1 = k_0 e^{-\frac{E_a}{R T_1}}\) Now, we need to find k0 (the pre-exponential factor) using the given values.
02

Solve for k0

Rearrange the equation to solve for k0: k0_=\(\frac{k_1}{e^{-\frac{E_a}{R T_1}}}\) Using the given values: k0_= \(\frac{2.75 \times 10^{-2} \mathrm{s^{-1}}}{e^{\frac{-75.5 \times 10^3 J \text{mol}^{-1}}{(8.314 J \text{mol}^{-1}K^{-1})(293.15K)}}\) Calculate k0.
03

Calculate k at the new temperature

Use the Arrhenius equation and plug in the calculated k0 and the new temperature T2: \(k_2=k_0 e^{-\frac{E_a}{R T_2}}\) Calculate k2, which is the rate constant at 60潞C for reaction (a). (b) We are given the same initial k and temperature values, but now: Ea_= \(125 \times 10^3 J \text{mol}^{-1}\) Repeat steps 1-3 using this new Ea value to find the rate constant at 60潞C for reaction (b). (c) To answer the assumptions made for parts (a) and (b):
04

Assumption 1

The reaction follows first-order kinetics, as mentioned in the problem.
05

Assumption 2

The activation energy (Ea) and pre-exponential factor (k0) are constant over the temperature range considered.
06

Assumption 3

The temperature change does not affect the reaction mechanism, and the rate-determining step remains the same.

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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 rate depends linearly on the concentration of a single reactant. This means the rate of the reaction can be expressed as:
  • \( ext{Rate} = k[A] \)
In this equation, \( A \) represents the concentration of the reactant, and \( k \) is the rate constant. Such reactions are characterized by a constant half-life that doesn't depend on the initial concentration of the reactants.
This makes them predictable and easy to study in laboratory settings.
A typical example of a first-order reaction is radioactive decay.
  • Easy prediction with fixed half-lives
  • Simplifies mathematical treatment due to linear relationships
Rate Constant
The rate constant, denoted as \( k \), plays a crucial role in determining the speed of a chemical reaction. It is a proportionality constant in the rate law equations. In the context of a first-order reaction, its unit is inverse time, such as \( ext{s}^{-1} \).
The value of the rate constant indicates how fast or slow a reaction proceeds. Higher values of \( k \) correspond to faster reactions and vice versa.
The rate constant is temperature dependent and can be calculated using the Arrhenius equation:
  • \( k = A e^{- rac{E_a}{RT}} \)
Where:
  • \( A \) is the pre-exponential factor
  • \( E_a \) is the activation energy
  • \( R \) is the gas constant
  • \( T \) is the temperature in Kelvin
Activation Energy
Activation energy, often symbolized as \( E_a \), is the minimum energy that reacting molecules need to undergo a reaction. It is a threshold energy barrier that must be overcome for reactants to transform into products.
In practical terms, activation energy affects the rate of reaction; the lower the energy, the faster the reaction can proceed.
Typically measured in kilojoules per mole (kJ/mol), activation energy can be determined through experimental observations of reaction rates at different temperatures.
It's a critical factor in the Arrhenius equation, influencing how \( k \) changes with temperature.
Many reactions require the input of activation energy to proceed, even if the overall process is exothermic.
Pre-Exponential Factor
The pre-exponential factor, often denoted as \( A \), is a term within the Arrhenius equation that embodies the frequency of collisions and the probability that the collisions have the correct orientation to lead to a reaction.
It is also known as the frequency factor or the Arrhenius constant.
Usually, \( A \) has similar units as the rate constant and is temperature-independent over the range considered in many reactions.
Although it might seem like a constant with little significance, \( A \) greatly influences the rate of reaction by dictating the general number of effective collisions per unit time.
  • Represents the likelihood of successful collisions
  • A higher \( A \) indicates more frequent and oriented collisions leading towards a reaction

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Most popular questions from this chapter

Heterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure 14.24, are subject to "poisoning," which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such compounds might act as poisons.

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \quad \text { (slow) } \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \text { (fast) } \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

Which of the following linear plots do you expect for a reaction \(A \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4]

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{H}^{+}(a q)\). In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Platinum nanoparticles of diameter \(\sim 2 \mathrm{~nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \dot{A}=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}\). (b) Estimate how many platinum atoms are on the surface of a \(2.0\)-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(2.8 \mathrm{~A}\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0\)-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?
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