/*! 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 105 At \(25^{\circ} \mathrm{C}\), ra... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 105

At \(25^{\circ} \mathrm{C}\), raw milk sours in \(6.0 \mathrm{~h}\) but takes \(60 \mathrm{~h}\) to sour in a refrigerator at \(5{ }^{\circ} \mathrm{C}\). Estimate the activation energy in \(\mathrm{kJ} / \mathrm{mol}\) for the reaction that leads to the souring of milk.

Short Answer

Expert verified
Approximately \( 73.93 \, \text{kJ/mol} \).

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation is used to relate the rate constants of a reaction at different temperatures and is expressed as: \[ k = A e^{-\frac{E_a}{RT}} \] where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. For our purposes, we will use the transformed version involving a ratio of rate constants, which will allow us to solve for the activation energy.
02

Apply the Arrhenius Equation for Two Temperatures

Use the two given conditions: at \( 25^{\circ}C \) and \( 5^{\circ}C \). First, convert these temperatures to Kelvin: \[ T_1 = 25^{\circ}C + 273 = 298 \text{ K} \] \[ T_2 = 5^{\circ}C + 273 = 278 \text{ K} \] The rate constants for the reactions are the inverses of the time for souring, hence: \[ k_1 = \frac{1}{6 \text{ h}} \] \[ k_2 = \frac{1}{60 \text{ h}} \] Using the equation: \[ \ln\left(\frac{k_1}{k_2}\right) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
03

Calculate the Ratio of Rate Constants

Calculate the ratio \( \frac{k_1}{k_2} \): \[ \frac{k_1}{k_2} = \frac{1/6}{1/60} = 10 \] Substitute into the logarithmic expression: \[ \ln\left(10\right) = \frac{E_a}{R} \left(\frac{1}{278} - \frac{1}{298}\right) \]
04

Solve for Activation Energy \(E_a\)

The gas constant \( R \) is \( 8.314 \, \text{J/mol路K} \). Substitute the known values: \[ \ln(10) = \frac{E_a}{8.314} \left(\frac{1}{278} - \frac{1}{298}\right) \] Calculate \( \ln(10) \approx 2.302 \) and the temperature differences: \[ \frac{1}{278} - \frac{1}{298} \approx 0.000259 \] Then: \[ 2.302 = \frac{E_a}{8.314} \times 0.000259 \] Rearrange for \( E_a \): \[ E_a = \frac{2.302 \times 8.314}{0.000259} \approx 73926.6 \text{ J/mol} \] Convert Joules to kilojoules: \[ E_a \approx 73.93 \, \text{kJ/mol} \]
05

Finalize the Answer

We estimate the activation energy \( E_a \) for the souring of milk using the given temperature conditions to be approximately \( 73.93 \, \text{kJ/mol} \).

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Key Concepts

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

Arrhenius equation
The Arrhenius equation is a critical concept in chemistry for understanding how temperature affects the rate of chemical reactions. This equation is expressed as:\[ k = A e^{-\frac{E_a}{RT}} \]where:
  • k is the rate constant.
  • A represents the pre-exponential factor, which is often related to the frequency of collisions and the probability that the collisions are successful.
  • Ea is the activation energy, the minimum energy required for a reaction to occur.
  • R is the universal gas constant.
  • T denotes the temperature in Kelvin.
This formula highlights the exponential relationship between the rate constant and temperature, as governed by the activation energy. Essentially, as temperature increases, the rate constant typically increases, suggesting a faster reaction. To estimate or solve for the activation energy, we can rearrange and utilize the equation considering changes in temperature and observe the variance in rate constants.
rate constants
Rate constants are pivotal in describing how quickly a reaction occurs. These constants, denoted by k, provide insight into the rate of a given reaction at specific conditions.The value of a rate constant varies with temperature and is influenced by the activation energy and the pre-exponential factor in the Arrhenius equation. In the souring of milk problem, the rate constants are inherently tied to how long it takes for the milk to sour at distinct temperatures. The formulas:\[ k_1 = \frac{1}{6 \text{ h}} \] \[ k_2 = \frac{1}{60 \text{ h}} \] illustrate how rate constants can be derived from the reciprocal of time. By knowing k at multiple temperatures, we can engage the Arrhenius equation for further exploration, including approximating activation energy. Understanding the manipulation and calculation of rate constants is fundamental to mastering kinetics.
temperature conversion
Temperature conversion is a necessary preliminary step in many kinetic calculations, as the Arrhenius equation requires temperature inputs in Kelvin. Kelvin units are used because this scale starts at absolute zero, providing an appropriate measure for thermodynamic equations. In our context, converting Celsius to Kelvin is straightforward:
  • For 25掳颁: \[ T_1 = 25 + 273 = 298 \text{ K} \]
  • For 5掳颁: \[ T_2 = 5 + 273 = 278 \text{ K} \]
Knowing these conversions allows for consistent and correct application of the Arrhenius equation, ensuring accurate calculations of rate constants and activation energies.
kinetics calculation
Kinetics calculations help us quantify how reactions progress over time, focusing on aspects like the rate of reaction and activation energy. In our milk souring example, we use the Arrhenius equation and observed data to estimate the activation energy for the process.

Steps Involved

  • Determine rate constants from given data: The inverse of the time it takes for the reaction to complete at each temperature gives us k.
  • Apply the modified Arrhenius equation: \[ \ln\left(\frac{k_1}{k_2}\right) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
  • Solve for Ea: Rearrange to find the activation energy, which involves using the natural logarithm of the ratio of rate constants and the inverse of the Kelvin temperatures.
This step-by-step approach allows us to thoroughly understand and compute the necessary kinetic parameters, providing insights into the reaction鈥檚 energetic requirements.

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Most popular questions from this chapter

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\) (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?\) (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

A reaction \(A+B \longrightarrow C\) obeys the following rate law: Rate \(=k[A]^{3}\). (a) If [B] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B}\) ? What is the overall reaction order? (c) What are the units of the rate constant?

(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

The addition of NO accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (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?

In solution, chemical species as simple as \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right]\) of a solution containing an acidcatalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\) concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.
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