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Chapter 14: Problem 87

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) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) At 25掳C, the catalyst increases the rate by about \(1.01 \times 10^7\) times. (b) At 125掳C, it increases the rate by about \(1.74 \times 10^5\) times.

Step by step solution

01

Understanding the Arrhenius Equation

The Arrhenius equation is a formula that gives the rate constant \(k\) of a reaction as a function of temperature \(T\) and activation energy \(E_a\). It is given by \[ k = A e^{-E_a / RT} \]where \(A\) is the pre-exponential factor (or collision factor), \(E_a\) is the activation energy, \(R\) is the universal gas constant (\(8.314 \, \text{J/mol K}\)), and \(T\) is the temperature in Kelvin.
02

Converting Temperature to Kelvin

To use the Arrhenius equation, temperatures must be in Kelvin. Convert the given temperatures:- For \(25^{\circ} \text{C}\), \[ T_1 = 25 + 273.15 = 298.15 \text{ K} \]- For \(125^{\circ} \text{C}\), \[ T_2 = 125 + 273.15 = 398.15 \text{ K} \]
03

Calculate the Rate Constant Ratio at Each Temperature

Using the Arrhenius equation, the ratio of the rate constants for the catalyzed \((k_{cat})\) and uncatalyzed \((k_{uncat})\) reactions is:\[ \frac{k_{cat}}{k_{uncat}} = \frac{A e^{-E_{cat} / RT}}{A e^{-E_{uncat} / RT}} = e^{(E_{uncat} - E_{cat}) / RT} \]For:- \(E_{uncat} = 95,000 \text{ J/mol}\)- \(E_{cat} = 55,000 \text{ J/mol}\)
04

Calculate the Factor Increase at 25掳C

With \(T_1 = 298.15 \text{ K}\), the ratio is:\[ \frac{k_{cat}}{k_{uncat}} = e^{(95,000 - 55,000) / (8.314 \times 298.15)} = e^{40,000 / 2481.911} \approx e^{16.12} \]Evaluating this gives approximately \(1.01 \times 10^7\).
05

Calculate the Factor Increase at 125掳C

With \(T_2 = 398.15 \text{ K}\), the ratio is:\[ \frac{k_{cat}}{k_{uncat}} = e^{(95,000 - 55,000) / (8.314 \times 398.15)} = e^{40,000 / 3310.9991} \approx e^{12.08} \]Evaluating this gives approximately \(1.74 \times 10^5\).

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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 fundamental formula in chemistry that describes how the rate constant \( k \) of a reaction depends on temperature \( T \) and activation energy \( E_a \). It is written as: \[ k = A e^{-E_a / RT} \]where:
  • \( A \) is the pre-exponential factor, also known as the collision factor.
  • \( E_a \) is the activation energy, which is the energy barrier that reactants must overcome to form products.
  • \( R \) is the universal gas constant (8.314 J/mol路K).
  • \( T \) is the temperature in Kelvin.
This equation implies that as temperature increases or activation energy decreases, the rate constant \( k \) increases, leading to a faster reaction rate. Consequently, this helps in predicting how changes in conditions affect reaction rates.
Reaction rate
The reaction rate is a measure of how quickly a chemical reaction occurs. In essence, it illustrates the speed at which reactants turn into products. The rate can be influenced by various factors:
  • Concentration of reactants: Higher concentration often leads to a faster reaction.
  • Temperature: Increasing temperature generally increases the rate as molecules gain energy and collide more frequently.
  • Catalysts: These lower the activation energy, thus increasing the reaction rate without being consumed in the process.
  • Surface area: More exposed surface area can enhance the rate of reaction due to increased opportunities for collisions.
Understanding these factors can help predict and control reaction behaviors in practical applications.
Catalyst effect
Catalysts play a crucial role in increasing the reaction rate by lowering the activation energy of a reaction. For example, in the original exercise, a catalyst reduced the activation energy from 95 kJ/mol to 55 kJ/mol. This significant reduction means that less energy is required for the reactants to convert into products. By using the Arrhenius equation, the dramatic increase in the reaction rate can be calculated. For a given temperature, the rate of the catalyzed reaction can be several orders of magnitude faster than the uncatalyzed reaction. Catalysts are invaluable in both industrial and biological processes as they accelerate reactions while remaining unchanged, thus allowing them to be used repeatedly.
Temperature conversion
Converting temperature to Kelvin is essential when using the Arrhenius equation since it requires absolute temperature rather than Celsius. The conversion is straightforward: simply add 273.15 to the Celsius temperature.For instance, if the temperature is 25掳C, converting it to Kelvin means:\[ T = 25 + 273.15 = 298.15 \, \text{K} \]Likewise, for 125掳C:\[ T = 125 + 273.15 = 398.15 \, \text{K} \]This conversion ensures accurate calculations in the Arrhenius equation, which helps predict how the reaction rate varies with temperature, emphasizing the exponential effect temperature has on rate constants and subsequently, on reaction rates.

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Most popular questions from this chapter

(a) The gas-phase decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right), \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), At \(300^{\circ} \mathrm{C}\) the half-life for this process is two and a half days. What is the rate constant at this temperature?

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g),\) at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?(\mathbf{b})\) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol} ?\) (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C} ?\)

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?

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(\mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g)\)

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?
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