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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
The catalyst increases the rate of the reaction by a factor of approximately 243.27 at \(25^{\circ}\mathrm{C}\) and by a factor of approximately 44.21 at \(125^{\circ}\mathrm{C}\).

Step by step solution

01

Convert Temperatures to Kelvin

Convert both given temperatures to Kelvin by adding 273.15 to the temperatures in Celsius: \(T_1 = 25 + 273.15 = 298.15\,\text{K}\) \(T_2 = 125 + 273.15 = 398.15\,\text{K}\) Step 2: Calculate the rate constant of the uncatalyzed reaction and the catalyzed reaction at both temperatures using the Arrhenius equation. Note that the pre-exponential factor \(A\) remains the same, as the collision factor does not change.
02

Use Arrhenius Equation

For both reactions, the rate constant k can be calculated using the Arrhenius equation: \( k = Ae^{\frac{-E_a}{RT}} \) where \(A\) is the pre-exponential factor, \(R\) is the gas constant (8.314 J/mol K), and \(T\) is the temperature in Kelvin. For the uncatalyzed reaction: \( k_1 = Ae^{\frac{-E_a}{RT_1}} = Ae^{\frac{-95 \times 10^3}{8.314 \times 298.15}} \) For the catalyzed reaction: \( k_2 = Ae^{\frac{-E_a'}{RT_1}} = Ae^{\frac{-55 \times 10^3}{8.314 \times 298.15}} \) Repeat the calculations for \(T_2 = 398.15\,\text{K}\). Step 3: Calculate the factor by which the rate of the reaction increases due to the catalyst at each temperature.
03

Calculate Rate Increase Factor

Divide the rate constant of the catalyzed reaction (\(k_2\)) by the rate constant of the uncatalyzed reaction (\(k_1\)) at each temperature. This gives the factor by which the rate of the reaction increases due to the catalyst. At \(T_1\), the factor is: \( \frac{k_2}{k_1} = \frac{Ae^{\frac{-55 \times 10^3}{8.314 \times 298.15}}}{Ae^{\frac{-95 \times 10^3}{8.314 \times 298.15}}} \) At \(T_2\), the factor is: \( \frac{k_2}{k_1} = \frac{Ae^{\frac{-55 \times 10^3}{8.314 \times 398.15}}}{Ae^{\frac{-95 \times 10^3}{8.314 \times 398.15}}} \) Step 4: Simplify the expressions to obtain the final results.
04

Simplify Expressions

To find the rate increase factor, simplify the expressions by canceling A and applying properties of exponentials: At \(T_1\): \( \frac{k_2}{k_1} = e^{\frac{(-55 + 95) \times 10^3}{8.314 \times 298.15}} \) At \(T_2\): \( \frac{k_2}{k_1} = e^{\frac{(-55 + 95) \times 10^3}{8.314 \times 398.15}} \) Step 5: Evaluate the expressions to find the factors by which the rate of the reaction increases due to the catalyst at each temperature.
05

Evaluate Expressions

For \(T_1 = 298.15\,\text{K}\): \( \frac{k_2}{k_1}= e^{\frac{40 \times 10^3}{8.314 \times 298.15}} \approx 243.27 \) For \(T_2 = 398.15\,\text{K}\): \( \frac{k_2}{k_1}= e^{\frac{40 \times 10^3}{8.314 \times 398.15}} \approx 44.21 \) Conclusion: The catalyst increases the rate of the reaction by a factor of approximately 243.27 at 25°C and by a factor of approximately 44.21 at 125°C.

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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 used in chemistry to describe how the rate of a chemical reaction depends on temperature. It is represented as:\[k = Ae^{-\frac{E_a}{RT}}\]Here, \(k\) is the rate constant that indicates the speed of the reaction, \(A\) is the pre-exponential factor which relates to the frequency of collisions and geometry of molecular orientations, \(E_a\) is the activation energy - the minimum energy required for a reaction to occur, \(R\) is the universal gas constant \(8.314 \text{ J/mol K}\), and \(T\) is the temperature in Kelvin.
When the activation energy \(E_a\) is lower, this means that less energy is required for the reaction to happen. Thus, using a catalyst to lower \(E_a\) can significantly speed up the reaction. Since the Arrhenius equation is exponential, small changes in activation energy or temperature can lead to large changes in reaction rates. This is why keeping the pre-exponential factor \(A\) constant is helpful for comparing different scenarios. However, in real scenarios, \(A\) might change based on the specific catalyst and reaction environment.
Catalyst Effect
A catalyst is an agent that increases the rate of a chemical reaction without being consumed in the process. Catalysts work by lowering the activation energy \(E_a\), which is the energy needed to start a reaction, and thus increasing the reaction rate.
In the example problem, the activation energy of the uncatalyzed reaction is \(95 \text{ kJ/mol}\) and upon adding a catalyst it decreases to \(55 \text{ kJ/mol}\). This results in a faster reaction because lower energy is needed to reach the transition state.
Catalysts do not alter the thermodynamics of a reaction (i.e., the equilibrium position remains the same), but they do influence how quickly equilibrium is reached. They achieve this by providing an alternative reaction pathway with a lower energy barrier, allowing more reactant molecules to come together with sufficient energy at given temperature conditions. This is crucial in both industrial applications and biological systems where specific, efficient pathways must be controlled and optimized.
Reaction Rate
The reaction rate is a measure of how quickly a reactant is consumed or a product is formed in a chemical reaction. Several factors can influence reaction rates:
  • Temperature - An increase in temperature generally increases the reaction rate.
  • Concentration - Higher concentrations of reactants typically lead to faster reaction rates.
  • Catalysts - The presence of a catalyst lowers the activation energy needed, thus speeding up the rate.
  • Surface Area - More exposed surface area can accelerate reactions, as seen with powdered vs. solid forms of reactants.
The concept of reaction rate is important for understanding the efficiency and practicality of chemical processes. In our example, the reaction rate is significantly accelerated by a catalyst, as shown by the calculated factors (e.g., a factor of 243.27 at 25°C). Such enhancement allows for processes to proceed swiftly and with less energy input, which is especially favorable in industrial contexts where efficiency is paramount. This demonstrates the importance of tuning conditions such as temperature and catalyst usage to control the speed of chemical synthesis.

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Most popular questions from this chapter

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) 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 \(2.0 \mathrm{~L} .(\mathbf{a})\) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?\) (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 \(70{ }^{\circ} \mathrm{C}\) ?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.025 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? (b) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is \(10 \mathrm{kPa}\) per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

As shown in Figure 14.23 , the first step in the heterogeneous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed explanation for the "sticking" of ethylene to a metal surface is the interaction of the electrons in the \(\mathrm{C}-\mathrm{C} \pi\) bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene?

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O}\). At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1} .\) Calculate the half- life at this temperature. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

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