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A biological catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ} / \mathrm{mol}\) to \(206 \mathrm{~kJ} / \mathrm{mol}\). Calculate by what factor the rate constant, \(k,\) would increase at \(25^{\circ} \mathrm{C}\). Assume that the frequency factors \((A)\) are the same for the uncatalyzed and catalyzed reactions.

Short Answer

Expert verified
The rate constant increases by a factor of approximately 37.9.

Step by step solution

01

Understanding the Arrhenius Equation

The Arrhenius equation describes the temperature dependence of reaction rates and is expressed as: \[ k = A e^{-\frac{E_a}{RT}} \]where \(k\) is the rate constant, \(A\) is the frequency 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

Calculate Rate Constant Ratio with Catalyst

Given the activation energy for the uncatalyzed reaction \(E_a = 215\, \text{kJ/mol}\) and for the catalyzed reaction \(E'_a = 206 \, \text{kJ/mol}\), we need to compare the rate constants. The formula for the ratio of rate constants is:\[\frac{k'}{k} = \frac{A e^{-\frac{206000}{RT}}}{A e^{-\frac{215000}{RT}}} = e^{-\frac{206000-215000}{RT}} = e^{\frac{9000}{RT}}\]
03

Convert Temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin:\[ T = 25 + 273.15 = 298.15\, K \]
04

Calculate Using Arrhenius Equation

Substitute \(R = 8.314\, \text{J/mol K}\) and \(T = 298.15\, K\) into the equation:\[\frac{k'}{k} = e^{\frac{9000}{8.314 \times 298.15}}\]This becomes:\[\frac{k'}{k} = e^{3.63}\]Calculating \(e^{3.63}\) gives approximately 37.9.

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Key Concepts

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

Activation Energy
Activation energy is a crucial concept in understanding chemical reactions. It refers to the minimum energy required to initiate a reaction. Imagine activation energy as the hurdle a chemical must overcome to transform into new substances. In essence, it represents the energy barrier to a reaction.

In a reaction with high activation energy, molecules need more energy to collide and transform effectively. This means that fewer molecules will have sufficient energy to react, slowing down the reaction rate. Conversely, a low activation energy implies that even less energetic molecules can successfully react, leading to a faster reaction.
  • Catalysts serve the important purpose of lowering this energy barrier, allowing more molecules to overcome it and thus increasing the reaction rate.
  • In the given exercise, the biological catalyst reduced the activation energy from 215 kJ/mol to 206 kJ/mol, facilitating a quicker reaction.
Understanding activation energy is key to controlling reaction speeds and designing efficient chemical processes.
Arrhenius Equation
The Arrhenius Equation is a vital formula in chemistry that quantifies the effect of temperature on reaction rates. It is expressed as:\[ k = A e^{\frac{-E_a}{RT}} \]where:
  • \(k\) is the rate constant, a measure of the speed of the reaction.
  • \(A\) is the frequency factor, accounting for the number of times that reactants approach each other per unit time.
  • \(E_a\) is the activation energy, the energy barrier previously mentioned.
  • \(R\) is the universal gas constant \(8.314 \, \text{J/mol K}\).
  • \(T\) is the temperature in Kelvin, which influences molecular collisions.
The equation beautifully illustrates how even small changes in temperature or activation energy can significantly impact the rate constant, \(k\). Lower activation energy, as often achieved by catalysts, dramatically increases \(k\), thereby speeding up the reaction.
Rate Constants
Rate constants are fundamental parameters in the study of reaction kinetics. In simple terms, the rate constant \(k\) indicates how fast a reaction proceeds. It is involved in the rate law of a reaction, which directly relates the reaction rate to the concentration of reactants.

The value of \(k\) can be influenced by various factors, such as temperature and the presence of a catalyst. According to the Arrhenius equation, when activation energy decreases, as it does when a catalyst is present, the rate constant increases.
  • In the exercise above, the factor by which the rate constant increased due to the catalyst can be calculated with the formula: \(\frac{k'}{k} = e^{\frac{9000}{RT}}\).
  • Given the reduced activation energy from 215 kJ/mol to 206 kJ/mol, the calculation showed that the rate constant increased by approximately 37.9 at 25°C.
Knowing the rate constant allows chemists to predict how a reaction will proceed over time, enabling better control and optimization of chemical processes.

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Most popular questions from this chapter

Make an Arrhenius plot and calculate the activation energy for the gas-phase reaction \(2 \mathrm{NOCl}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) \begin{tabular}{lc} \hline\(T(\mathrm{~K})\) & Rate Constant \(\left(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline \(400 .\) & \(6.95 \times 10^{-4}\) \\ \(450 .\) & \(1.98 \times 10^{-2}\) \\ \(500 .\) & \(2.92 \times 10^{-1}\) \\ \(550 .\) & 2.60 \\ \(600 .\) & 16.3 \\ \hline \end{tabular}

Suppose a reaction rate constant has been measured at two different temperatures, \(T_{1}\) and \(T_{2}\), and its values are \(k_{\perp}\) and \(k_{2}\), respectively. (a) Write the Arrhenius equation at each temperature. (b) By combining these two equations, derive an expression for the ratio of the two rate constants, \(k_{1} / k_{2}\). Use this expression to answer the next four questions.

For the reaction $$ 2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) $$ make qualitatively correct plots of the concentrations of \(\mathrm{NO}_{2}(\mathrm{~g}), \mathrm{NO}(\mathrm{g}),\) and \(\mathrm{O}_{2}(\mathrm{~g})\) versus time. Draw all three graphs on the same axes; assume that you start with \(\mathrm{NO}_{2}(\mathrm{~g})\) at a concentration of \(1.0 \mathrm{~mol} / \mathrm{L}\). Explain how you would determine, from these plots, (a) the initial rate of the reaction. (b) the final rate (that is, the rate as time approaches infinity).

Experiments show that the reaction of nitrogen dioxide with fluorine $$ 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{FNO}_{2}(\mathrm{~g}) $$ has the rate law $$ \text { Rate }=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{F}_{2}\right] $$ and the reaction is thought to occur in two steps: $$ \begin{array}{l} \text { Step } 1: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \\ \text { Step } 2: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g}) \end{array} $$ (a) Show that the sum of this sequence of reactions gives the balanced equation for the overall reaction. (b) Which step is rate-determining?

Cyclopropane isomerizes to propene when heated. Rate constants for the reaction cyclopropane \(\longrightarrow\) propene are \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470.0^{\circ} \mathrm{C}\) and \(1.02 \times 10^{-3} \mathrm{~s}^{-1}\) at \(510.0^{\circ} \mathrm{C}\) (a) Calculate the activation energy, \(E_{\mathrm{a}},\) for this reaction. (b) Calculate how long it takes at \(500 .{ }^{\circ} \mathrm{C}\) for the concentration of cyclopropane to drop from \(0.10 \mathrm{M}\) to \(0.023 \mathrm{M}\).

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