Problem 79 - Chemistry The Molecular Science

Question

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.

Step-by-Step Solution

Answer

The rate constant increases by a factor of approximately 37.9.

Step 1: 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.

Step 2: 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:\[\frac{k'}{k} = \frac{A e^{-\frac{206000}{RT}}}{A e^{-\frac{215000}{RT}}} = e^{-\frac{206000-215000}{RT}} = e^{\frac{9000}{RT}}\]

Step 3: Convert Temperature to Kelvin

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

Step 4: 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.

Key Concepts

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.