Problem 81 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) At \(25^{\circ} \mathrm{C}\) (298.15 K), the catalyst increases the rate of the reaction by a factor of approximately 148.14. (b) At \(125^{\circ} \mathrm{C}\) (398.15 K), the catalyst increases the rate of the reaction by a factor of approximately 19.44.

Step 1: Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we add 273.15 to the Celsius values. (a) For 25°C: \(T_1 = 25 + 273.15 = 298.15K\) (b) For 125°C: \(T_2 = 125 + 273.15 = 398.15K\)

Step 2: Calculate the rate constant ratio

We will determine the ratio of the rate constants in the presence and absence of a catalyst, which can be represented as: \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{Ae^{-\frac{E_a^{catalyzed}}{RT}}}{Ae^{-\frac{E_a^{uncatalyzed}}{RT}}} \] Since the collision factor (A) remains the same, it cancels out: \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-\frac{E_a^{catalyzed}}{RT}}}{e^{-\frac{E_a^{uncatalyzed}}{RT}}} \] Also, we need to substitute the values of activation energies given in the problem: Uncatalyzed activation energy (\(E_a^{uncatalyzed}\)) = 95 kJ/mol = 95,000 J/mol Catalyzed activation energy (\(E_a^{catalyzed}\)) = 55 kJ/mol = 55,000 J/mol

Step 3: Calculate the factor at each temperature

(a) For 25°C (298.15K) \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-\frac{55,000}{8.314 \times 298.15}}}{e^{-\frac{95,000}{8.314 \times 298.15}}} \] \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} \approx 148.14 \] (b) For 125°C (398.15K) \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-\frac{55,000}{8.314 \times 398.15}}}{e^{-\frac{95,000}{8.314 \times 398.15}}} \] \[ \frac{k_{catalyzed}}{k_{uncatalyzed}} \approx 19.44 \]

Step 4: Interpret the results

(a) At 25°C, the rate of the reaction will increase by a factor of approximately 148.14 with the addition of a catalyst. (b) At 125°C, the rate of the reaction will increase by a factor of approximately 19.44 with the addition of a catalyst.

Key Concepts

Catalyst

A catalyst is a substance that can significantly speed up the rate of a chemical reaction without being consumed in the process. It works by lowering the activation energy required for the reaction to occur, which allows more reactant molecules to have enough energy to reach the transition state.

In the given exercise, we saw that introducing a catalyst to a reaction lowered the activation energy from 95 kJ/mol to 55 kJ/mol. This considerable reduction in activation energy indicates that the catalyst provides an alternative pathway for the reaction, one that requires less energy. As a result, even at the same temperature, more molecules have sufficient energy to surpass the lowered activation energy barrier, thus increasing the rate of the reaction.

It is critical to note that while catalysts accelerate reaction rates, they do not affect the thermodynamics of the reaction, such as the equilibrium position, which remains unchanged. By providing a more efficient reaction path, catalysts are invaluable in many industrial processes, allowing for faster production rates and lower energy costs.

Rate Constant

The rate constant, often denoted by 'k', is a proportionality constant in the rate equation of a chemical reaction that provides the relationship between the reactant concentrations and the reaction rate at a given temperature. It reflects the rate at which a reaction proceeds and is influenced by factors including temperature, activation energy, and the presence of a catalyst.

In the exercise, we calculated the rate constant ratio between catalyzed and uncatalyzed reactions at different temperatures. As shown in the solution, the catalyst significantly increased the rate constant, which in turn increases the reaction rate. Since the rate constant is higher in the presence of a catalyst, it means that for every successful collision between reactant molecules, the reaction is much more likely to proceed, leading to faster reaction times.

Understanding how the rate constant varies with conditions like temperature and the presence of a catalyst is crucial in controlling the rates of reactions, which is especially important when designing industrial processes or in lab experiments.

Arrhenius Equation

The Arrhenius equation mathematically expresses the relationship between the rate constant 'k', the activation energy of the reaction 'Ea', and the temperature 'T'. It is given by the formula: \[ k = Ae^{-\frac{E_a}{RT}} \]where 'A' is known as the frequency factor or pre-exponential factor, 'Ea' is the activation energy, 'R' is the gas constant, and 'T' is the absolute temperature in Kelvin.

The Arrhenius equation is pivotal because it shows that even small changes in activation energy or temperature can have massive impacts on the reaction rate. As seen in the exercise, by exponentiating the negative ratio of activation energy over the product of the gas constant and temperature (\(-\frac{E_a}{RT}\)), the resulting rate constant 'k' can differ greatly with or without a catalyst. Moreover, it explains why reactions speed up with increasing temperature – as 'T' rises, the value of \(e^{-\frac{E_a}{RT}}\) increases, leading to a higher 'k' and thus a faster reaction rate.

The equation also demonstrates the profound effect of catalysts on reaction rates by significantly reducing the activation energy, 'Ea', thereby increasing the rate constant, 'k'. This relation allows chemists to predict and control the speed of chemical reactions effectively.