Problem 64 - Chemistry: Principles and Reactions

Question

Cold-blooded animals decrease their body temperature in cold weather to match that of their environment. The activation energy of a certain reaction in a cold-blooded animal is \(65 \mathrm{~kJ} / \mathrm{mol} .\) By what percentage is the rate of the reaction decreased if the body temperature of the animal drops from \(35^{\circ} \mathrm{C}\) to \(22^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: The rate of the reaction decreases by approximately \(34.045\%\) when the body temperature of the animal drops from \(35^{\circ}\mathrm{C}\) to \(22^{\circ}\mathrm{C}\).

Step 1: Understand the Arrhenius equation and given information

The Arrhenius equation is given by \(k=A e^{-\frac{E_{a}}{RT}}\), where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_{a}\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. We are given the activation energy \(E_a = 65 \mathrm{~kJ/mol}\) and we are asked to find the percentage decrease in the rate of reaction when the temperature drops from \(35^{\circ}\mathrm{C}\) to \(22^{\circ}\mathrm{C}\).

Step 2: Convert the temperatures to Kelvin

We need to work with temperatures in Kelvin. To convert Celsius to Kelvin, we add 273.15. So, \(35^{\circ}\mathrm{C} = 35 + 273.15 = 308.15\mathrm{K}\) and \(22^{\circ}\mathrm{C} = 22 + 273.15 = 295.15\mathrm{K}\).

Step 3: Calculate the rate constants at both temperatures

Using the Arrhenius equation, we can find the rate constant at both temperatures. However, to avoid dealing with the pre-exponential factor \(A\), we can take ratios of the rate constants as follows: \(\frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a}}{R T_2}}}{A e^{-\frac{E_{a}}{R T_1}}}\) Where \(k_1\) is the rate constant at \(T_1 = 308.15\mathrm{K}\) and \(k_2\) is the rate constant at \(T_2 = 295.15\mathrm{K}\). Notice that \(A\) cancels out in this equation.

Step 4: Solve for the rate constant ratio

First, we convert the activation energy to \(\mathrm{J/mol}\): \(E_a = 65\mathrm{~kJ/mol} \times 1000 = 65000\mathrm{~J/mol}\). Now, we plug the given values into the equation and use the gas constant \(R = 8.314\mathrm{J/(mol\cdot K)}\): \(\frac{k_2}{k_1} = \frac{e^{-\frac{65000}{8.314 \cdot 295.15}}}{e^{-\frac{65000}{8.314 \cdot 308.15}}} = 0.65955\)

Step 5: Calculate the percentage decrease in the rate of reaction

Now that we have the ratio of the rate constants, we can find the percentage decrease in the rate of reaction as follows: Percentage decrease \(= \frac{k_1 - k_2}{k_1} \times 100 = \frac{k_1 - 0.65955k_1}{k_1} \times 100 = (1 - 0.65955) \times 100 = 34.045\%\) The rate of the reaction decreases by approximately \(34.045\%\) when the body temperature of the animal drops from \(35^{\circ}\mathrm{C}\) to \(22^{\circ}\mathrm{C}\).

Key Concepts

Activation Energy

Activation energy (\(E_a\)) is a fundamental concept in chemical kinetics, referring to the minimum amount of energy required for a chemical reaction to occur. It's like the initial push needed to start a reaction, much like pushing a boulder up a hill before it can roll down. The higher the activation energy, the slower the reaction tends to be, as fewer molecules have the necessary energy to overcome this barrier at a given temperature.

In the context of our exercise, the activation energy for the reaction occurring in a cold-blooded animal is given as 65 kJ/mol. This value is crucial as it directly influences the rate at which the reaction will proceed. In biological systems, activation energy is paramount as it determines the speed of metabolic reactions, which in turn affects the organism's overall function. Knowing how activation energy relates to reaction rates allows us to understand how variations in temperature, like those experienced by cold-blooded animals in different climates, can affect the chemistry of life.

Chemical Kinetics

Chemical kinetics involves the study of reaction rates and the factors that affect them, such as temperature, pressure, concentration, and the presence of catalysts. It helps us understand the speed at which chemical reactions occur and how they can be controlled or optimized. The Arrhenius equation, featured in our exercise, is a key equation in chemical kinetics. It shows the relationship between the rate constant (\(k\)) and the temperature (\(T\)). As the equation suggests, even a slight change in temperature can have a significant impact on the reaction rate, due to the exponential dependence on the ratio of activation energy to temperature (\(E_a/T\)).

Our exercise demonstrates this relationship by comparing the rate of a biochemical reaction at two different body temperatures of a cold-blooded animal. The result shows how sensitive reaction rates can be to temperature changes, emphasizing the importance of temperature regulation in living organisms, and underlining how chemical kinetics can predict and explain these effects.

Temperature Conversion

Temperature conversion is a fundamental step in many scientific calculations, especially when using the Arrhenius equation. Since this equation requires the temperature in Kelvin, converting from Celsius or Fahrenheit to Kelvin is necessary. The conversion allows for consistent measurements across different scales and avoids errors in calculations such as determining reaction rates. It's quite straightforward: to go from Celsius to Kelvin, one must add 273.15 to the Celsius temperature.

In our exercise, body temperatures were initially given in Celsius and had to be converted to Kelvin before we could apply the Arrhenius equation. This step is crucial; without it, our subsequent calculations would be incorrect. Temperature conversion ensures that all temperature-dependent calculations reflect the absolute thermal energy present in the system, which is essential for accuracy in scientific work.