Problem 94 - Chemistry The Central Science In Si Units

Question

At a temperature of \(700 \mathrm{~K},\) the forward and reverse rate constants for the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) are \(k_{f}=1.8 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(k_{r}=0.063 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) What is the value of the equilibrium constant \(K_{c}\) at \(700 \mathrm{~K} ?(\mathbf{b})\) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of \(k_{f}=0.097 M^{-1} \mathrm{~s}^{-1}\) and \(k_{r}=2.6 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(800 \mathrm{~K}\) ?

Step-by-Step Solution

Answer

(a) \( K_c = 0.02857 \) at 700 K; (b) The forward reaction is endothermic.

Step 1: Understanding Equilibrium Constant

The equilibrium constant \( K_c \) can be found using the relationship between the forward \( k_f \) and reverse \( k_r \) rate constants: \( K_c = \frac{k_f}{k_r} \). This relationship holds for reactions in dynamic equilibrium.

Step 2: Calculating \( K_c \) at 700 K

Substitute the given values for \( k_f \) and \( k_r \) at 700 K into the equation. \[ K_c = \frac{1.8 \times 10^{-3} \text{ M}^{-1} \text{ s}^{-1}}{0.063 \text{ M}^{-1} \text{ s}^{-1}} \]Calculate the result to find \( K_c \).

Step 3: Solving for \( K_c \) at 700 K

Perform the division: \[ K_c = \frac{1.8 \times 10^{-3}}{0.063} = 0.02857 \] This is the equilibrium constant at 700 K.

Step 4: Comparing Rate Constants at Different Temperatures

Compare the equilibrium constants calculated at different temperatures to infer the reaction's thermodynamics. At 800 K, the constants are \( k_f = 0.097 \text{ M}^{-1} \text{ s}^{-1} \) and \( k_r = 2.6 \text{ M}^{-1} \text{ s}^{-1} \).

Step 5: Calculate \( K_c \) at 800 K

For 800 K, use the same formula: \[ K_c = \frac{0.097}{2.6} = 0.03731 \]. Compare this with the value at 700 K to analyze thermodynamic favorability.

Step 6: Analyzing Reaction Nature Based on \( K_c \) Values

If \( K_c \) increases with temperature, the forward reaction is endothermic since it becomes more favorable as temperature rises. Conversely, if \( K_c \) decreases, the forward reaction is exothermic.

Key Concepts

Dynamic Equilibrium

In a chemical reaction, dynamic equilibrium refers to the state where the rates of the forward and reverse reactions are equal. This doesn't mean the concentrations of reactants and products are equal, but rather, their levels remain constant over time. At this point, no net change occurs in the concentration of reactants and products.
Understanding dynamic equilibrium is crucial when calculating the equilibrium constant, because this concept implies that the relationship between the rate constants for forward and reverse reactions defines the equilibrium constant (\( K_c \)).

• At equilibrium, \( K_c = \frac{k_f}{k_r} \)
• Dynamic equilibrium allows one to predict the position of equilibrium and how it will respond to changes in conditions such as temperature and pressure.

This concept is foundational for determining how a reaction behaves under various conditions and helps in predicting the outcomes of changes in systemic parameters.

Rate Constants

Rate constants, denoted as \( k_f \) and \( k_r \), dictate how quickly reactions occur. The forward rate constant \( k_f \) applies to the reaction transforming reactants into products, while the reverse rate constant \( k_r \) describes the transformation from products back to reactants. These values are crucial for understanding how fast a reaction approaches equilibrium.
In our exercise, we determined the equilibrium constant using these rate constants:

• \( k_f = 1.8 \times 10^{-3} \text{ M}^{-1} \text{ s}^{-1} \)
• \( k_r = 0.063 \text{ M}^{-1} \text{ s}^{-1} \)

By taking the ratio \( \frac{k_f}{k_r} \), we evaluate the equilibrium constant \( K_c \). Rate constants are temperature-dependent, changing as temperature varies, which helps in analyzing reaction kinetics under different conditions.

Endothermic Reaction

An endothermic reaction is characterized by the absorption of heat. In these reactions, energy is taken in from the surroundings, typically leading to a drop in temperature.
For a forward endothermic reaction, the equilibrium constant \( K_c \) increases with an increase in temperature. This is because higher temperatures provide the necessary energy to break bonds and push the reaction towards the products. Considering the equilibrium constants calculated at different temperatures in our exercise, if \( K_c \) increases as the temperature rises from 700 K to 800 K, we conclude the reaction favors the formation of products more at higher temperatures, indicating endothermic behavior.

• Examples include photosynthesis and melting ice.
• Typical sign: Absorption of energy makes the surroundings feel cooler.

Recognizing endothermic reactions is crucial, especially in industrial processes where maintaining specific temperature settings is vital for optimal production.

Exothermic Reaction

Exothermic reactions release energy in the form of heat. These reactions typically result in an increase in the temperature of the surroundings.
Contrary to endothermic reactions, an exothermic reaction sees a decrease in \( K_c \) as temperature increases. This is because the reaction is already releasing energy, and an increase in temperature pushes the equilibrium back towards the reactants.
If we see a decrease in \( K_c \) with an increase in temperature from 700 K to 800 K, it indicates that the reaction is exothermic.

• Examples include combustion and respiration.
• Typical sign: The surroundings become warmer, and sometimes light is also emitted.

Understanding whether a reaction is exothermic or endothermic is crucial in predicting the reaction's response to changes in temperature and in designing processes that either harness or mitigate the released heat.