Problem 12 - Chemistry The Science in Context

Question

Suppose at \(298 \mathrm{K}\) the reaction \(\mathrm{C}(g) \rightleftharpoons \mathrm{D}(g)\) has a forward rate constant of \(5 / \mathrm{s}\) and a reverse rate constant of 10/s. What is the value of the equilibrium constant of the reaction at \(298 \mathrm{K} ?\)

Step-by-Step Solution

Answer

Answer: The equilibrium constant (K) of the reaction at 298 K is 1/2.

Step 1: Identify the given values

We are given the following values: - Forward rate constant (k_forward) = 5/s - Reverse rate constant (k_reverse) = 10/s - Temperature (T) = 298 K

Step 2: Calculate the equilibrium constant

Using the relationship between the rate constants and the equilibrium constant, we can find the equilibrium constant (K) of the reaction: K = k_forward / k_reverse Plug in the given values: K = (5/s) / (10/s)

Step 3: Simplify the expression

We can simplify the expression by dividing the constants: K = 5/10 K = 1/2

Step 4: Write the final answer

The equilibrium constant (K) of the reaction at 298 K is 1/2.

Key Concepts

Rate Constants

In chemical reactions, rate constants are crucial for understanding how fast a reaction proceeds. For any given reaction, there are typically two rate constants to consider: the forward rate constant

• The forward rate constant ( \( k_{\text{forward}} \) ) describes how quickly the reactants transform into products. A larger value indicates a faster forward reaction.
• The reverse rate constant ( \( k_{\text{reverse}} \) ) corresponds to the speed at which the products revert back into reactants. Similarly, a larger reverse rate constant denotes a quicker reverse process.

Rate constants are temperature-dependent. This means they can change if the temperature varies. The units of rate constants can also provide insights into the reaction order. In this exercise, both constants are expressed as per second (\(/ \text{s}\)). This implies a first-order reaction where the rate depends on the concentration of one reactant.

Reaction Rate

The reaction rate gives a measure of the speed at which reactants are converted into products in a chemical reaction. It is determined by the concentration of reactants, the rate constant, and in some cases, the presence of a catalyst. Reaction rates can change based on environmental factors, such as:

• Temperature: Typically, increasing temperature results in a faster reaction rate.
• Concentration: Higher concentrations of reactants can often speed up reactions.
• Catalysts: These substances can significantly boost the reaction rate without being consumed in the process.

Moreover, the reaction rate is a critical factor in reaching chemical equilibrium. If both the forward and reverse reactions occur at the same pace, the system achieves a state of balance, and the concentration of reactants and products remains constant.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions happen at identical rates. This means the concentrations of reactants and products remain constant over time, even though the reactions continue to occur. At equilibrium, the equilibrium constant (\( K \)) describes the ratio of the product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. In this exercise, \( K \) is expressed as a ratio of the rate constants:\[K = \frac{k_{\text{forward}}}{k_{\text{reverse}}} \]When \( K > 1 \), the products are favored at equilibrium. Conversely, if \( K < 1 \), the reactants are favored. In our example, \( K = 1/2 \), indicating that at equilibrium, the reactants are more prevalent than the products. Achieving chemical equilibrium is essential in processes where constant concentrations over time are desired, such as in industrial synthesis and biochemical pathways.