Problem 68
Question
How would the value of the equilibrium constant for a one-step reaction calculated as \(k_{\mathrm{f}} / k_{\mathrm{r}}\) compare with the value calculated from the concentrations of all substances present at equilibrium?
Step-by-Step Solution
Verified Answer
The value of the equilibrium constant, K, calculated as \(k_f / k_r\) (ratio of forward and reverse rate constants) is equivalent to the value calculated from the concentrations of all substances present at equilibrium. This is confirmed by the equality \(\frac{k_{f}}{k_{r}} = \frac{[\text{C}]^c_{eq} [\text{D}]^d_{eq}}{[\text{A}]^a_{eq} [\text{B}]^b_{eq}}\).
1Step 1: Understanding the Equilibrium Constant
The equilibrium constant, K, is a measure of the extent to which a chemical reaction proceeds to completion. It is the ratio of the rate constants of the forward reaction (kf) and the reverse reaction (kr). The equilibrium constant can be calculated using the formula:
\[K = \frac{k_{f}}{k_{r}}\]
where K is the equilibrium constant, kf is the forward rate constant, and kr is the reverse rate constant.
2Step 2: Writing the Reaction Quotient (Q) Expression
For a balanced chemical reaction, the ratio of the concentrations of the products to the concentrations of the reactants can be expressed as the reaction quotient, Q. It is a measure of the state of a reaction at any given moment. If the reaction is a one-step reaction represented as:
\[\text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD}\]
the reaction quotient, Q, can be calculated using the formula:
\[Q = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b}\]
where [A], [B], [C], and [D] are the concentrations of the substances A, B, C, and D, respectively, and a, b, c, and d are their stoichiometric coefficients.
3Step 3: Understanding the Relationship between Q and K
At equilibrium, Q becomes equal to K. This means that the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium is equal to the ratio of the forward and reverse rate constants. Thus, we can write:
\[K = \frac{[\text{C}]^c_{eq} [\text{D}]^d_{eq}}{[\text{A}]^a_{eq} [\text{B}]^b_{eq}}\]
where [A]eq, [B]eq, [C]eq, and [D]eq are the concentrations of the substances A, B, C, and D at equilibrium, respectively.
4Step 4: Comparing the two ways of calculating K
Now that we have the expressions for K from both kf/kr and the concentrations at equilibrium, we can compare the two ways of calculating K. The value of K, calculated as kf/kr, should be equal to the value calculated from the concentrations at equilibrium. To visualize this, we can equate the two expressions for K:
\[\frac{k_{f}}{k_{r}} = \frac{[\text{C}]^c_{eq} [\text{D}]^d_{eq}}{[\text{A}]^a_{eq} [\text{B}]^b_{eq}}\]
This equation confirms that the value of the equilibrium constant K calculated using the forward and reverse rate constants (kf/kr) is equivalent to the value calculated using the concentrations of all the substances present at equilibrium.
Key Concepts
Chemical EquilibriumReaction Quotient (Q)Rate ConstantsStoichiometry
Chemical Equilibrium
Imagine a see-saw perfectly balanced with children of equal weight on both sides. Similarly, in a chemical reaction, this state of balance is called chemical equilibrium. It occurs when the rate of the forward reaction (where reactants form products) equals the rate of the reverse reaction (where products revert to reactants). The important thing to note is, even though the rates are equal, the actual amounts of reactants and products do not have to be the same.
At equilibrium, no net change in the concentrations of reactants and products occurs, yet molecules are still moving back and forth between being reactants and products. This dynamic process is essential for understanding how reactions behave when they're left undisturbed over time and is crucial for fields like pharmacology, environmental science, and engineering.
At equilibrium, no net change in the concentrations of reactants and products occurs, yet molecules are still moving back and forth between being reactants and products. This dynamic process is essential for understanding how reactions behave when they're left undisturbed over time and is crucial for fields like pharmacology, environmental science, and engineering.
Reaction Quotient (Q)
The reaction quotient (Q) acts as a 'snapshot' of a reaction at any point in time. Think of it as a progress report of the reaction, telling you how far along the reaction is at that moment.
To calculate Q, you use an expression similar to that for the equilibrium constant (K), but rather than using the concentrations at equilibrium, you use the current concentrations of the reactants and products. If Q = K, the reaction is at equilibrium. If Q > K, the reaction will shift left, favoring reactants. Conversely, if Q < K, the reaction will shift right, favoring products. Understanding Q can help predict the direction of the reaction's shift to reach equilibrium.
To calculate Q, you use an expression similar to that for the equilibrium constant (K), but rather than using the concentrations at equilibrium, you use the current concentrations of the reactants and products. If Q = K, the reaction is at equilibrium. If Q > K, the reaction will shift left, favoring reactants. Conversely, if Q < K, the reaction will shift right, favoring products. Understanding Q can help predict the direction of the reaction's shift to reach equilibrium.
Rate Constants
When it comes to chemical reactions, speed matters. The rate constants, kf (forward) and kr (reverse), are like the accelerator and brake pedals of a car that control the speed of the forward and reverse reactions, respectively.
These constants are determined experimentally and reflect how quickly a reaction proceeds. Factors such as temperature, pressure, and the presence of a catalyst can influence these constants. Understanding the rate constants gives us insight into the dynamics of a reaction and ultimately, allows us to calculate the equilibrium constant (K). Therefore, they're a big deal in figuring out the potential outcome and efficiency of a chemical process.
These constants are determined experimentally and reflect how quickly a reaction proceeds. Factors such as temperature, pressure, and the presence of a catalyst can influence these constants. Understanding the rate constants gives us insight into the dynamics of a reaction and ultimately, allows us to calculate the equilibrium constant (K). Therefore, they're a big deal in figuring out the potential outcome and efficiency of a chemical process.
Stoichiometry
Let's talk about the recipe of a reaction, known as stoichiometry. It involves the precise calculation of reactant and product quantities in a chemical equation. Think of it as a baker measuring ingredients to get a perfect cake.
The coefficients in a chemical equation represent the stoichiometry of the reaction. They tell you the proportions of reactants needed to form products. When a chemical reaction is at equilibrium, the stoichiometry is crucial for understanding the relationship between the concentration of each substance and the equilibrium constant (K). Mastery of stoichiometry is not just academic; it's a critical tool for any chemical engineer, pharmacist, or environmental scientist looking to predict the outcomes of reactions in real-world situations.
The coefficients in a chemical equation represent the stoichiometry of the reaction. They tell you the proportions of reactants needed to form products. When a chemical reaction is at equilibrium, the stoichiometry is crucial for understanding the relationship between the concentration of each substance and the equilibrium constant (K). Mastery of stoichiometry is not just academic; it's a critical tool for any chemical engineer, pharmacist, or environmental scientist looking to predict the outcomes of reactions in real-world situations.
Other exercises in this chapter
Problem 66
Write the balanced chemical equation for the reaction that goes with the equilibrium constant \(K_{\mathrm{eq}}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}
View solution Problem 67
An 8.00-L reaction vessel at \(491^{\circ} \mathrm{C}\) contains \(0.650\) mole of \(\mathrm{H}_{2}, 0.275\) mole of \(\mathrm{I}_{2}\), and \(3.00\) moles of H
View solution Problem 69
Suppose a reaction is at equilibrium and you then disturb the equilibrium by adding reactants. What happens to the value of \(K_{\text {eq }}\) ? Explain your a
View solution Problem 70
State Le Châtelier's principle using the words undo and partially.
View solution