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Consider the following reaction, which takes place in a single elementary step: $$2 \mathrm{~A}+\mathrm{B} \underset{k_{-1}}{\frac{k_{1}}{\longrightarrow}} \mathrm{A}_{2} \mathrm{~B}$$ If the equilibrium constant \(K_{\mathrm{c}}\) is 12.6 at a certain temperature and if \(k_{\mathrm{r}}=5.1 \times 10^{-2} \mathrm{~s}^{-1},\) calculate the value of \(k_{\mathrm{f}}\).

Short Answer

Expert verified
The forward rate constant (\(k_{f}\) or \(k_{1}\)) for the given reaction is approximately \(6.31 \times 10^{-2} \mathrm{s}^{-1}\)

Step by step solution

01

Understand Relationship Between Constants

From the equilibrium constant and rate constants, we have the equation \(K_{c}=\frac{k_{\mathrm{1}}}{k_{-1}}\), where \(k_{1}\) is the forward rate constant, \(k_{-1}\) is the reverse rate constant, and \(K_{c}\) is the equilibrium constant.
02

Relate Reaction Rate and Equilibrium Constants

Given that \(k_{\mathrm{r}}=k_{\mathrm{1}}+k_{\mathrm{-1}}=5.1 \times 10^{-2} \mathrm{~s}^{-1}\), we need to use this relationship to solve for the reverse reaction rate constant \(k_{-1}\). This will involve a bit of algebraic manipulation.
03

Calculate Reverse Rate Constant

By rearranging the equation introduced in the previous step, we can find \(k_{-1}\). Substituting the value of \(k_{1}\) from the equation \(K_{c}=\frac{k_{\mathrm{1}}}{k_{-1}}\) into \(k_{r}=k_{1}+k_{-1}\), we get \(k_{-1}=\frac{k_{\mathrm{r}}}{K_{c}+1}\). Now we substitute the given values into this equation.
04

Calculate the Forward Rate Constant

With the reverse rate constant (\(k_{-1}\)) calculated, it can be used to find the forward rate constant (\(k_{1}\)) using the equation \(K_{c}=\frac{k_{\mathrm{1}}}{k_{-1}}\), by rearranging to \(k_{\mathrm{1}}=K_{c} \times k_{-1}\) and substituting the values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant \(K_{c}\) is a critical value in chemical kinetics that provides insight into the balance between reactants and products at equilibrium in a reversible chemical reaction. An equilibrium constant is unique for every reaction and depends solely on temperature. For the given reaction \((2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{A}_{2} \mathrm{~B})\), the equilibrium constant is given as \(K_{c} = 12.6\), reflecting the product ratio concentrations to reactant concentrations when the rates are equal, indicating balance.Key points to remember about equilibrium constants include:
  • A large \(K_{c}\) (greater than 1) suggests that at equilibrium, the reaction favours product formation.
  • A small \(K_{c}\) (less than 1) implies that reactants are predominant at equilibrium.
  • The units of \(K_{c}\) depend on the nature of the reaction, but it is dimensionless if considering activities.
This value provides vital information to chemists, helping predict whether a reaction will yield more products or retain higher amounts of reactants.
Rate Constants
Rate constants are the numerical values that define the speed of a reaction under given conditions. For a chemical reaction, there are distinct rate constants for the forward and reverse reactions. These constants help in determining how fast the reactants turn into products and vice versa. The overall speed or rate of a chemical reaction is directly proportional to the product of these rate constants and concentrations. Understanding this relationship helps in:
  • Developing better catalysts.
  • Predicting the behaviour of reactions under different conditions.
  • Engineering processes for maximal efficiency and yield.
Rate constants are affected by factors such as:
  • Temperature: Increasing the temperature generally increases the rate constants.
  • Pressure: Significant mainly in gaseous reactions.
Overall, they play an essential role in understanding kinetic properties and behaviour of a reaction.
Forward Rate Constant
The forward rate constant, \(k_{1}\), for a given reaction indicates the rate at which reactants are converted to products. In the context of chemical equilibrium and kinetics, \(k_{1}\) is crucial for determining how quickly the reaction progresses from reactants to equilibrium products.For the reaction \((2 \mathrm{~A} + \mathrm{B} \rightarrow \mathrm{A}_{2} \mathrm{~B})\), \(k_{1}\) is utilized to relate to the equilibrium constant and reverse rate constant. The equation used is \(K_{c} = \frac{k_{1}}{k_{-1}}\), which allows chemists to derive the forward rate constant if the equilibrium constant and reverse rate constant are known.By determining \(k_{1}\), one can:
  • Evaluate the speed of product formation.
  • Understand the influence of changing conditions on the reaction speed.
The understanding of \(k_{1}\) helps in optimizing reaction conditions for desired outcomes, particularly in industrial setups.
Reverse Rate Constant
The reverse rate constant, \(k_{-1}\), provides insight into the speed at which products revert back to reactants in a reversible reaction. It is an integral part of understanding the full scope of a reaction's kinetics, as it balances the forward rate constant.For the given reversible reaction, the equilibrium constant equation \(K_{c} = \frac{k_{1}}{k_{-1}}\) helps us express the relationship between the forward and reverse reactions. Calculating \(k_{-1}\) is a critical step, where the offshoot of an equation like \(k_{r} = k_{1} + k_{-1}\) assists in simplifying calculations and achieving accurate values.The reverse rate constant influences:
  • Predictive modeling of how long a reaction takes to reach equilibrium.
  • Understanding reversibility and energy dynamics in reactions.
In the context of equilibrium and reaction rate analysis, \(k_{-1}\) is just as essential as \(k_{1}\) to grasp the big picture of a reaction's behaviour.

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Most popular questions from this chapter

List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it?

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