Chapter 6: Problem 29
\(\Delta G^{\circ}\) for the reaction: \(\mathrm{X}+\mathrm{Y} \rightleftharpoons \mathrm{C}\) is \(-4.606\) kcal at \(1000 \mathrm{~K}\). The equilibrium constant for the reverse mode of the reaction is (a) 100 (b) 10 (c) \(0.01\) (d) \(0.1\)
Short Answer
Expert verified
(c) 0.01
Step by step solution
01
- Understanding the Gibbs Free Energy Change
The Gibbs free energy change \(\Delta G^\circ\) for a reaction at standard conditions is related to the equilibrium constant \(K\) of the reaction by the following equation: \(\Delta G^\circ = -RT \ln K\), where \(R\) is the gas constant and \(T\) is the temperature in Kelvin. To find the equilibrium constant \(K\) for the reverse reaction, we need to plug in the given values into the equation.
02
- Converting \(\Delta G^\circ\) to SI Unit Joules
First, we need to convert the given Gibbs free energy change from kcal to J since the gas constant \(R\) is typically given in J/(mol\cdot K). The conversion factor is that 1 kcal equals 4184 J. So the given \(\Delta G^\circ\) will be \(\Delta G^\circ = -4.606 \times 4184 \;\text{J/mol}\).
03
- Calculating the Equilibrium Constant \(K\) for the Reverse Reaction
Using the equation \(\Delta G^\circ = -RT \ln K\), we can rearrange it to solve for \(K\) as follows: \(K = e^{-(\Delta G^\circ/(RT))}\). We need to insert the values for \(\Delta G^\circ\) in J/mol, the gas constant \(R = 8.314 \text{J/(mol\cdot K)}\), and the temperature \(T = 1000 \text{K}\). After the calculation, we can determine the value of \(K\) for the reverse reaction.
04
- Finding the Equilibrium Constant for the Reverse Reaction
The equilibrium constant for the reverse reaction is simply the reciprocal of the equilibrium constant for the forward reaction. Knowing the equilibrium constant for the forward reaction as \(K_{forward}\), we can find the reverse equilibrium constant using \(K_{reverse} = 1/K_{forward}\).
05
- Determining the Correct Answer
After performing the calculations, if the determined \(K_{reverse}\) matches one of the provided options (a) 100, (b) 10, (c) 0.01, or (d) 0.1, that option is the correct answer.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thermodynamics
Thermodynamics is a branch of physical science that deals with the relationships between heat and other forms of energy. In particular, it describes how thermal energy is converted to and from other forms of energy and how it affects matter. The key concept here is the flow of energy and how it governs the physical and chemical changes in matter.
Understanding thermodynamics is crucial when studying reactions, such as the one involving substances X, Y, and C. The Gibbs Free Energy, named after Josiah Willard Gibbs, is a thermodynamic potential that can be used to predict the direction of a chemical reaction and determine whether it will occur spontaneously under constant pressure and temperature conditions.
Understanding thermodynamics is crucial when studying reactions, such as the one involving substances X, Y, and C. The Gibbs Free Energy, named after Josiah Willard Gibbs, is a thermodynamic potential that can be used to predict the direction of a chemical reaction and determine whether it will occur spontaneously under constant pressure and temperature conditions.
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction proceeds forward and backward at the same rate, resulting in no net change in the concentration of reactants and products over time. At equilibrium, the reactants and products are present in ratios that don't change, but they are not necessarily equal to each other.
For the reaction \[\begin{equation}\mathrm{X} + \mathrm{Y} \rightleftharpoons \mathrm{C},\end{equation}\]the point at which the rates of the forward and reverse reactions are equal is where equilibrium is established. Understanding how equilibrium can be shifted by changing conditions such as temperature or concentration—known as Le Chatelier's principle—is a fundamental aspect of physical chemistry and is crucial for predicting how a system will respond to such changes.
For the reaction \[\begin{equation}\mathrm{X} + \mathrm{Y} \rightleftharpoons \mathrm{C},\end{equation}\]the point at which the rates of the forward and reverse reactions are equal is where equilibrium is established. Understanding how equilibrium can be shifted by changing conditions such as temperature or concentration—known as Le Chatelier's principle—is a fundamental aspect of physical chemistry and is crucial for predicting how a system will respond to such changes.
Equilibrium Constant Calculation
The equilibrium constant (\[\begin{equation}K\end{equation}\]) for a chemical reaction is a dimensionless number that provides a measure of the extent of the reaction at equilibrium. For the reaction given in the exercise, the equilibrium constant can be calculated using the expression related to the Gibbs Free Energy change (\[\begin{equation}\Delta G^\circ\end{equation}\]):\[\begin{equation}K = e^{-(\Delta G^\circ/(RT))}\end{equation}\]where \[\begin{equation}R\end{equation}\] is the gas constant and \[\begin{equation}T\end{equation}\] is the temperature in Kelvin. Converting \[\begin{equation}\Delta G^\circ\end{equation}\] to the appropriate units and solving this equation gives us the equilibrium constant. It’s important for students to practice these calculations, as they are central to many areas in physical chemistry.
Physical Chemistry
Physical chemistry is the study of how matter behaves on a molecular and atomic level and how chemical reactions occur. It combines principles of physics and chemistry to understand the physical properties of molecules, the forces that act upon them, and the chemical transformations they undergo. One of the central themes in physical chemistry is the use of thermodynamics to describe the energy changes that accompany chemical reactions.
The concepts of chemical equilibrium and the calculation of the equilibrium constant are rooted in physical chemistry. Problems like the one presented with substances X, Y, and C involving Gibbs Free Energy are typical in this field, demonstrating the practical application of physical chemistry in understanding and predicting the behavior of chemical systems.
The concepts of chemical equilibrium and the calculation of the equilibrium constant are rooted in physical chemistry. Problems like the one presented with substances X, Y, and C involving Gibbs Free Energy are typical in this field, demonstrating the practical application of physical chemistry in understanding and predicting the behavior of chemical systems.