Problem 79
Question
For a reaction: \(\mathrm{A} \rightleftharpoons \mathrm{B}\), carried out at \(27^{\circ} \mathrm{C}\), the ratio of equilibrium concentrations of product to reactant changes by a factor of \(\mathrm{e}^{4}\) for every (a) \(1.2\) kcal rise in \(\Delta G^{\circ}\) (b) \(1.2 \mathrm{kcal}\) fall in \(\Delta G^{\circ}\) (c) \(2.4 \mathrm{kcal}\) rise in \(\Delta G^{\circ}\) (d) \(2.4\) kcal fall in \(\Delta G^{\circ}\)
Step-by-Step Solution
Verified Answer
The correct option is (a) since a 1.2 kcal rise in \(\Delta G^\circ\) changes the ratio of equilibrium concentrations of product to reactant by a factor of \(e^4\) at 300.15 K.
1Step 1: Understanding the Relationship Between Equilibrium Constant and Gibbs Free Energy Change
The equilibrium constant, K, of a reaction is related to the standard Gibbs free energy change, \(\Delta G^\circ\), by the equation \(\Delta G^\circ = -RT \ln K\) where R is the gas constant and T is the temperature in Kelvin. Change in \(\Delta G^\circ\) can affect the equilibrium constant by the same multiplicative factor.
2Step 2: Converting Temperature to Kelvin
First, convert the temperature from Celsius to Kelvin because the Gibbs free energy equation requires temperature in Kelvin. The conversion is \(T(K) = T(\degree C) + 273.15\). For \(27\degree C\), the temperature in Kelvin would be \(27 + 273.15 = 300.15 K\).
3Step 3: Calculating the Factor Change in Equilibrium Constant
The equation for the factor change in the equilibrium constant is \(e^{-\Delta G^\circ / (RT)}\). When \(\Delta G^\circ\) increases, the equilibrium constant decreases, and when \(\Delta G^\circ\) decreases, the equilibrium constant increases.
4Step 4: Determining the Effect of a Rise or Fall in \(\Delta G^\circ\)
To find the factor of change for an increase or decrease in \(\Delta G^\circ\), plug the values of \(\Delta G^\circ\) (converted to joules or an equivalent unit) and the calculated Kelvin temperature into the equation from the previous step. For an increase in \(\Delta G^\circ\), the ratio of product to reactant decreases, and for a decrease, the ratio increases.
5Step 5: Converting kcal to Joules
To use the Gibbs free energy equation, convert kcal to joules using the conversion factor \(1 kcal = 4184 J\). We can then use these converted values to calculate the change in equilibrium constant.
6Step 6: Solving for Each Given Case
Calculate the change in the equilibrium constant ratio (the factor change) for each of the given increases or decreases in \(\Delta G^\circ\). The correct option results in a factor change of the equilibrium constant equal to \(e^4\) for the stated change in \(\Delta G^\circ\).
Key Concepts
Chemical EquilibriumThermodynamics in ChemistryStandard Gibbs Free Energy ChangeEquilibrium Constant Calculation
Chemical Equilibrium
Chemical equilibrium is a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction, meaning there's no net change in the concentration of reactants and products over time. It's essential to visualize this not as a static situation, but as one where reactions are continuously happening but balance each other out.
At equilibrium, the system's properties become constant, though the individual molecules are very much active. Understanding the concept of equilibrium is crucial for predicting how a reaction will proceed under a set of conditions, and is fundamental in fields like chemistry, biochemistry, and environmental science.
For instance, if we look at a simple equilibrium reaction such as \(\text{A} \rightleftharpoons \text{B}\), we say that the system is at equilibrium when the rate at which A converts to B is the same as the rate at which B converts back to A. The equilibrium constant (K) for this reaction tells us the ratio of product concentration to reactant concentration at equilibrium, providing a numerical value that represents where the balance point lies.
At equilibrium, the system's properties become constant, though the individual molecules are very much active. Understanding the concept of equilibrium is crucial for predicting how a reaction will proceed under a set of conditions, and is fundamental in fields like chemistry, biochemistry, and environmental science.
For instance, if we look at a simple equilibrium reaction such as \(\text{A} \rightleftharpoons \text{B}\), we say that the system is at equilibrium when the rate at which A converts to B is the same as the rate at which B converts back to A. The equilibrium constant (K) for this reaction tells us the ratio of product concentration to reactant concentration at equilibrium, providing a numerical value that represents where the balance point lies.
Thermodynamics in Chemistry
Thermodynamics in chemistry deals with the energy and work related to chemical reactions and changes in matter. It's a branch of physical sciences that is not only essential for chemistry students but also for those in physics, biology, and engineering disciplines.
The key concept here is the Gibbs free energy (G), a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure, which is very relevant in determining the spontaneity of reactions. If the change in Gibbs free energy (\(\text{Δ}G\)) is negative, the process is spontaneous, while a positive \(\text{Δ}G\) indicates a non-spontaneous process.
This concept also explains how different factors such as temperature and pressure affect a chemical system, helping us understand how to control a reaction's direction and rate.
The key concept here is the Gibbs free energy (G), a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure, which is very relevant in determining the spontaneity of reactions. If the change in Gibbs free energy (\(\text{Δ}G\)) is negative, the process is spontaneous, while a positive \(\text{Δ}G\) indicates a non-spontaneous process.
This concept also explains how different factors such as temperature and pressure affect a chemical system, helping us understand how to control a reaction's direction and rate.
Standard Gibbs Free Energy Change
The standard Gibbs free energy change (\(\text{Δ}G^\circ\)) refers to the change in Gibbs free energy as reactants transform into products under standard conditions (1 bar of pressure, the concentrations of 1 M, etc.). It's a valuable predictor of whether a reaction will occur spontaneously under these standard conditions.
A key equation connects \(\text{Δ}G^\circ\) to the equilibrium constant (K): \(\text{Δ}G^\circ = -RT \text{ln} K\), where R is the universal gas constant and T is the temperature in Kelvin. This equation implies that the sign and magnitude of \(\text{Δ}G^\circ\) directly affect the equilibrium position of a reaction: a negative \(\text{Δ}G^\circ\) is associated with a larger K value, indicating products are favored at equilibrium, while a positive \(\text{Δ}G^\circ\) is associated with a smaller K, indicating reactants are favored.
Understanding how to calculate and interpret \(\text{Δ}G^\circ\) is crucial for any chemist or student of science as it assists in anticipating the behavior of chemical systems.
A key equation connects \(\text{Δ}G^\circ\) to the equilibrium constant (K): \(\text{Δ}G^\circ = -RT \text{ln} K\), where R is the universal gas constant and T is the temperature in Kelvin. This equation implies that the sign and magnitude of \(\text{Δ}G^\circ\) directly affect the equilibrium position of a reaction: a negative \(\text{Δ}G^\circ\) is associated with a larger K value, indicating products are favored at equilibrium, while a positive \(\text{Δ}G^\circ\) is associated with a smaller K, indicating reactants are favored.
Understanding how to calculate and interpret \(\text{Δ}G^\circ\) is crucial for any chemist or student of science as it assists in anticipating the behavior of chemical systems.
Equilibrium Constant Calculation
The equilibrium constant (K) calculation is pivotal in understanding chemical equilibria. It quantifies the ratio of concentrations of products to reactants at equilibrium, raised to the power of their stoichiometric coefficients.
To calculate K, one must be familiar with the reaction quotient (Q), an expression with the same form as K but applied to non-equilibrium conditions. When a system is at equilibrium, Q equals K. The relationship between \(\text{Δ}G^\circ\) and K provides a way to calculate the equilibrium constant from the standard Gibbs free energy change using the formula \(\text{Δ}G^\circ = -RT \text{{ln}} K\).
For example, if we know the \(\text{Δ}G^\circ\) for a reaction, we can calculate the equilibrium constant at any temperature, which then allows us to predict the concentrations of reactants and products at equilibrium. This ability is fundamental in designing chemical processes and understanding natural phenomena.
To calculate K, one must be familiar with the reaction quotient (Q), an expression with the same form as K but applied to non-equilibrium conditions. When a system is at equilibrium, Q equals K. The relationship between \(\text{Δ}G^\circ\) and K provides a way to calculate the equilibrium constant from the standard Gibbs free energy change using the formula \(\text{Δ}G^\circ = -RT \text{{ln}} K\).
For example, if we know the \(\text{Δ}G^\circ\) for a reaction, we can calculate the equilibrium constant at any temperature, which then allows us to predict the concentrations of reactants and products at equilibrium. This ability is fundamental in designing chemical processes and understanding natural phenomena.
Other exercises in this chapter
Problem 78
A definite mass of a monoatomic ideal gas at 1 bar and \(27^{\circ} \mathrm{C}\) expands against \(\begin{array}{llll}\text { vacuum } & \text { from } & 1.2 \m
View solution Problem 78
One mole of an ideal monoatomic gas undergoes adiabatic free expansion from 2 to \(20 \mathrm{dm}^{3}, 300 \mathrm{~K}\). The value of \(\Delta S\) for the gas
View solution Problem 80
Given the following entropy values (in \(\mathrm{J} / \mathrm{K}-\mathrm{mol}\) ) at \(298 \mathrm{~K}\) and 1 atm \(\mathrm{H}_{2}(\mathrm{~g})\) \(=130.6, \ma
View solution Problem 81
The \(\Delta G\) in the process of melting of ice at \(-15^{\circ} \mathrm{C}\) is (a) less than zero (b) greater than zero (c) equal to zero (d) none of these
View solution