Problem 87
Question
Decide whether each of the following statements is true or false. If false, rewrite it to make it true. (a) The entropy of a substance increases on going from the liquid to the vapor state at any temperature. (b) An exothermic reaction will always be spontaneous. (c) Reactions with a positive \(\Delta_{r} H^{\circ}\) and a positive \(\Delta_{\mathrm{r}} S^{\circ}\) can never be product-favored. (d) If \(\Delta_{r} G^{\circ}\) for a reaction is negative, the reaction will have an equilibrium constant greater than 1.
Step-by-Step Solution
Verified Answer
(a) True, (b) False, depends on
ΔG
, (c) False, possible at high temperature, (d) True.
1Step 1: Understand Entropy Change from Liquid to Vapor
Entropy (
S
) is a measure of disorder or randomness in a system. When a substance transitions from a liquid to a vapor, molecules move more freely and randomness increases. Therefore, the entropy increases when going from a liquid to a vapor state at any temperature. This statement is true.
2Step 2: Analyze Spontaneity of Exothermic Reactions
An exothermic reaction releases energy (heat), but spontaneity is determined by Gibbs free energy (
ΔG
) and not just enthalpy (
ΔH
). While many exothermic reactions are spontaneous, they aren't always, because the entropy change (
ΔS
) also plays a role. Thus, this statement is false. We can correct it: 'An exothermic reaction has the potential to be spontaneous, but isn't always.'
3Step 3: Evaluate Reactions with Positive ΔH and ΔS
Reactions with positive
ΔH
(endothermic) and positive
ΔS
(increase in entropy) can sometimes be product-favored, particularly at high temperatures. At higher temperatures, the entropy term
(TΔS)
often dominates, potentially leading to a negative
ΔG
. Therefore, the statement is false. It can be corrected to: 'Reactions with positive
ΔH
and positive
ΔS
can be product-favored at high temperatures.'
4Step 4: Correlate Negative ΔG with Equilibrium Constant
If
ΔG^
{°}
is negative, the reaction is spontaneous in the forward direction, suggesting that at equilibrium, the concentration of products is greater than reactants. This means the equilibrium constant
K
will be greater than 1. Thus, this statement is true.
Key Concepts
EntropySpontaneity in Chemical ReactionsEquilibrium Constant
Entropy
Entropy, often symbolized as \( S \), is a fundamental concept in thermodynamics that represents the amount of disorder or randomness in a system. When a substance transitions from a liquid to a vapor state, its molecules gain freedom to move. This is because in the vapor state, molecules are not held closely together, unlike in the liquid state where they are more confined.
As a result, the randomness or disorder of the system increases, thereby increasing entropy. This increase in entropy is true for all temperatures as it reflects the universal behavior of molecular systems undergoing phase changes from liquid to vapor.
As a result, the randomness or disorder of the system increases, thereby increasing entropy. This increase in entropy is true for all temperatures as it reflects the universal behavior of molecular systems undergoing phase changes from liquid to vapor.
- This concept helps us understand why vaporization involves an increase in entropy.
- In practical terms, when water boils and changes into steam, the steam has higher entropy than the liquid water.
Spontaneity in Chemical Reactions
The spontaneity of chemical reactions isn't determined just by whether the reaction is exothermic or endothermic. Instead, it is evaluated using Gibbs free energy, denoted as \( \Delta G \). A reaction is deemed spontaneous if \( \Delta G \) is negative.
Gibbs free energy combines two important thermodynamic quantities: enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)), as summarized in the equation \( \Delta G = \Delta H - T\Delta S \), where \( T \) is the temperature in Kelvin.
Gibbs free energy combines two important thermodynamic quantities: enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)), as summarized in the equation \( \Delta G = \Delta H - T\Delta S \), where \( T \) is the temperature in Kelvin.
- An exothermic reaction typically releases heat, reflected by a negative \( \Delta H \), but this alone does not ensure spontaneity.
- Both \( \Delta H \) and \( \Delta S \) contribute to whether \( \Delta G \) is negative, dictating if a reaction proceeds without external energy.
Equilibrium Constant
In chemical thermodynamics, the equilibrium constant \( K \) provides information about the ratio of concentrations of products to reactants at equilibrium. If the standard Gibbs free energy change \( \Delta G^{\circ} \) is negative, it means the forward reaction is spontaneous and product-favored.
This implicates that at equilibrium, the concentration of products exceeds that of the reactants, making \( K \) greater than 1. This concept is pivotal in determining the extent to which a reaction will proceed.
This implicates that at equilibrium, the concentration of products exceeds that of the reactants, making \( K \) greater than 1. This concept is pivotal in determining the extent to which a reaction will proceed.
- A negative \( \Delta G^{\circ} \) indicates that achieving equilibrium results in more products than reactants.
- Conversely, a positive \( \Delta G^{\circ} \) suggests a lower concentration of products relative to reactants, as it corresponds to \( K \) being less than 1.
Other exercises in this chapter
Problem 85
Mercury vapor is dangerous because breathing it brings this toxic element into the lungs. We wish to estimate the vapor pressure of mercury at two different tem
View solution Problem 86
Explain why each of the following statements is incorrect. (a) Entropy increases in all spontaneous reactions. (b) Reactions with a negative free energy change
View solution Problem 88
Under what conditions is the entropy of a pure substance \(0 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol} ?\) Could a substance at standard conditions at \(25^{\c
View solution Problem 91
Write a chemical equation for the oxidation of \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g}) \mathrm{by} \mathrm{O}_{2}(\mathrm{g})\) to form \(\mathrm{CO}_{2}(\m
View solution