Lecture 12: Part 2 - Chemical Reaction and Spontaneity Study Notes
Lecture 12: Part 2 - Chemical Reaction and Spontaneity
I. Enthalpy and Entropy Driven Processes
Concept Exploration:
Enthalpy (Delta H), entropy (Delta S), and temperature (T) are key factors in determining Gibbs free energy (Delta G), hence influencing the spontaneity of chemical reactions.
Enthalpy (Delta H): This term quantifies the heat change of a system at constant pressure. It is an indicator of the stability of reactants versus products.
Negative Delta H (Exothermic): Heat is released by the system to the surroundings. This often leads to more stable products (lower energy state) and is generally favorable for spontaneity. Examples include combustion reactions.
Positive Delta H (Endothermic): Heat is absorbed by the system from the surroundings. This often leads to less stable products (higher energy state) and is generally unfavorable for spontaneity unless compensated by a large increase in entropy. Examples include melting ice.
Entropy (Delta S): This term is a quantitative measure of the disorder or randomness of a system. It reflects the number of possible microstates for a given macroscopic state.
Positive Delta S (Increased Disorder): The system becomes more disordered, or there are more ways to arrange its components (e.g., a solid turning into a gas, or a single molecule breaking into two). This increase in disorder is generally favorable for spontaneity.
Negative Delta S (Decreased Disorder): The system becomes more ordered. This is generally unfavorable for spontaneity. Example: Freezing water from liquid to solid involves a decrease in disorder.
Temperature (T): Measured in Kelvin (K), temperature is a scaling factor for the entropy term in the Gibbs free energy equation. It quantifies the amount of thermal energy available to disperse energy across microstates. The higher the temperature, the more significant the impact of entropy on spontaneity.
Spontaneity: A spontaneous process is one that occurs without continuous external intervention once started. It does not necessarily mean fast; it only refers to the direction of a reaction under a given set of conditions. Chemical systems naturally tend towards states of lower potential energy (enthalpy-driven) and higher disorder (entropy-driven). Gibbs free energy combines these two tendencies.
The value of Delta G indicates the favorability of a reaction:
Delta G is less than 0: Process is spontaneous. The system lowers its free energy, meaning it can proceed without external work being done on it.
Delta G is greater than 0: Process is non-spontaneous. The system would need energy input to proceed in the indicated direction.
Delta G equals 0: The system is at equilibrium. There is no net change in the system, as the forward and reverse reaction rates are equal.
Key Equation:
Delta G equals Delta H minus T multiplied by Delta S
Derivation/Logic: This equation stems from the second law of thermodynamics, which states that for a spontaneous process, the total entropy of the universe must increase (Delta S of the universe is greater than 0).
We know that Delta S of the universe equals Delta S of the system plus Delta S of the surroundings.
For a process occurring at constant temperature and pressure, the heat exchanged with the surroundings (q surroundings) is equal to the negative of the heat exchanged by the system (negative q of the system). Since pressure is constant, q of the system equals Delta H of the system.
The entropy change of the surroundings is defined as Delta S of the surroundings equals q of the surroundings divided by T, which also equals negative Delta H of the system divided by T.
Substituting this into the universe entropy equation (and dropping the 'system' subscript for simplicity, as it's understood we refer to the system): Delta S of the universe equals Delta S minus Delta H divided by T.
Multiplying by negative T to relate it to free energy: negative T multiplied by Delta S of the universe equals negative T multiplied by Delta S plus Delta H.
We define Gibbs Free Energy change as Delta G equals negative T multiplied by Delta S of the universe.
Therefore, we arrive at the fundamental equation: Delta G equals Delta H minus T multiplied by Delta S.
For spontaneity, Delta S of the universe is greater than 0, which means negative T multiplied by Delta S of the universe is less than 0, hence Delta G is less than 0. This means a spontaneous process is characterized by a decrease in the system's Gibbs free energy.
Example: Evaporation of Water (Detailed explanation in Section IV).
Hess’ Law: The enthalpy change for a reaction is independent of the pathway taken to convert reactants to products. This is a direct consequence of enthalpy being a state function, meaning it only depends on the initial and final states of the system, not the intermediate steps.
Protein Folding Problem: Exploring the thermodynamic aspects of protein folding. (Detailed explanation in Section IX).
II. Gibbs Free Energy
Rate Principles: It is crucial to distinguish between spontaneity and reaction rate. Gibbs free energy (Delta G) determines the spontaneity (if a reaction can occur), not the rate (how fast it occurs). A reaction can be highly spontaneous (Delta G is less than 0) but proceed very slowly (e.g., diamond turning into graphite at room temperature). Conversely, a non-spontaneous reaction (Delta G is greater than 0) can be forced to occur by continuous energy input.
Gibbs Free Energy criterion states:
A spontaneous process occurs if Delta G is less than 0; this results from the interplay between enthalpy and entropy at a given temperature.
Detailed Balance Principle:
At equilibrium: the rates of forward and reverse reactions are equivalent, leading to constant compositions. When a system reaches equilibrium, the net change in concentrations of reactants and products is zero. This doesn't mean the reactions stop; rather, the forward reaction rate becomes exactly equal to the reverse reaction rate, and at this point, Delta G equals 0.
III. Enthalpy and Entropy in Chemical Reactions
Understanding Conditions for Spontaneity: The signs of Delta H and Delta S determine how temperature influences spontaneity.
Enthalpy Driven Processes (If Delta H is less than 0):
If Delta H is less than 0 (exothermic) and Delta S is greater than 0 (increase in disorder): Delta G equals Delta H minus T multiplied by Delta S will always be negative, as both terms are favorable for spontaneity. The process is spontaneous at all temperatures.
If Delta H is less than 0 (exothermic) and Delta S is less than 0 (decrease in disorder): Delta G will be negative only if the magnitude of the favorable enthalpy term (absolute value of Delta H) is greater than the magnitude of the unfavorable entropy term (absolute value of T multiplied by Delta S). This occurs at low temperatures where T is small. The process is spontaneous at low temperatures.
Entropy Driven Processes (If Delta S is greater than 0):
If Delta S is greater than 0 (increase in disorder) and Delta H is greater than 0 (endothermic): Delta G equals Delta H minus T multiplied by Delta S. For Delta G to be negative, the favorable entropy term (T multiplied by Delta S) must be larger than the unfavorable enthalpy term (Delta H). This can only happen at high temperatures. The process is spontaneous at high temperatures.
If Delta S is less than 0 (decrease in disorder) and Delta H is greater than 0 (endothermic): Delta G equals Delta H minus T multiplied by Delta S. Both terms are unfavorable for spontaneity (Delta H is positive and negative T multiplied by Delta S is positive because Delta S is negative). Delta G will always be positive. The process is non-spontaneous at all temperatures.
IV. Example: Vaporization of Liquid Water
General Reaction: This reaction represents the physical process where liquid water turns into water vapor.
H2O (liquid) yields H2O (gas)
Enthalpy of Vaporization (Delta H of vaporization):
Delta H of vaporization equals +44 kilojoules per mole
Explanation: Vaporization is an endothermic process, meaning it requires energy input (heat) to overcome intermolecular forces holding water molecules together in the liquid phase, allowing them to move freely as a gas. The positive value of +44 kilojoules per mole confirms that heat is absorbed by the system.
Vaporization is an endothermic process (Delta H of vaporization is greater than 0) and leads to an increase in disorder (Delta S of vaporization is greater than 0 as gas is more disordered than liquid). According to the spontaneity conditions (Case 2 from Section III, with Delta H is greater than 0 and Delta S is greater than 0), it is driven to spontaneity by elevated temperatures above the boiling point (373 K).
At boiling point conditions (373 K):
At the boiling point, liquid and gas phases are in dynamic equilibrium. This means there is no net change in the system's free energy:
Delta G of vaporization equals 0
Substituting Delta G of vaporization equals 0 into the Gibbs free energy equation:
0 equals Delta H of vaporization minus T at boiling point multiplied by Delta S of vaporization
Calculate entropy of vaporization (Delta S of vaporization):
Start with the equilibrium condition: 0 equals Delta H of vaporization minus T at boiling point multiplied by Delta S of vaporization
Rearrange the equation to solve for Delta S of vaporization: T at boiling point multiplied by Delta S of vaporization equals Delta H of vaporization
Divide by T at boiling point: Delta S of vaporization equals Delta H of vaporization divided by T at boiling point
Calculation:
Given Delta H of vaporization equals +44 kilojoules per mole and T at boiling point equals 373 Kelvin.
Convert Delta H of vaporization to Joules: 44 kilojoules per mole equals 44000 joules per mole.
Substitute values: Delta S of vaporization equals 44000 joules per mole divided by 373 Kelvin
Perform the division: Delta S of vaporization is approximately 117.96 joules per mole Kelvin
Rounded to a practical value: Delta S of vaporization equals 118 joules per mole Kelvin
Interpretation: The positive value of Delta S of vaporization indicates an increase in disorder, as expected when a liquid transitions to a gas at its boiling point.
V. Relationship of Delta G with Temperature
Observations: Let's analyze the spontaneity of water vaporization (Delta H of vaporization equals +44 kilojoules per mole and Delta S of vaporization equals +118 joules per mole Kelvin) at different temperatures using Delta G equals Delta H minus T multiplied by Delta S.
Case 1: T is less than T boiling point (e.g., 300 Kelvin or 27 degrees Celsius):
Delta G of vaporization equals 44000 joules per mole minus (300 Kelvin multiplied by 118 joules per mole Kelvin)
Delta G of vaporization equals 44000 joules per mole minus 35400 joules per mole
Delta G of vaporization equals +8600 joules per mole, which is greater than 0
Conclusion: Vaporization is non-spontaneous below the boiling point.
Case 2: T equals T boiling point (373 Kelvin):
Delta G of vaporization equals 44000 joules per mole minus (373 Kelvin multiplied by 118 joules per mole Kelvin)
Delta G of vaporization equals 44000 joules per mole minus 44014 joules per mole (Slight difference due to previous rounding of Delta S of vaporization)
Delta G of vaporization is approximately 0
Conclusion: At the boiling point, the system is at equilibrium.
Case 3: T is greater than T boiling point (e.g., 400 Kelvin or 127 degrees Celsius):
Delta G of vaporization equals 44000 joules per mole minus (400 Kelvin multiplied by 118 joules per mole Kelvin)
Delta G of vaporization equals 44000 joules per mole minus 47200 joules per mole
Delta G of vaporization equals -3200 joules per mole, which is less than 0
Conclusion: Vaporization is spontaneous above the boiling point.
Summary:
T is less than T boiling point: Delta G of vaporization is greater than 0 (non-spontaneous)
T equals T boiling point: Delta G of vaporization equals 0 (equilibrium)
T is greater than T boiling point: Delta G of vaporization is less than 0 (spontaneous)
Slope of Delta G vs T:
To find how Delta G changes with respect to temperature, we can take the derivative of the Gibbs free energy equation, assuming Delta H and Delta S are approximately constant over the temperature range:
Start with: Delta G equals Delta H minus T multiplied by Delta S
Differentiate with respect to T: the derivative of Delta G with respect to T equals the derivative with respect to T of (Delta H minus T multiplied by Delta S)
Since Delta H is treated as a constant, and Delta S is treated as a constant: the derivative of Delta G with respect to T equals 0 minus Delta S
Thus, for vaporization: the derivative of Delta G with respect to T equals negative Delta S of vaporization
Conceptual Meaning: This means a plot of Delta G versus T will yield a straight line with a slope equal to the negative of the entropy change. Since Delta S of vaporization is positive for vaporization, the slope (negative Delta S of vaporization) will be negative, indicating that as temperature increases, Delta G decreases, making the process more spontaneous.
Conclusion:
If Delta H and Delta S of vaporization are constant, Delta G has a linear relationship with T. This linear approximation is useful for understanding general trends, though in reality Delta H and Delta S do change slightly with temperature.
VI. Introduction to Enthalpy (H)
Definition:
Enthalpy (H) represents the heat content of a system at constant pressure. It is an extensive property and a state function. For chemical reactions and physical changes occurring in open containers (constant atmospheric pressure), the heat exchanged is precisely the change in enthalpy (Delta H).
Differential Form of Enthalpy: This rigorous mathematical definition stems from the First Law of Thermodynamics (the differential of E equals the differential of q plus the differential of w).
Recall the definition of enthalpy: H is defined as E plus P multiplied by V (Equation 1)
Taking the total differential of (1):
the differential of H equals the differential of E plus the differential of (P multiplied by V)
Applying the product rule for the differential of (P multiplied by V) (the differential of (P multiplied by V) equals P multiplied by the differential of V plus V multiplied by the differential of P):
the differential of H equals the differential of E plus P multiplied by the differential of V plus V multiplied by the differential of P
Connecting to constant pressure processes: From the First Law, for an infinitesimal process, the differential of E equals the differential of q plus the differential of w. For pressure-volume work, the differential of w equals negative P multiplied by the differential of V. Substituting this into the First Law: the differential of E equals the differential of q minus P multiplied by the differential of V. At constant pressure (the differential of P equals 0), the term V multiplied by the differential of P becomes zero. Substituting the differential of E equals the differential of q minus P multiplied by the differential of V into the differential of H equation (at constant P):
the differential of H equals (the differential of q minus P multiplied by the differential of V) plus P multiplied by the differential of V plus 0
the differential of H equals the differential of q
This confirms that at constant pressure, the change in enthalpy (differential of H or Delta H) is equal to the heat absorbed or released (differential of q or q at constant pressure).
Measurement of enthalpy changes occurs via calorimetry:
Delta H equals q at constant pressure, which equals C at constant pressure multiplied by Delta T
q at constant pressure: Heat absorbed or released at constant pressure.
C at constant pressure: Heat capacity at constant pressure. This is the amount of heat required to raise the temperature of a substance (or a calorimeter) by one degree Celsius (or Kelvin) at constant pressure.
Delta T: The observed change in temperature of the system or calorimeter.
Calorimetry: This experimental technique uses an insulated device (calorimeter) to measure temperature changes, allowing the calculation of heat changes (Delta H) of reactions or processes.
VII. Thermochemistry and Hess' Law
Key Principle:
Hess's Law states the total enthalpy change of a reaction is equal to the sum of the enthalpy changes for each individual reaction step. This is a direct consequence of enthalpy being a state function.
Example:
For an overall reaction that can be broken down into steps, the enthalpy changes of the steps add up:
If Reaction 1 has Delta H for reaction 1 and Reaction 2 has Delta H for reaction 2, then for the overall reaction formed by adding Reaction 1 and Reaction 2, the enthalpy change is: Delta H for the reaction equals Delta H for reaction 1 plus Delta H for reaction 2
Hypothetical Reaction to Prove Hess's Law:
Objective: Use Hess's Law to find the enthalpy of formation of carbon monoxide (CO(gas)) from its elements, C (solid) plus 1/2 O2 (gas) yield CO(gas), using the following reactions:
C (solid) plus O2 (gas) yield CO2 (gas), with Delta H1 equals -394 kilojoules
CO(gas) plus 1/2 O2 (gas) yield CO2 (gas), with Delta H2 equals -283 kilojoules
Step 1: Reaction 1 provides C (solid) on the reactant side and CO2 (gas) on the product side. We will use this reaction as is. Its Delta H is -394 kilojoules.
Step 2: Reaction 2 contains CO(gas) on the reactant side, but for the formation of CO(gas), we need it on the product side. Therefore, we reverse Reaction 2. When a reaction is reversed, the sign of its Delta H is also reversed.
CO2 (gas) yields CO(gas) plus 1/2 O2 (gas), with Delta H for the reversed reaction 2 equals +283 kilojoules
Step 3: Add Reaction 1 and the reversed Reaction 2, and sum their respective enthalpy changes:
C(solid) + O2(gas) yields CO2(gas) (Delta H1 = -394 kilojoules)
CO2(gas) yields CO(gas) + 1/2 O2(gas) (Delta H for the reversed reaction 2 = +283 kilojoules)
C(solid) + O2(gas) + CO2(gas) yields CO2(gas) + CO(gas) + 1/2 O2(gas)
Step 4: Cancel identical species appearing on both sides of the arrow (reactants and products):
CO2(gas) cancels out completely.
O2(gas) on the left and 1/2 O2(gas) on the right simplifies to 1/2 O2(gas) remaining on the left.
Resulting Overall Reaction (Formation of CO):
C(solid) plus 1/2 O2(gas) yield CO(gas)
Overall Enthalpy Change:
Overall Delta H equals Delta H1 plus Delta H for the reversed reaction 2, which is (-394 kilojoules) + (283 kilojoules), resulting in -111 kilojoules
This value, -111 kilojoules, represents the standard enthalpy of formation of carbon monoxide (standard enthalpy of formation of CO(gas)).
VIII. Enthalpy of Elements
Standard State:
By definition, the enthalpy of formation (standard enthalpy of formation) of elements in their standard states equals zero. The standard state of an element is its most stable form at 1 atmosphere pressure and a specified temperature (usually 25 degrees Celsius or 298.15 Kelvin). For example, the standard state of carbon is graphite, not diamond, and for oxygen, it's gaseous diatomic oxygen, O2(gas).
Example Reactions: (These were already used in conjunction with Hess' Law above to illustrate calculation details).
IX. Protein Folding
Concept:
Protein folding is viewed as a biological phase transition. Just as water transitions between liquid and solid, a polypeptide chain transitions from a disordered, denatured ensemble of states to a highly ordered, functional three-dimensional structure (native state). This 'phase transition' is governed by thermodynamic forces, aiming for a state of lower free energy.
Driving Forces:
Evaluate whether folding is driven enthalpically (Delta H is less than 0) or entropically (Delta S is greater than 0). Protein folding involves a significant decrease in the conformational entropy of the polypeptide chain itself (it becomes much more ordered). However, folding is spontaneous (Delta G is less than 0), implying that other favorable contributions must outweigh this unfavorable entropic term.
Hydrophobic Effect:
Nonpolar amino acids migrate to the protein’s interior in folding, contrasting with polar amino acids that interact with water, contributing to entropic changes. This is often considered the primary driving force for protein folding.
Before Folding (Unfolded State): Nonpolar (hydrophobic) amino acid side chains are exposed to the aqueous solvent. Water molecules around these nonpolar groups become highly ordered to minimize their contact with the nonpolar surface. This ordered arrangement of water molecules represents a decrease in entropy of the solvent.
During Folding (Collapsed State): The protein folds in a way that sequesters these nonpolar side chains into its core, away from water. This allows the previously ordered water molecules surrounding the hydrophobic groups to return to a more disordered, bulk-like liquid state.
Contribution to Spontaneity: The increase in entropy of the solvent (due to the disruption of ordered water structures around hydrophobic groups) is a major favorable entropic contribution (Delta S of the solvent is greater than 0) that helps drive protein folding, often compensating for the unfavorable decrease in conformational entropy of the protein itself.
X. Driving Forces in Protein Folding
Iceberg Model:
Describes the unfavorable ordering of water around a nonpolar solute. When nonpolar regions of a protein are exposed to water, water molecules form cage-like, highly structured arrangements (clathrate-like structures, often called "icebergs") around them. This structured water has lower entropy.