Enthalpy: Comprehensive Notes

Enthalpy

Constant Pressure and Enthalpy (H)

• Most lab reactions occur at constant pressure (1 atm) in closed thermodynamic systems.
• Enthalpy (H) is used to express heat changes at constant pressure.
• Enthalpy is a state function, meaning the change in enthalpy ($\Delta H$) depends only on the initial and final states, not the path taken.
• For a chemical reaction, $\Delta H$ is calculated by comparing the enthalpy of the final state to the initial state.

Change in Enthalpy and Heat Transfer

• The change in enthalpy ($\Delta H$) equals the heat transferred into or out of the system at constant pressure.
• To find the enthalpy change of a reaction ($\Delta H<em>{RXN}$), subtract the enthalpy of the reactants from the enthalpy of the products: $\Delta H</em>{RXN} = H<em>{products} - H</em>{reactants}$.
• A positive $\Delta H_{RXN}$ indicates an endothermic process (heat absorbed).
• A negative $\Delta H_{RXN}$ indicates an exothermic process (heat released).
• Enthalpy cannot be measured directly; only $\Delta H$ can be measured for fast and spontaneous processes.

Standard Heat of Formation ($\Delta H_f^\circ$)

• The standard enthalpy of formation ($\Delta H_f^\circ$) is the enthalpy change required to produce one mole of a compound from its elements in their standard states.
• Standard state refers to the most stable physical state of an element or compound at 298 K and 1 atm.
• By definition, the $\Delta H_f^\circ$ of an element in its standard state is zero.
• $\Delta H_f^\circ$ values for many substances are tabulated.

Standard Heat of Reaction ($\Delta H_{RXN}^\circ$)

• The standard enthalpy of a reaction ($\Delta H_{RXN}^\circ$) is the enthalpy change accompanying a reaction carried out under standard conditions.
• Calculated by taking the difference between the sum of the standard heats of formation for the products and the sum of the standard heats of formation of the reactants:
  $\Delta H<em>{RXN}^\circ = \sum \Delta H</em>{f(products)}^\circ - \sum \Delta H_{f(reactants)}^\circ$

Hess's Law

• Enthalpy is a state function; the pathway is irrelevant to the change in enthalpy.
• Hess's Law: Enthalpy changes of a reaction are additive.
• When thermochemical equations (chemical equations with energy changes) are added, the corresponding heats of reaction are also added.
• Any reaction can be described as breaking down reactants into component elements, then forming products from these elements.
• The enthalpy change for the reverse of any reaction has the same magnitude but the opposite sign as the forward reaction:
  $\Delta H<em>{(\text{reactants to elements})} = -\Delta H</em>{(\text{elements to reactants})}$
• $\Delta H<em>{RXN} = \Delta H</em>{(\text{reactants to elements})} + \Delta H<em>{(\text{elements to products})}$, which is another way of writing $\Delta H</em>{RXN}^\circ = \sum \Delta H<em>{f(products)}^\circ - \sum \Delta H</em>{f(reactants)}^\circ$

Example: Phase Change of $Br_2$

• $Br<em>2(l) \rightarrow Br</em>2(g)$; $\Delta H_{RXN}^\circ = 31 \frac{kJ}{mol}$
• The enthalpy change for this phase change is the heat of vaporization ($\Delta H_{vap}^\circ$).
• As long as initial and final states are under standard conditions, $\Delta H<em>{RXN}^\circ = \Delta H</em>{vap}^\circ$ irrespective of the pathway.

Example: Calculating $\Delta H$ using Hess's Law

Given:

• A: $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(l)$; $\Delta H_A = -890.4 \frac{kJ}{mol}$
• B: $C(s, graphite) + O<em>2(g) \rightarrow CO</em>2(g)$; $\Delta H_B = -393.5 \frac{kJ}{mol}$
• C: $H<em>2(g) + \frac{1}{2}O</em>2(g) \rightarrow H<em>2O(l)$; $\Delta H</em>C = -285.8 \frac{kJ}{mol}$

Calculate $\Delta H$ for:

• D: $C(s, graphite) + 2H<em>2(g) \rightarrow CH</em>4(g)$

Steps:

1. Reverse equation A to move $CH_4$ to the product side (Equation E):
   • E: $CO<em>2(g) + 2H</em>2O(l) \rightarrow CH<em>4(g) + 2O</em>2(g)$; $\Delta H_E = 890.4 \frac{kJ}{mol}$
2. Keep equation B as is (Equation F):
   • F: $C(s, graphite) + O<em>2(g) \rightarrow CO</em>2(g)$; $\Delta H_F = -393.5 \frac{kJ}{mol}$
3. Multiply equation C by 2 (Equation G):
   • G: $2(H<em>2(g) + \frac{1}{2}O</em>2(g) \rightarrow H<em>2O(l))$; $\Delta H</em>G = 2 \times -285.8 \frac{kJ}{mol}$
4. Calculate $\Delta H_D$ from E + F + G:
   • $\Delta H<em>D = \Delta H</em>E + \Delta H<em>F + \Delta H</em>G$
   • $\Delta H_D = 890.4 + (-393.5) + 2(-285.8) = -74.7 \frac{kJ}{mol}$

State Functions

• Hess's law applies to any state function, including entropy and Gibbs free energy.

Bond Dissociation Energy

• Hess's law can be expressed in terms of bond enthalpies (bond dissociation energies).
• Bond dissociation energy is the average energy required to break a particular type of bond between atoms in the gas phase.
• Bond dissociation is an endothermic process.
• Units: kJ/mol of bonds broken.
• Bond enthalpies are averages from measurements of the same bond in many different compounds.
• Bond formation is the opposite of bond breaking and has the same magnitude of energy, but is negative (exothermic).
• Atoms form bonds to become more stable, often completing an octet.
• $\Delta H<em>{RXN}^\circ = \sum \Delta H</em>{(\text{bonds broken})} - \sum \Delta H_{(\text{bonds formed})} = \text{Total energy absorbed} - \text{Total energy released}$

Example: Calculating $\Delta H$ using Bond Dissociation Energies

• Reaction: $C(s) + 2H<em>2O(g) \rightarrow CH</em>4(g)$; $\Delta H = ?$
• Given: Bond dissociation energies of H-H = 436 kJ/mol, C-H = 415 kJ/mol, $\Delta H_f$ of C(g) = 715 kJ/mol.
• $CH<em>4$ is formed from free elements in their standard states (C(s) and $H</em>2(g)$). Thus, $\Delta H<em>{RXN} = \Delta H</em>f$

Steps:

1. $C(s) \rightarrow C(g)$; $\Delta H_1$
2. $2(H<em>2(g) \rightarrow 2H(g))$; $2 \times \Delta H</em>2$
3. $C(g) + 4H(g) \rightarrow CH<em>4(g)$; $\Delta H</em>3$

• $\Delta H<em>f = \Delta H</em>1 + 2 \times \Delta H<em>2 + \Delta H</em>3$
• $\Delta H<em>1 = \Delta H</em>f$ of C(g) = 715 kJ/mol.
• $\Delta H<em>2$ = bond enthalpy of $H</em>2$ = 436 kJ/mol.
• $\Delta H_3$ = -4 \times (bond energy of C-H) = -4 \times 415 kJ/mol = -1660 kJ/mol (negative because energy is released).
• $\Delta H_{RXN} = 715 + 2(436) - 1660 = -73 \frac{kJ}{mol}$

Standard Heat of Combustion ($\Delta H_{COMB}^\circ$)

• The standard heat of combustion ($\Delta H_{COMB}^\circ$) is the enthalpy change associated with the combustion of a fuel.
• Combustion reactions are ideal for enthalpy change measurements because they are spontaneous and fast.
• Most combustion reactions involve atmospheric oxygen ($O_2$), but other oxidants can be used (e.g., diatomic fluorine).
• Example: Hydrogen gas combusting with chlorine gas to form gaseous hydrochloric acid.
• Reactions in the $CH<em>4(g)$ example are combustion reactions with $O</em>2(g)$ as the oxidant; the enthalpy change listed for each reaction is the $\Delta H_{COMB}$.

Glycolysis

• The glycolytic pathway is a combustion reaction using glucose and oxygen to produce carbon dioxide and water.
• $C<em>6H</em>{12}O<em>6 + 6O</em>2 \rightarrow 6CO<em>2 + 6H</em>2O$
• The heat of combustion is found similarly to Hess's Law.
• Given numerous reactions and pathways, the overall enthalpy of the reaction can be determined.