Thermochemistry: PV Work, Internal Energy, Heat Transfer, and Endo/Exothermic Processes

Thermochemistry: Energy Changes in Chemical Reactions

Calculating the Amount of PV Work

In chemical reactions involving gases, the system may do work on the surroundings by expanding or contracting against an external pressure. This work is called PV work and is calculated using the following formula:

• PV Work Formula: $w = -P riangle V$ where w is work (in L·atm or J), P is external pressure, and ΔV is the change in volume.

• Sign Convention: If the system expands ($riangle V > 0$), work is done by the system (w is negative). If the system contracts ($riangle V < 0$), work is done on the system (w is positive).

• Unit Conversion: $1 ext{ L·atm} = 101 ext{ J}$

Example: If a system expands from 12.0 L to 14.5 L at 5.0 atm: $w = -(5.0 ext{ atm})(2.5 ext{ L}) = -12.5 ext{ L·atm}$ $-12.5 ext{ L·atm} imes 101 ext{ J/L·atm} = -1.3 imes 10^3 ext{ J} = -1.3 ext{ kJ}$

Additional info: The negative sign indicates energy leaves the system as work.

Calculating Internal Energy Change ($riangle E$) for a Reaction

The internal energy change ($riangle E$) of a reaction is related to the enthalpy change ($riangle H$), pressure, and volume change:

• Relationship: $riangle H = riangle E + P riangle V$ Rearranged: $riangle E = riangle H - P riangle V$

• Application: For a reaction at constant pressure, calculate $P riangle V$ and subtract from $riangle H$ to find $riangle E$.

Example: For $N_2(g) + 3H_2(g) ightarrow 2NH_3(g)$, $riangle H = -92.2 ext{ kJ}$, $P = 40.0 ext{ atm}$, $riangle V = -1.12 ext{ L}$: $P riangle V = (40.0 ext{ atm})(-1.12 ext{ L}) = -44.8 ext{ L·atm}$ $-44.8 ext{ L·atm} imes 101 ext{ J/L·atm} = -4520 ext{ J} = -4.52 ext{ kJ}$ $riangle E = -92.2 ext{ kJ} - (-4.52 ext{ kJ}) = -87.7 ext{ kJ}$

Additional info: $riangle E$ is often similar in magnitude to $riangle H$ for reactions where $P riangle V$ is small.

Calculating the Amount of Heat Released in a Reaction

The heat released or absorbed in a chemical reaction can be calculated using the balanced thermochemical equation and the amount of reactant.

• Thermochemical Equation: Example: $C_3H_8(g) + 5O_2(g) ightarrow 3CO_2(g) + 4H_2O(g)$, $riangle H^ ext{o} = -2044 ext{ kJ}$

• Calculating Heat for a Given Mass: Convert grams to moles, then use the equation: $q = ext{moles} imes rac{ riangle H^ ext{o}}{ ext{mol}}$

Example: 5.00 g of $C_3H_8$ ($ext{molar mass} = 44.09 ext{ g/mol}$): $5.00 ext{ g} imes rac{1 ext{ mol}}{44.09 ext{ g}} = 0.113 ext{ mol}$ $0.113 ext{ mol} imes rac{2044 ext{ kJ}}{1 ext{ mol}} = 231 ext{ kJ}$

Additional info: The amount of heat released is proportional to the amount of reactant used.

Classifying Endothermic and Exothermic Processes

Chemical and physical processes can be classified as endothermic (heat absorbed) or exothermic (heat released) based on the direction of heat flow between the system and surroundings.

• Endothermic Process: The system absorbs heat from the surroundings. Example: Melting ice ($H_2O(s) ightarrow H_2O(l)$), evaporation of sweat.

• Exothermic Process: The system releases heat to the surroundings. Example: Freezing water ($H_2O(l) ightarrow H_2O(s)$), dissolving MgSO$_4$ in water to produce a hand warmer.

• Sign of $riangle H$: Endothermic: $riangle H > 0$ Exothermic: $riangle H < 0$

Example:

• Freezing water: Exothermic, $riangle H < 0$

• Evaporating sweat: Endothermic, $riangle H > 0$

• Dissolving MgSO$_4$ in water: Exothermic, $riangle H < 0$

Classifying Endo- and Exothermic Processes: Additional Examples

• Condensation from shower on mirror: Exothermic, $riangle H < 0$

• Burning charcoal: Exothermic, $riangle H < 0$

• CO$_2$ subliming: Endothermic, $riangle H > 0$

Cold Pack Chemistry

Cold packs and hot packs use salts that dissolve in water to absorb or release heat. The salt with the most positive $riangle H$ will absorb the most heat and cause the greatest temperature decrease.

Salt Dissolution	ΔH (kJ)
LiF(aq) + F(aq)	+5.5
LiCl(aq) + Cl(aq)	-37.1
BaSO4(aq) + SO4(aq)	+26.3
CaCl2(aq) + 2Cl(aq)	-81.8

Additional info: BaSO4 would result in the greatest temperature decrease when dissolved in water for a cold pack, due to its large positive enthalpy of dissolution.