Thermochemistry

Thermochemistry Chapter 6

Energy and change • The study of the energy involved in a change is THERMODYNAMICS • In thermodynamics, the universe is divided into System (what you are studying) and Surroundings (the rest of the universe!)

Systems • A system may be OPEN • both matter and energy can be exchanged with the surroundings

Systems • A system may be CLOSED • only energy can be exchanged with the surroundings

Systems • A system may be ISOLATED • neither matter nor energy can be exchanged with the surroundings

Systems may beopen, closed, or isolated

Internal energy • The internal energy E of a system is all the energy (both kinetic & potential) contained in the system • Chemists are especially interested in • thermal energy (energy of random molecular motion) • chemical energy (energy stored in chemical bonds and intermolecular forces)

First Law of Thermodynamics • The energy a system contains is its internal energy, E • Energy can move in or out of a system as heat (q) and/or work (w). • Energy is conserved, so all energy lost or gained by a system must be accounted for as heat and/or work:

E is a state function • A state function (aka function of state) is a property whose value depends only on the state of the system, not how it achieved that state • The state of the system is specified by the pressure, temperature, and composition of the system. • P, V, & T are state functions. E is a state function. • q and w are NOT state functions.

The elevation gain is the same by either path: a state function • How ∆E is distributed between q and w depends on which path you take: q and w are not state functions

Path 1: short circuit the battery with a wrench. Lots of heat, maybe even sparks and a fire, but no work! ∆E the same by either path fully charged battery fully discharged battery Path 2: connect the battery to a motor. Some heat, but also some work.

Heat • Heat is energy transferred because of a temperature difference • A system does not contain heat; it contains ENERGY • Heat is just a form by which energy is transferred • The other form by which energy is transferred is work Whether q or w, energy IN is positive, energy OUT is negative!

Heat • The amount of energy transferred as heat, q, is related to • the amount of matter gaining/losing energy (m) • the specific heat of matter gaining/losing energy (c) • the amount of the temperature change (∆T) James Prescott Joule

Heat capacity and specific heat • Specific heat capacity = c • Intrinsic capacity to gain/lose energy as heat • unit = J/g K or J/g °C • Molar specific heat capacity = J/mol K or J/mol °C • Heat capacity of a system = C • Quantity of energy to change temperature by 1 °C • unit = J/K or J/°C

Examples • How much heat energy (in kJ) is required to raise the temperature of 237 g ice water from 4.0 °C to 37.0 °C? cwater = 4.18 J/g °C • How much heat energy (in kJ) is required to raise the temperature of 2.50 kg Hg from –20.0 °C to –6.0 °C? cHg = 28.0 J/mol °C

Determination of specific heat • Heat sample of metal to temperature of boiling water • Measure temperature of measured amount of water in insulated beaker • Put hot metal in cold water and measure final temperature • All energy is assumed to stay in the insulated container • qmetal + qwater = 0 or qmetal = – qwater

Examples • When 1.00 kg Pb (specific heat = 0.13 J/g °C) at 100.0 °C is added to some water at 28.5 °C, the final temperature is 35.2 °C. What is the mass of the water? • 100.0 g Cu (specific heat 0.385 J/g °C) at 100.0 °C is added to 50.0 g water at 26.5 °C. What is the final temperature of the mixture?

Work • Work is done when a force acts for a distance: w = F x d • Energy is the capacity to do work • kinetic energy is the energy of motion • the random motion of molecules (thermal energy) is kinetic • potential energy is stored energy, that can do work when it is released • the energy in chemical bonds is potential energy

Pexternal   Pgas Pressure-volume work • Imagine a gas trapped in a cylinder with a moveable lid (a piston) • If the piston is not moving, how does the gas pressure inside compare to the external pressure? • If the piston is not moving, Pgas = Pexternal

Pexternal   Pgas Pressure-volume work • What if Pgas increases, so Pgas > Pexternal?

Pexternal   Pgas Pressure-volume work • The piston rises and the gas expands, until once again Pgas = Pexternal • The gas has moved the piston some distance against the opposing Pext • The gas has done work!

Pexternal   Pgas Pressure-volume work • How much work did the gas do? ∆h

Pexternal   Pgas Pressure-volume work • How much work did the gas do? • When the gas did work by expanding, it lost energy ∆h

Pexternal   Pgas Pressure-volume work • Now what if we increase the external pressure, so Pexternal > Pgas ?

Pressure-volume work • The gas is compressed until once again Pgas = Pexternal • The piston has been moved some distance against the opposing Pgas • This time work was done on the gas • How much work? Pexternal  ∆h  Pgas

Pressure-volume (P∆V) work • When the gas expands, it expends energy to do work = Pext∆V • ∆V is positive • energy leaves the system (the gas) • When the gas is compressed, it gains the energy used to compress it = Pext∆V • ∆V is negative • energy enters the system (the gas)

Sign conventions for work • When energy goes into the system as work, w is positive • When energy leaves the system as work, w is negative w = –P∆V when a gas expands ∆V is + and w is – (energy leaves the system) when a gas is compressed ∆V is – and w is + (energy enters the system)

Units • w = –P∆V • work is in joules (J) • P is in atm and V is in L, so P∆V is in atm L • The relationship between J and atm L is 1 atm L = 101.325 J

Example • What is the work done on a gas (in J) when the gas is compressed from an initial volume of 35.0 L to a final volume of 23.5 L under a constant pressure of 0.987 atm?

Calorimetry • The process for measuring the amount of heat energy exchanged by system and surroundings is CALORIMETRY • Assume total energy is constant: heat lost by system is gained by surroundings qsystem + qsurroundings = 0 qsystem = – qsurroundings

If the reaction releases energy, qrxn is negative (the reaction loses energy) The calorimeter gains that energy and gets hotter The reaction is EXOTHERMIC qrxn = – qcalorimeter If the reaction absorbs energy, qrxn is positive (the reaction gains energy) The calorimeter loses that energy and gets colder The reaction is ENDOTHERMIC qrxn = – qcalorimeter Heat of reaction, qrxn

Bomb calorimeter • Combustion reaction occurs in sample compartment • Reaction releases energy to water in calorimeter • qrxn = – qcalorimeter = – C∆T • C = heat capacity of calorimeter

Examples • The combustion of 1.013 g vanillin (C8H8O3) in a bomb calorimeter with heat capacity = 4.90 kJ/°C causes the temperature to rise from 24.89 °C to 30.09 °C. What is the heat of combustion of vanillin, in kJ/mol? • Combustion of 1.176 g benzoic acid (HC7H5O2, heat of combustion –26.42 kJ/g) causes the temperature in a bomb calorimeter to increase by 4.96 °C. What is C for that calorimeter?

Coffee cup calorimeter • Reactants (usually in aqueous solution) mix and react in the insulated cup • Reaction releases energy to (or absorbs energy from) liquid in which it is dissolved • qrxn = – qcal = – (mc∆T)cal • Calorimeter = liquid in the cup (cup assembly is usually ignored)

Example • 100.0 mL 1.00 M AgNO3 (aq) and 100.0 mL 1.00 M NaCl (aq), both initially at 22.4 °C, are mixed in a coffee cup calorimeter. The temperature rises to 30.2 °C. What is the heat of reaction, in kJ per mol AgCl, for the reaction: Ag1+ (aq) + Cl1– (aq)  AgCl (s)

Example • 100.0 mL 1.020 M HCl and 50.0 mL 1.988 M NaOH, both initially at 24.52 °C, are mixed in a coffee cup calorimeter. If qneutralization = –56 kJ/mol H2O, what will be the final temperature in the mixture? • Hint: this is a limiting reactant problem

Energy and Enthalpy • ∆E = q + w • w = –P∆V • At constant volume, ∆V = 0 so w = 0 and ∆E = qV • We define H = E + PV • H is called enthalpy • At constant pressure, ∆H = ∆E + P∆V • But ∆E = qP – P∆V, so ∆H = qP – P∆V + P∆V • ∆H = qP

∆E and ∆H • At constant volume, measured heat = ∆E • At constant pressure, measured heat = ∆H • Constant pressure is the more common lab condition • ∆H is the heat of reaction for a chemical change carried out at constant pressure

How different are ∆E and ∆H? • C (s) + 1/2 O2 (g)  CO (g) ∆H = – 110.5 kJ • ∆H = ∆E + ∆(PV) or ∆E = ∆H – ∆(PV) • ∆(PV) = ∆(nRT) = ∆ngasRT (solids are not significant) • ∆ngas = 1/2 mol • At 298 K, • ∆E is – 111.7 kJ, only 1% difference from ∆H

Enthalpy • Like E, absolute values of H cannot be measured, but changes in H can be measured: ∆H = qP • ∆H is part of the chemical reaction stoichiometry • ∆H is extensive (depends on amount of material) • ∆H is directional (sign reverses if direction of reaction reverses) • ∆H is a state function • Value of ∆H same whether reaction occurs in a single step or a series of steps, if final result is the same

Reactions add together • Look at these reactions: C (s) + O2 (g)  CO (g) + 1/2 O2 (g) CO (g) + 1/2 O2 (g)  CO2 (g) • The reactions add to give C (s) + O2 (g)  CO2 (g) • The CO and 1/2 O2 terms on both sides cancel

Heats of reaction add, too • DH values add just like reactions: ∆H (kJ/mol) • C (s) + O2 (g)  CO (g) + 1/2 O2 (g) –111 kJ/mol • CO (g) + 1/2 O2 (g)  CO2 (g)–283 kJ/mol • C (s) + O2 (g)  CO2 (g) –394 kJ/mol

Hess’ Law • Germain Hess discovered that heats of reaction add together in 1840 • The additivity of ∆H values is called Hess’ Law • Hess’ Law establishes enthalpy as a state function • State functions depend only on the state of the system • It doesn’t matter how the system achieved that state

Hess’ Law • Hess’ Law allows us to calculate ∆H for a reaction from measured ∆H values for other reactions • The rules of the game • Multiply reaction by a factor to get desired coefficients • Multiply ∆H by the same factor • Reverse a reaction to get a substance on the desired side • If you reverse the reaction, reverse the sign of ∆H

Example • Given these heats of reaction, • H2 (g) + 1/2 O2 (g)  H2O (l) ∆H = –286 kJ • 2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (l) ∆H = –3123 kJ • 2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (l) ∆H = –2602 kJ • Find ∆H for this reaction: C2H2 (g) + 2 H2 (g) ––> C2H6 (g)

Example • First look at the target reaction: C2H2 (g) + 2 H2 (g)  C2H6 (g) • You want 1 C2H2 on the left. We need to multiply the reaction with C2H2 by 1/2: • 1/2 [2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (l)] • C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (l) • Multiply ∆H by 1/2: ∆H = 1/2[–2602 kJ] = –1301 kJ

Example • Another look at the target reaction: C2H2 (g) + 2 H2 (g)  C2H6 (g) • You need 2 H2 on the left. We need to multiply the reaction with H2 by 2: • 2 [H2 (g) + 1/2 O2 (g)  H2O (l)] • 2 H2 (g) + O2 (g)  2 H2O (l) • Multiply ∆H by 2: ∆H = 2[–286 kJ] = –572 kJ

Example • The target reaction: C2H2 (g) + 2 H2 (g)  C2H6 (g) • Finally, you want 1 C2H6 on the right. Multiply the reaction with C2H6 by 1/2, and reverse it: • 1/2 [2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (l)] • reverse C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (l) • 2 CO2 (g) + 3 H2O (l)  C2H6 (g) + 7/2 O2 (g) • Multiply ∆H by 1/2, and reverse its sign: ∆H = 1/2[+3123 kJ] = +1561 kJ

Example • Now you have • C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (l) ∆H = –1301 kJ • 2 H2 (g) + O2 (g)  2 H2O (l) ∆H = –572 kJ • 2 CO2 (g) + 3 H2O (l)  C2H6 (g) + 7/2 O2 (g) ∆H = +1561 kJ • Now add the reactions and cancel items identical on both sides • The total is C2H2 (g) + 2 H2 (g)  C2H6 (g) ∆H = –312 kJ

Formation reactions • To organize collections of ∆H values for reactions, chemists defined a formation reaction • A formation reaction is the equation to form one mole of compound from its elements in their most stable state • The formation reaction for H2O is H2 (g) + 1/2 O2 (g)  H2O (l) • Notice that the reactants are elements in their most common, stable form • Notice that only one mole of product forms

Thermochemistry