Energy & Chemistry

Energy & Chemistry 2H2(g) + O2(g) → 2H2O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 → 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O → 4 OH- H2/O2 Fuel Cell Energy

Energy & Chemistry ENERGYis the capacity to do work (w) or transfer heat (q). HEATis the thermal energy that can be transferred from an object at one temperature to an object at another temperature – Net transfer of thermal energy stops when the two objects reach the same temperature. Other forms of energy — • Radiant (light)— energy in light, microwaves, and radio waves • Thermal (kinetic and potential) — results from atomic and molecular motion – Temperature of an object is a measure of the thermal energy content • Chemical — results from the particular arrangement of atoms in a chemical compound; radiant and thermal energy produced in this reaction due to energy released during the breaking and reforming of chemical bonds • Nuclear — radiant and thermal energy released when particles in the nucleus of the atoms are rearranged • Electrical— due to the flow of electricallychargedparticles

Potential & Kinetic Energy Kinetic energy • Potential energy – Energy storedin an object because of the relative positions or orientations of its components – PE = Fd = mad = mgh = work • Kinetic energy – Energy due to the motion of an object – KE = ½ mv2

Potential & Kinetic Energy

Internal Energy (E) • PE + KE = Internal energy (E or U) • Internal Energy of a chemical system depends on • number of particles • type of particles • temperature • The higher the T the higher the internal energy • So, use changes in T (∆T) to monitor changes in E (∆E).

heat transfer in (endothermic), +q heat transfer out (exothermic), -q w transfer in (+w) w transfer out (-w) Internal Energy (E) SYSTEM ∆E = q + w

Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. • The total energy is unchanged in a chemical reaction. • If PE of products is less than reactants, the difference must be released as KE. Energy Change in Chemical Processes Potential Energy of system dropped. Kinetic energy increased. Therefore, you often feel a Temperature increase.

Thermodynamics -Enthalpy • Thermodynamics – the science of heat (energy) transfer. Heat transfers until thermal equilibrium is established. ∆T measures energy transferred. • SYSTEM – The object under study • 1. Open system — can exchange both matter and energy with its surroundings • 2. Closed system — can exchange energy but not matter with its surroundings • 3. Isolated system — exchanges neither energy nor matter with the surroundings; total energy of the system plus the surroundings is constant • SURROUNDINGS – Everything outside the system

heat energy transferred work done by the system energy change FIRST LAW OF THERMODYNAMICS ∆E = q + w Energy is conserved!

The First Law of Thermodynamics • Exothermic reactions generate specific amounts of heat. • This is because the potential energies of the products are lower than the potential energies of the reactants.

The First Law of Thermodynamics • There are two basic ideas of importance for thermodynamic systems. • Chemical systems tend toward a state of minimum potential energy. • Chemical systems tend toward a state of maximum disorder. • The first law is also known as the Law of Conservation of Energy. • Energy is neither created nor destroyed in chemical reactions and physical changes.

State of a System • The state of the system is a complete description of the system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter Some examples of state functions are: • T (temperature), P (pressure), V (volume), E (change in energy), H (change in enthalpy – the transfer of heat), and S (entropy) • Examples of non-state functions are: • n (moles), q (heat), w (work) ∆H along one path = ∆H along another path • This equation is valid because ∆H is a STATE FUNCTION • These depend only on the state of the system and not how it got there. • V, T, P, energy — and your bank account! • Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.

Standard States and Standard Enthalpy Changes • Thermochemical standard state conditions • The thermochemical standard T = 298.15 K. • The thermochemical standard P = 1.0000 atm. • Be careful not to confuse these values with STP. • Thermochemical standard states of matter • For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. • For gases the standard state is the gas at 1.00 atm of pressure. • For gaseous mixtures the partial pressure must be 1.00 atm. • For aqueous solutions the standard state is 1.00 M concentration. • ∆Hfo = standard molar enthalpy of formation • the enthalpy change when 1 mol of compound is formed from elements under standard conditions.

State Function • The properties of a system that depend only on the state of the system are called state functions. • State functions are always written using capital letters. • The value of a state function is independent of pathway. • An analog to a state function is the energy required to climb a mountain taking two different paths. • E1 = energy on the first floor of Heldenfels • E1 = mgh1 • E2 = energy on the fourth floor of Heldenfels • E2 = mgh2 • E = E2-E1 = mgh2 – mgh1 = mg(h)

ENTHALPY Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆H where H = enthalpy and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = Hfinal - Hinitial

T(system) goes down T(surr) goes up Directionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.

T(system) goes up T (surr) goes down Directionality of Heat Transfer • Heat always transfers from hotter object to cooler one. • ENDOthermic: heat transfers from SURROUNDINGSto theSYSTEM.

ENTHALPY ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC

USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) → H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

USING ENTHALPY Example of HESS’S LAW— Making liquid H2O from H2 + O2 involves twoexothermic steps. Liquid H2O H2O vapor H2 + O2 gas Making liquid H2O from H2 involves two steps. H2(g) + 1/2 O2(g) → H2O(g) + 242 kJ H2O(g) → H2O(l) + 44 kJ H2(g) + 1/2 O2(g) → H2O(l) + 286 kJ If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H2(g) + 1/2 O2(g) → H2O(g) ∆H˚ = -242 kJ 2H2(g) + O2(g) → 2H2O(g) ∆H˚ = -484 kJ H2O(g) → H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ H2(g) + 1/2 O2(g) → H2O(l) ∆H˚ = -286 kJ

Standard Molar Enthalpies of Formation, Hfo • The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements. • The symbol for standard molar enthalpy of formation is Hfo. • The standard molar enthalpy of formationfor MgCl2 is:

Standard Molar Enthalpies of Formation, Hfo • Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 and Appendix K in the text. • Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. • Example 15-4: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for whichHorxn = -1281 kJ. P in standard state is P4 Phosphoric acid in standard state is H3PO4(s)

Standard Molar Enthalpies of Formation, Hfo H2(g) + ½ O2(g) → H2O(g) ∆Hf˚ (H2O, g)= -241.8 kJ/mol C(s) + ½ O2(g) → CO(g)∆Hf˚ of CO = - 111 kJ/mol By definition, ∆Hfo= 0 for elements in their standard states. Use ∆H˚’s to calculate enthalpy change for H2O(g) + C(graphite) → H2(g) + CO(g)

Standard Molar Enthalpies of Formation, Hfo • Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.

Standard Molar Enthalpies of Formation, Hfo • Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.

Hess’s Law – Enthalpies of Formation and Reaction • Standard enthalpies of reaction (ΔH°xn) – The enthalpy that occurs when a reaction is carried out with all reactants and products in their standard states – General reaction: aA + bB  cC + dD where A, B, C, and D are chemical substances and a, b, c, and d are their stoichiometric coefficients – Magnitude of ΔH°rxn is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient , minus the sum of the standard enthalpies of formation of the reactants, multiplied by their coefficients ΔH°rxn= [cΔH°f(C) + dΔH°f(D)]  [aΔH°f(A) + bΔH°f(B)] or ΔH°rxn= mΔH°f(product)  nΔH°f(reactants) where  means “sum of” and m and n are the stoichiometric coefficients of each of the products and reactants, respectively

Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

Enthalpies of Formation • Standard enthalpies of formation –Themagnitude of ΔH for a reaction depends on the physical states of the reactants and products, the pressure of any gases present, and the temperature at which the reaction is carried out. – A specific set of conditions under which enthalpy changes are measured are used to ensure uniformity of reaction conditions and data. – Standard conditions: a pressure of 1 atmosphere (atm) for gases and a concentration of 1 M for species in solution (1 mol/L), each pure substance must be in its standard state, its most stable form at a pressure of 1 atm at a specified temperature. – Enthalpies of formation measured under standard conditions are called standard enthalpies of formation (ΔH°f). – Standard enthalpy of formation of any element in its standard state is zero.

Enthalpies of Reaction  Hess’s law allows the calculation of the enthalpy change for any conceivable chemical reaction by using a relatively small set of tabulated data • Enthalpy of combustion, ΔHcomb—enthalpy change that occurs when a substance is burned in excess oxygen • Enthalpy of fusion, ΔH fus—enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance • Enthalpy of vaporization, ΔHvap—the enthalpy change that accompanies the vaporization of 1 mol of a substance • Enthalpy of solution, ΔHsoln—enthalpy change when a specified amount of solute dissolves in a given quantity of solvent • Enthalpy of formation, ΔHf—enthalpy change for the formation of 1 mol of a compound from its component elements

Hess’s Law • Hess’s Law of Heat Summation, Hrxn = H1 +H2 +H3 + ..., states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. • Hess’s Law is true because H is a state function. • If we know the following Ho’s

Hess’s Law • For example, we can calculate the Ho for reaction [1] by properly adding (or subtracting) the Ho’s for reactions [2] and [3]. • Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product. • Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product. • Each reaction can be doubled, tripled, or multiplied by half, etc. • The Ho values are also doubled, tripled, etc. • If a reaction is reversed the sign of the Ho is changed.

Hess’s Law • Given the following equations and Hovalues calculate Ho for the reaction below.

Hess’s Law • Calculate the H o298 forthe following reaction:

Hess’s Law • Application of Hess’s Law and more algebra allows us to calculate the Hfofor a substance participating in a reaction for which we know Hrxno , if we also know Hfofor all other substances in the reaction. • Given the following information, calculate Hfo for H2S(g).

HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity? Thermal energy cannot be measured easily. Temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry is the set of techniques employed to measure enthalpy changes in chemical processes using calorimeters.

Specific heat capacity = heat lost or gained by substance (J) (mass, g) (T change, K) Specific Heat Capacity How much energy is transferred due to Temperature difference? The heat (q) “lost” or “gained” is related to a) sample mass b) change in T and c) specific heat capacity

Table of specific heat capacities • Heat capacity of an object depends on both its massand itscomposition. • Molar heat capacity (Cp) — amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; units of Cp are J/(mol•°C) • Specific heat (Cs) — amount of energy needed to increase the temperature of 1 g of a substance by 1°C, units are J/(g•°C) • The quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change are related in two ways: q = nCpΔT where n = number of moles of substance q = mCsΔT where m = mass of substance in grams Aluminum

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?

Heat/Energy TransferNo Change in State q transferred = (sp. ht.)(mass)(∆T)

Heat Transfer • Use heat transfer as a way to find specific heat capacity, Cp • 55.0 g Fe at 99.8 ˚C • Drop into 225 g water at 21.0 ˚C • Water and metal come to 23.1 ˚C • What is the specific heat capacity of the metal?

Heating/Cooling Curve for Water Note that T is constant as ice melts or water boils

Thermochemical equations are a balanced chemical reaction plus the H value for the reaction. For example, this is a thermochemical equation. The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles. 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions. Thermochemical Equations

This is an equivalent method of writing thermochemical equations. H < 0 designates an exothermic reaction. H > 0 designates an endothermic reaction Thermochemical Equations

Thermochemical Equations • Write the thermochemical equation for the reaction for CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq) 50.0mL of 0.400 M CuSO4 at 23.35 oC Tfinal 25.23oC 50.0mL of 0.600 M NaOH at 23.35 oC Density final solution = 1.02 g/mL CH2O = 4.184 J/goC

Gas Laws Standard Temperature and Pressure • At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: • 1. Temperature (expressed in K) • 2. Volume (expressed in liters) • 3. Amount (expressed in moles) • 4. Pressure (given in atmospheres) • • These variables are not independent — if the values of any three of these quantities are known, the fourth can be calculated. • Standard temperature and pressure is given the symbol STP. • It is a reference point for some gas calculations. • Standard P  1.00000 atm or 101.3 kPa • Standard T  273.15 K or 0.00oC • Gas laws must use the Kelvin scale to be correct. • Relationship between Kelvin and centigrade.

Boyle’s Law: The Volume-Pressure Relationship • V  1/P or V= k (1/P) or PV = k • P1V1 = k1 for one sample of a gas. • P2V2 = k2 for a second sample of a gas. • k1 = k2 for the same sample of a gas at the same T. • Mathematically we write Boyle’s Law as P1V1 = P2V2 • This relationship between pressure and volume is known as Boyle’slaw which states that at constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

Charles’ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale Charles’s and Gay-Lussac’s findings can be stated as: At constant pressure, the volume of a fixed amount of a gas is directly proportional to its absolute temperature (in K). This relationship is referred to as Charles’s law and is stated mathematically as V = (constant) [T (in K)] or V  T (in K, at constant P). Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

Charles’ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale 35 30 25 20 Volume (L) 15 Gases liquefy 10 before reaching 0K 5 • Gay-Lussac showed that a plot of V versus T was a straight line that could be extrapolated to –273.15ºC at zero volume, a theoretical state. • The slope of the plot of V versus T varies for the same gas at different pressures, but the intercept remains constant at –273.15ºC. • Significance of the invariant T intercept in plots of V versus T was recognized by Thomson (Lord Kelvin), who postulated that –273.15ºC was the lowest possibletemperature that could theoretically be achieved, and he called it absolute zero (0 K). 0 0 50 100 150 200 250 300 350 400 Temperature (K) absolute zero = -273.15 0C

The Combined Gas Law Equation • Boyle’s and Charles’ Laws combined into one statement is called the combined gas law equation. • Useful when the V, T, and P of a gas are changing.

Energy & Chemistry