Thermochemistry: Chemical Energy Chapter 8 Thermochemistry: Chemical Energy

Energy Energy – capacity to supply heat or do work Energy = Heat + Work E = q + w 2 types of Energy Potential Energy Kinetic Energy

Energy Two fundamental kinds of energy. Law of Conservation of Energy Potential energy is stored energy. Kinetic energy is the energy of motion. Law of Conservation of Energy Energy can be converted from one kind to another but never destroyed

Energy Units Conversions SI Unit – Joule (J) Additional units Calorie (Cal) – food calorie calorie (cal) – scientific calorie Conversions 1 cal = 4.184 J 1000 cal = 1 Cal

Energy and Chemical Bonds Chapter 6 Kept a careful accounting of atoms as they rearranged themselves Reactions also involve a transfer of energy

Energy and Chemical Bonds A chemical Potential - attractive forces in an ionic compound or sharing of electrons covalent compound Kinetic – (often in form of heat) occurs when bonds are broken and particles allowed to move To determine the energy of a reaction it is necessary to keep track of the energy changes that occur during the reaction

Internal Energy and State Functions In an experiment: Reactants and products are the system; everything else is the surroundings. Energy flow from the system to the surroundings has a negative sign (loss of energy). Energy flow from the surroundings to the system has a positive sign (gain of energy).

Internal Energy and State Functions Tracking energy changes Energy changes are measured from the point of view of the system (Internal Energy - IE) Change in Energy of the system – ΔE ΔE = Efinal - Einitial

Internal Energy and State Functions IE depends on Chemical identity, sample size, temperature, etc. Does not depend on the system’s history Internal Energy is a state function A function or property whose value depends only on the present state (condition) of the system, not on the path used to arrive at that condition

Expansion Work E = q + w In physics w = force (F) x distance (d) Force – energy that produces movement of an object In chemistry w = expansion work Force - the pressure that the reaction exerts on its container against atmospheric pressure hence it is negative Distance – change in volume of the reaction w = -PΔV

Energy and Enthalpy q = DE + PDV ΔE = q – PΔV The amount of heat exchanged between the system and the surroundings is given the symbol q. q = DE + PDV At constant volume (DV = 0): qv = DE At constant pressure: Energy due to heat and work but work minimal compared to heat energy qp = DE + PDV = DH Enthalpy change (heat of reaction): DH = Hproducts – Hreactants

The Thermodynamic Standard State ΔH = amount of energy absorbed or released in the form of heat DH = Hproducts – Hreactants Important factors States of matter Thermodynamic standard state – most stable form of a substance at 1 atm and at a specified temperature, usually 25oC; and 1 M concentration for all substances in solution DH – valid for the reaction as written including exact # of moles of substances N2H4(g) + H2(g)  2 NH3(g) + heat (188 kJ)

Enthalpies of Physical and Chemical Change

Enthalpies of Physical and Chemical Changes Enthalpies of Chemical Change: Often called heats of reaction (DHreaction). Endothermic: Heat flows into the system from the surroundings and DH has a positive sign. Unfavorable Process Exothermic: Heat flows out of the system into the surroundings and DH has a negative sign. Favorable process

Enthalpies of Physical and Chemical Changes Reversing a reaction changes the sign of DH for a reaction. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ Multiplying a reaction increases DH by the same factor. 3 C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l) DH = –6657 kJ

Problems How much heat (in kilojoules) is evolved or absorbed in each of the following reactions? Burning of 15.5 g of propane: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride: Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l) DH = +80.3 kJ

Determination of Heats of Reaction Experimentally – calorimetry Hess’s Law Standard Heat’s of Formation Bond Dissociation Energies

Calorimetry and Heat Capacity Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters: Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = DE. Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = DH.

Calorimetry and Heat Capacity

Calorimetry and Heat Capacity Heat capacity (C) is the amount of heat required to raise the temperature of an object or substance a given amount. Specific Heat: The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C. Molar Heat: The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C. C = q D T

Problems What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C? When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0°C in a calorimeter, the temperature of the solution increases to 33.9°C. Assume specific heat of solution is 4.184 J/(g–1·°C–1), and the density is 1.00 g/mL–1, calculate the heat absorbed or released for this reaction.

Hess’s Law Allows the enthalpy to be determined for: Reactions that occur too quickly or take too long to use calorimetry Reactions that are too dangerous Works like the Haber process in chapter 6 Take reactions for which the heat is known and manipulate them to give the desired reaction

Standard Heats of Formation Standard Heats of Formation (DH°f): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. The standard heat of formation for any element in its standard state is defined as being ZERO. DH°f = 0 for an element in its standard state

Standard Heats of Formation H2(g) + 1/2 O2(g)  H2O(l) DH°f = –286 kJ/mol 3/2 H2(g) + 1/2 N2(g)  NH3(g) DH°f = –46 kJ/mol 2 C(s) + H2(g)  C2H2(g) DH°f = +227 kJ/mol 2 C(s) + 3 H2(g) + 1/2 O2(g)  C2H5OH(g) DH°f = –235 kJ/mol

Standard Heats of Formation Calculating DH° for a reaction: DH° = Σ[DH°f (products) x moles] – Σ[DH°f (Reactants) x moles] For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. aA + bB cC + dD DH° = [cDH°f (C) + dDH°f (D)] – [aDH°f (A) + bDH°f (B)]

Problems Calculate DH° (in kilojoules) for the reaction of ammonia with O2 to yield nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid. Calculate DH° (in kilojoules) for the photosynthesis of glucose from CO2 and liquid water, a reaction carried out by all green plants.

Energy Calculations Other methods for calculating enthalpies Bond dissociation energies – measures the energy given off by the formation of bonds in the products and substracts the energy required to break bonds in the reactants

Why do chemical reactions occur? A chemical reaction will move from less stability to greater stability. Achieved by giving off more energy than is absorbed by the reactants This indicates that exothermic reactions occur by why do endothermic reactions occur? Gibb’s Free Energy DG = DH – TDS DH – enthalpy, T – temperature, DS - entropy

An Introduction to Entropy Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. (increases the degree of disorder) A spontaneous process is one that proceeds on its own without any continuous external influence. A nonspontaneous process takes place only in the presence of a continuous external influence.

An Introduction to Entropy

An Introduction to Entropy

An Introduction to Entropy The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. Entropy has units of J/K (Joules per Kelvin). DS = Sfinal – Sinitial Positive value of DS indicates increased disorder (favorable). Negative value of DS indicates decreased disorder (unfavorable).

Problems Predict whether DS° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate DS° for each: a. 2 CO(g) + O2(g)  2 CO2(g) b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g) c. C2H4(g) + Br2(g)  CH2BrCH2Br(l) d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)

An Introduction to Free Energy To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: Spontaneous process: Decrease in enthalpy (–DH). Increase in entropy (+DS). Nonspontaneous process: Increase in enthalpy (+DH). Decrease in entropy (–DS).

An Introduction to Free Energy Gibbs Free Energy Change (DG): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process. DG = DH – TDS DG < 0 Process is spontaneous (favorable) DG = 0 Process is at equilibrium DG > 0 Process is nonspontaneous (unfavorable)

Problems Which of the following reactions are spontaneous under standard conditions at 25°C? a. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) DG° = –55.7 kJ b. 2 C(s) + 2 H2(g)  C2H4(g) DG° = 68.1 kJ c. N2(g) + 3 H2(g)  2 NH3(g) DH° = –92 kJ; DS° = –199 J/K

An Introduction to Free Energy Equilibrium (DG° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? N2(g) + 3 H2(g)  2 NH3(g) DH° = –92.0 kJ DS° = –199 J/K Equilibrium is the point where DG° = DH° – TDS° = 0

Problem Benzene, C6H6, has an enthalpy of vaporization, DHvap, equal to 30.8 kJ/mol and boils at 80.1°C. What is the entropy of vaporization, DSvap, for benzene?

Optional Homework Text - 8.28, 8.32, 8.50, 8.52, 8.56, 8.58, 8.66, 8.70, 8.74, 8.82, 8.88, 8.90 Chapter 8 Homework from website

Required Homework Chapter 8 Assignment