Chapter 1: Fundamental Concepts -

Chapter 1: Fundamental Concepts -

Ch. 6: Energy and Thermochemistry Energy: Ability to do work Kinetic Energy: due to motion, mv2 Potential Energy: stored, due to position or composition Thermal Energy: movement of molecules; related to temperature Chemical Energy: positions of nuclei

and electrons (bonds) bond breaking energy is required bond making energy is released Heat and Temperature Heat is the flow of energy caused by a temperature difference; thermal energy being transferred. Temperature is a measure of the intensity of heat (thermal energy). Internal Energy -E = internal energy = PE + KE -Energy Changes When reactions occur there is an energy change E: E = Efinal - Einitial Efinal = energy of products Einitial = energy of reactants

reactants E products E = q + w where q = heat w = work sign convention: positive if system gains energy Measuring Thermal Energy 1. Units of Energy (see Table 6.1) SI: 1 J = 1 kgm2/s2 (mv2) memorize! 1 cal = 4.184 joule (exactly) 1 kcal = 1 Cal

2. Heat Capacity, C --amount of heat required to raise the temp of substance by 1 C units: energy/temp (J/C or kJ/C or cal/C) quantity of heat = (heat capacity) x t 3. Specific Heat, Cs Heat Capacity of specified mass of substance (1 gram) units: usually J/g C or cal/g C e.g. specific heat of water = 1.00 cal/g C = 4.18 J/g C quantity of heat = (specific heat) x mass x t 4. Molar Heat Capacity: Heat Capacity per mole of a substance e.g. molar heat capacity of water = 18.0 cal/mole C = J/g C x g x C Example Problems

Problem The temp of 250 g H2O (3 sig fig) is raised from 25.0 C to 30.0 C. How much heat energy is required? t = 30.0 - 25.0 = 5.0 C Amount of heat = (1.00 cal/g C) x (250 g) x (5.0 C) = 1,250 cal = 1.25 kcal = 1,250 cal x 4.184 J/cal = 5320 J = 5.3 kJ Problem Identify each energy change as primarily heat or work, and determine whether Esys is positive or negative. a. One billiard ball (the system) hits another one, and stops rolling. b. A book (the system) is dropped on the floor c. A father pushes his daughter on the swing (the daughter & swing are the system) a. work, negative b. work, negative c. work, positive

Thermal Energy Transfer Thermal energy flows from matter at higher temperature to matter at lower temp, until thermal equilibrium. qA = qB Example Problem A 3.35 g iron rod, initially at 22.7 C, is submerged into an unknown mass of H2O at 63.2 C, in an insulated container. The final temp of the mixture is 59.5 C. What is the mass of the water? (Cs iron = 0.449 J/g C, Cs water = 4.18 J/g C) m x CS,Fe x TFe = m x CS,H2O x TH2O TFe = 59.5 22.7 = 36.8 C for Fe TH2O = 59.5 63.2 = 3.7 C for H2O (3.35)(0.449)(36.8) = m(4.18)( 3.7) m = (3.35)(0.449)(36.8)/(4.18)(3.7) = 3.6 g Internal Energy and Enthalpy E = internal energy = KE + PE

H = enthalpy = E + PV (when P is constant) E = total energy change H = total heat change (see book for derivation) Bomb Calorimeter constant V; measures E Coffee-Cup Calorimeter constant P; measures H Internal energy and enthalpy are state functions. The energy change (E) and heat change (H) of a reaction depend only on the initial and final states of the system -- not on the specific pathway. Enthalpy Changes in Chemical Reactions

exothermic reaction reaction endothermic reaction heat is a product of the -- gives off heat to the surroundings -- system warms up heat is essentially a reactant --absorbs heat from the surroundings -- system cools off Enthalpy (H) -- Heat Content The total energy of a chemical system at constant pressure H = Hproducts - Hreactants endothermic reaction absorbed exothermic reaction released

H > 0 (positive) -- heat is H < 0 (negative) -- heat is Standard Heat of Reaction (H) H = the value of H for a reaction as written. H = the value of H for a reaction: Under standard conditions (temp = 25 C, pressure = 1 atm) With actual # moles specified by coefficients in balanced eqn e.g. reaction for the combustion of ethylene: C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l) H = -1411 kJ (very exothermic) i.e. 1411 kJ of heat energy are released in the reaction of 1 mole of C2H4 with 3 moles of O2 Problem If 10.0 g of C2H4 are burned, how much heat is produced? (10.0 g C2H4) x (1 mole C2H4/28.0 g C2H4) x (1411 kJ/mole C2H4)

= 504 kJ Manipulating Thermochemical Equations If reaction is reversed, change sign of H. If reaction is multiplied or divided by a factor, apply same factor to H. H for overall reaction = sum of H values for individual reactions. Problem Given the following thermochemical reactions: (eq 1) C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l) H = -1411 kJ (eq 2) C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(l) H = -1367 kJ Calculate H for the following reaction: C2H4(g) + H2O(l) --> C2H5OH(l) Example Problem, cont. Reverse 2nd reaction to put C2H5OH on product side then rewrite 1st equation and add them together. (eq 2)

2 CO2(g) + 3 H2O(l) --> C2H5OH(l) + 3 O2(g) H = + 1367 kJ (note the sign change!!!) (eq 1) C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l) H = -1411 kJ Net: C2H4(g) + H2O(l) --> C2H5OH(l) {note: 3 O2, 2 CO2, and 2 H2O cancel out} H = H1 + H2 = 1367 + (-1411) = -44 kJ Standard Heat of Formation Standard heat (enthalpy) of formation of a substance: Hf =

H for the formation of one mole of substance from the elements in their standard states a formation reaction H2(g) + 1/2 O2(g) --> H2O(l) Hf (liq water) = -286 kJ/mole Hf is a property of a substance -- see text for examples Practice writing formation reactions -- e.g. Na2SO4 2 Na(s) + 2 O2(g) + S(s) --> Na2SO4(s) Hf = -1385 kJ/mole Hess Law of Heat Summation Calculate H for a reaction from tabulated Hf values H = Hf (products) - Hf (reactants) Problem Determine H for the following reaction from Hf values. 2 H2O(l) + CaSO4(s) --> CaSO42H2O(s) H = Hf[CaSO42H2O(s)] - {Hf[CaSO4(s)] + 2 Hf[H2O(l)]}*

= (-2021.1) - {(-1432.7) + 2(-285.9)} = -16.6 kJ {*units: e.g., (2 moles) x (285.9 kJ/mole) = kJ} Summary two ways to get H for a reaction: By manipulating 2 or more given equations, then adding their Hs From tabulated Hf values using Hess Law Sample Problems Write a balanced chemical equation that represents the formation reaction for (NH4)3BO3. Given the following thermochemical equations, calculate the standard heat of formation (Hf) of Mg3N2(s) in kJ/mole. Mg3N2(s) + 3 H2(g) --> 3 Mg(s) + 2 NH3(g) H = 371 kJ

1/2 N2(g) + 3/2 H2(g) --> NH3(g) H = -46 kJ Sample Problem The specific heat of copper is 0.387 J/g C. The molar heat of fusion of water is 6.0 kJ/mole. If a copper rod weighing 225 g is heated to 80 C and then immersed in 100 g of ice at 0 C, how many grams of ice will melt?

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