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UMass Amherst CHEM 110 - Chemistry Chapter 5

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Chemistry Chapter 5Thermochemistry● Chemical change is associated with exchange of energy● Potential Energy: Energy associated with position○ If work is done to change the position of an object in a field, then its potentialenergy increases as a result of the change.○ If the change is in the opposite direction, the potential energy of the objectdecreases and the potential energy lost by the object could be used to do work.■ Gravitational energy■ Chemical energy● Electrons getting excited.■ Electrostatic energy● Hydrogen bonds and plus and minuses.● When oppositely charged objects come together due to attractiveforces, the potential energy decreases● Kinetic energy: Energy associated with motion○ Thermal energy■ Heat energy, enthalpy○ Mechanical energy■ Windmills. Turn physical things into energy.○ Electrical energy■ Electrons moving around. Electricity.○ KE = ½ m v2■ KE = Kinetic energy in joules■ m = mass in kg■ v = velocity● Enthalpy Change for a Reaction○ Breaking or forming chemical bonds leads to a change in enthalpy○ When the reactants and products in a chemical reaction are reversed, themagnitude of ΔH is identical but the sign of ΔH is reversed○ Enthalpies are state functions (does not depend on the path taken)○ More heat/enthalpy = less stable○ ΔHr = n x ΔHr^o■ ΔHr = Enthalpy of reaction■ n = moles■ ΔHr^o = Enthalpy of reaction under standard conditions.○ ΔHr = C x (E / n)■ Ex: When 8.83 grams of N2(g) react with sufficient O2(g), 20.9 kJ of energy areabsorbed. What is the value of H for the chemical equation given?● ΔHr = 1 x (20.9/.3151) = 66.3 kj■ ΔH = Enthalpy in kj■ C = Coefficient of molecule■ E = Energy absorbed or evolved in kj● Absorbed = Positive● Evolved = Negative■ n = moles of molecule● Exothermic:○ Releases energy○ Evolved = exothermic reaction○ ΔH = negative● Endothermic:○ Require energy○ ΔH = positive.● Hess’s Law○ If two or more chemical equations are added to give another chemical equation,the corresponding enthalpies of reaction can be added to obtain the desiredenthalpy of reaction● Standard Enthalpy of Formation○ The standard enthalpy of formation is the enthalpy change for formation of 1mole of a compound from its constituent elements in their standard state understandard conditions (1 atm, 25 oC)○ The ΔHf^o value for an element in its standard state is equal to 0 kJ/mol■ H2(g)■ N2(g)■ O2(g)■ F2(g)■ Cl2(g)■ Br2(l)■ I2(s)■ Li(s)■ Na(s)■ Ca(s)■ Mg(s)■ C(s)■ S8(s)■ P4○ ΔHr^o = ΔHf^o■ ΔH^o = Standard heat of formation■ ΔHf^o = standard enthalpy change○ ΔHr^o(reaction) = ΔHf^of(products) - ΔHf^of(reactants)■ Heat of reaction = (sum of heats of formation of products) - (sum of heatsof formation of reactants).● Standard Enthalpy of Reaction:○ Enthalpy of formation values, when combined with Hess’s law, are used tocalculate this● System: The object/set of objects or the reaction that is been studied● Surroundings: Anything other than the system that can exchange energy and/or matterwith the system● Universe: System + Surroundings● Open system: Matter and energy coming in and out of the system.● Closed system: No matter, but energy coming in or out of the system.● Isolated system: No matter or energy coming in or out of the system.● First Law of Thermodynamics○ The internal energy (U) of an isolated system is constant○ ΔU(universe) = 0○ ΔU = q + w○ w = ΔU - q○ q = ΔU + w■ ΔU = Internal energy of a system in joules● If positive:○ Non-spontaneous and○ Internal energy of the system increases.○ Energy is added to the system● If negative:○ Spontaneous○ Internal energy of the system decreases.○ Energy is removed from the system■ q = heat in joules● Absorbed(endothermic) = Positive● Evolved/generated(exothermic) = Negative■ w = work in joules● When work is done on the system w is positive● When work is done off/by the system w is negative.● Heat Transfer○ Heat transfer is associated with change in temperature● Specific Heat Capacity○ Specific heat capacity is the amount of energy required to raise the temperatureof 1 g of a substance by 1^oC○ q = m x Cspx T○ Csp= q / (m x ΔT)○ ΔT = q / (m x Csp)■ q = heat in joules● Heat gained = Heat lost○ q gained = -q lost■ m = mass in grams■ ΔT = change in temperature in Celsius(T2-T1)■ Csp = specific heat capacity●Specific heat is never less than zero.● Thermal Equilibrium○ Hot objects transfer heat to colder objects until both are at same temperature (astate of thermal equilibrium)● Energy and Phase changes:○ q = n x ΔH(fus)■ q = heat energy■ n = moles■ ΔH(fus) = heat of fusion○ q = n x ΔH(vap)■ q = heat energy■ n = moles■ ΔH(vap) = heat of vaporization● Energy Units○ SI unit of energy is Joule○ 1 Joule is defined as the energy required to accelerate 1 Kg object using a forceof 1 newton over a distance of 1 meter○ 1 Calorie (cal) is defined as the energy needed to raise the temperature of 1g ofwater by 1 degree Celsius○ 1 cal = 4.184 J○ 1 Cal =


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