Calorimetry

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial

How much heat is given off when an 869 g iron bar cools from 94oC to 5oC? s of Fe = 0.444 J/g • oC Dt = tfinal – tinitial = 5oC – 94oC = -89oC q = msDt = 869 g x 0.444 J/g • oC x –89oC = -34,000 J

Problem 6.33 A 6.22 kg piece of copper metal is heated from 20.5 ºC and 324.3 ºC. Calculate the heat absorbed (in kJ) by the metal.

Heating Curve for Water Ice Ice + water Water Water + steam ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = 4.184 J/g˚C http://www.kentchemistry.com/links/Matter/HeatingCurve.htm

Calculate the energy to heat 17. 9g of ice from 0 Calculate the energy to heat 17.9g of ice from 0.00˚C to liquid water at 100.˚C. (ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = 4.184 J/g˚C)

Constant-Volume Calorimetry qsys = qcal+ qrxn qsys = 0 qrxn = - qcal qcal = m x s x Dt qcal = Ccal x Dt Reaction at Constant V DH = qrxn DH ~ qrxn No heat enters or leaves!

Dt = tfinal – tinitial = 25.95oC – 20.28oC = 5.67oC A quantity of 1.435 g of naphthalene (C10H8), a pungent smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently the temperature of the water rose from 20.28ºC to 25.95ºC. If the heat capacity of the bomb plus water was 10.17 kJ/ºC, calculate the heat of combustion of naphthalene on a molar basis; that is find the molar heat of combustion. qcal = Ccal Dt Ccal= 10.17 kJ/oC Dt = tfinal – tinitial = 25.95oC – 20.28oC = 5.67oC qcal = Ccal Dt = 10.17 kJ/oC x 5.67oC = 57.66 kJ qrxn moles = 128.2 g mole x Dhcomb -57.66 kJ 1.435 g = = -5.151 x 103 kJ/mol

Problem 6.37 A 0.1375 g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3024 J/ºC. The temperature increases by 1.126ºC. Calculate the heat given off by the burning Mg in kJ/mol.

Constant-Pressure Calorimetry qsys = qcal + qrxn qsys = 0 qrxn = - (qcal) qcal = m x s x Dt qcal = Ccal x Dt Reaction at Constant P DH = qrxn No heat enters or leaves!

A lead (Pb) pellet having a mass of 26. 47 g at 89 A lead (Pb) pellet having a mass of 26.47 g at 89.98ºC was placed in a constant pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50ºC to 23.17ºC. What is the specific heat of the lead pellet? qPb + qwater = 0 qPb = - qwater qwater = ms∆t Dt = tfinal – tinitial = 23.17oC – 22.50oC = 0.67oC qwater = ms∆t = 100.0g x 4.184 J/goC x 0.67oC = 280.3 J

The heat lost be the pellet is equal to the heat gained by the water, so qPb = -280.3 J Dt = tfinal – tinitial = 23.17oC – 89.98oC = -66.81oC qPb = ms∆t -280.3 J= 26.47g x s x -66.81oC s = 0.158 J/goC

Problem 6.82 A 44.0 g sample of an unknown metal at 99.0ºC was placed in a constant pressure calorimeter containing 80.0 g of water at 24.0ºC. The final temperature of the system was found to be 28.4ºC. Calculate the specific heat capacity of the metal. (the specific heat of water is 4.184 J/gºC)

Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J Substance DHcombustion (kJ/g) Apple -2 Beef -8 Beer -1.5 Gasoline -34

Enthalpy of Formations

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (C, graphite) = 0 f DH0 (O2) = 0 f DH0 (C, diamond) = 1.90 kJ/mol f DH0 (O3) = 142 kJ/mol f

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

C (graphite) + 1/2O2 (g) CO (g) CO (g) + 1/2O2 (g) CO2 (g) C (graphite) + O2 (g) CO2 (g)

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ/mol rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ/mol rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ/mol rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ/mol 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1 kJ/mol x 2 rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ/mol rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ/mol rxn

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C6H6

Chemistry in Action: Bombardier Beetle Defense C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ? C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol H2O2 (aq) H2O (l) + ½O2 (g) DH0 = -94.6 kJ/mol H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol DH0 = 177 - 94.6 – 286 = -204 kJ/mol Exothermic!

Problem 6.62 From the following data C(graphite) + O2(g) CO2(g) ∆Hrxnº=-393.5 kJ/mol H2(g) + ½O2(g) H2O(l) ∆Hrxnº=-285.8 kJ/mol 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) ∆Hrxnº=-3119.6 kJ/mol calculate the enthalpy change for the reaction 2C(graphite) + 3H2(g) C2H6(g)

The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

The energy required to completely separate one mole of a solid ionic compound into gaseous ions is called lattice energy (U). The energy change associated with the hydration process is called the heat of hydration, ∆Hhyd ∆Hsoln = U + ∆Hhyd U = 788 kJ/mol + ∆Hhydr = -784 kJ/mol ∆Hsoln = 4 kJ/mol DH0 = 788 + (-784) = 4 kJ/mol rxn

Lattice energy (kJ/mol) Problem 6.133 From the following data, calculate the heat of solution for KI: NaCl NaI KCl KI Lattice energy (kJ/mol) 788 686 699 632 Heat of Solution (kJ/mol) 4.0 -5.1 17.2 ?