Unit 8: Thermochemistry

Kinetic Molecular Theory Collective hypotheses about the particulate nature of matter and the surrounding space Greeks - earliest written ideas on atoms Current view Matter comprised of microscopic particles - atoms Atoms combine to form molecules Many macroscopic phenomena can be traced to interactions on this level The molecules of a gas are in constant and random motion The temperature of a gas depends on its average kinetic energy

Temperature scales Measured with thermometer Defined with reference to various reference points Fahrenheit Celsius Kelvin Conversion formulas Fahrenheit to Celsius Celsius to Fahrenheit Celsius to Kelvin

Heat Vs. Temp Heat is the total E of molecular motion in a substance while temp is a measure of the avg E of molecular motion in a substance Heat E depends on the number of particles (the size or mass). Temperature does not (Videos)

EX: the temp of 20 mL of water might be the same as the temp of 1 L of water, but the liter of water has more heat because it has more water and thus more total thermal energy. Adding or removing heat (energy) will increase or decrease the temp. Higher temperatures mean that the molecules are moving, vibrating and rotating with more E

Heat (energy) moves from higher temperatures to lower temperatures. Temperature is not energy, but a measure of it (measure of hotness or coldness) Heat 50 degrees C 20 degrees C

Units of heat Metric units Joule (J) – SI unit calorie (cal) - energy needed to raise temperature of 1 g of water 1 degree Celsius kilocalorie (kcal, Calorie, Cal) - energy needed to raise temperature of 1 kg of water 1 degree Celsius English system British thermal unit (BTU) - energy needed to raise the temperature of 1 lb of water 1 degree Fahrenheit Mechanical equivalence 4.184 J = 1 cal

Land heats up and cools down faster than water Specific Heat a. . Some things heat up or cool down faster than others. Land heats up and cools down faster than water

b. Specific heat is the amount of heat required to raise the temperature of 1 g of a material by one degree (C or K). 1) C water = 4.184 J / g C 2) C sand = 0.664 J / g C This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

Why does water have such a high specific heat? Water molecules have strong intermolecular forces called hydrogen bonds; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

How to calculate changes in thermal energy Q = m x T x C Q Q = change in thermal energy m m = mass of substance  T T= change in temperature (Tf – Ti) Ccc c = specific heat of substance

rature of water are measured c. A calorimeter is used to help measure the specific heat of a substance. First, mass and its Q value, its mass, and its T, its C can be calculated rature of water are measured Then heated sample is put inside and heat flows into water This gives the heat lost by the substance T is measured for water to help get its heat gain

2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g) 6.2

DH = H (products) – H (reactants) Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ 6.4

Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ 6.4

Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ/mol ΔH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 mol x 6.01 kJ/mol = 12.0 kJ 6.4

Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DHreaction = -3013 kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.4

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 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f 6.6

Practice Problem How much heat is produced by the combustion of 1.00 lb (454 g) of propane (C3H8)? The ΔH for the combustion of propane is -2220 kJ mol-1. Convert the grams of propane to moles of propane and then find the amount of heat produced:

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 = S DH0 (products) f - S DH0 (reactants) f 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.) 6.6

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 DH0 (products) f = S DH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ -6535 kJ 2 mol = - 3267 kJ/mol C6H6 6.6

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ 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 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6

DHsoln = Hsoln - Hcomponents 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? 6.7

Energy Diagrams Activation energy (Ea) for the forward reaction Activation energy (Ea) for the reverse reaction (c) Delta H 50 kJ/mol 300 kJ/mol 150 kJ/mol 100 kJ/mol -100 kJ/mol +200 kJ/mol

Phase Changes The boiling point The normal boiling point 11.8

The critical temperature (Tc) The critical pressure (Pc) 11.8

Where vaporization occurs? Can you find… The Triple Point? Critical pressure? Critical temperature? Where fusion occurs? Where vaporization occurs? Melting point (at 1 atm)? Boiling point (at 6 atm)? Carbon Dioxide

The melting point of a solid or the freezing point H2O (s) H2O (l) The melting point of a solid or the freezing point 11.8

H2O (s) H2O (g) 11.8

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance. 11.8

11.8

Sample Problem How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C? Step 1: Heat the ice Q=mcΔT Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ Step 2: Convert the solid to liquid ΔH fusion Q = 2.0 mol x 6.01 kJ/mol = 12 kJ Step 3: Heat the liquid Q=mcΔT Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ

Now, add all the steps together Sample Problem How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C? Step 4: Convert the liquid to gas ΔH vaporization Q = 2.0 mol x 44.01 kJ/mol = 88 kJ Step 5: Heat the gas Q=mcΔT Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ Now, add all the steps together 0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ