Thermochemistry

THERMOCHEMISTRY THERMOCHEMISTRY, is the study of the heat released or absorbed by chemical and physical changes. 1N = 1Kg.m/s 2, 1J = 1Nm =1Kg.m 2 /s 2 1cal = J,

Temperature, T & Heat, q Temperature is defined as the degree of hotness Conversion of a temperature from Fahernheit scale(t  F ) to the Celsius scale (t  C ) Conversion of a temperature from Fahernheit scale(t  F ) to the Celsius scale (t  C ) t  C = (5/9)(t° F - 32) t  C = (5/9)(t° F - 32)

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

system The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothingSYSTEM SURROUNDINGS 6.2

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 = mass x specific heat = ms Heat (q) absorbed or released: q = ms  t q = C  t  t = t final - t initial Specific heats of some common substances

Heat capacity (C) for water is J/g.°C, heat capacity for 500 g of water is C = 500 x = 2090 J / °C = 2.09 KJ/°C This sample absorbs 2.09 KJ of heat for each degree that temperature increase. By raising temperature from 20 to 25°C, the heat absorbed by the sample q is, q = 2.09 ( ) = 10.4 KJ Calorimeter is used to measure the heat changes accompanying the chemical reactions

Measuring  H in the lab by bomb calorimetry

Heats of Some Typical Reactions Measured at Constant Pressure

in an open system at constant pressure, i.e. H is a measure of heat Note that the standard enthalpy, ΔH f o, of formation for a pure element in its standard state is zero.

 H = H final - H initial heat in products reactants Heat released or absorbed or absorbed Enthalpy Change,  H

Find  H for the reaction C (diamond) + O 2(g)  CO 2(g) if T i = o C, T f = o C, for a g sample of diamond, with m = 1560 g H 2 O q = mc  T = 1560 g x 4.18 J o C -1 g -1 x ( ) o C = 8.22 x 10 3 J = 8.22 kJ  H per mole ( g) = (8.22 kJ/0.25 g) x ( g/mol) = -395 kJ/mol

example... CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g)  H = kJ/mol reactants  products + heat i.e. less H in products than reactants i.e. heat is released!!!

H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ

H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = kJ 266 g P 4 1 mol P g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = C q = ms  t = 869 g x J/g 0 C x –89 0 C= -34,000 J

Standard enthalpy of formation (  H 0 ) 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.  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f

Standard Enthalpies of Formation of Some Inorganic Substances at 25  C

The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn a A + b B c C + d D H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (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.

Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = (2x-296.1) = 86.3 kJ rxn

Benzene (C 6 H 6 ) 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 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x– x–187.6 ] – [ 2x49.04 ] = kJ kJ 2 mol = kJ/mol C 6 H 6

The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.  H soln = H soln - H components Heats of Solution of Some Ionic Compounds

Hess’s law Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. For example: C + O 2  CO 2 This reaction can occur as 2 steps C + ½O 2  CO  H  = – kJ CO + ½O 2  CO 2  H  = – kJ C + CO + O 2  CO + CO 2  H  = – kJ I.e. C + O 2  CO 2  H  = – kJ Hess’s law allows us to add equations. We add all reactants, products, &  H  values.

Hess’s law: We may need to manipulate equations further: 2Fe + 1.5O 2  Fe 2 O 3  H  =?, given Fe 2 O 3 + 3CO  2Fe + 3CO 2  H  = – kJ CO + ½ O 2  CO 2  H  = – kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 1.5O 2  Fe 2 O 3 3CO O 2  3CO 2  H  = – kJ 2Fe + 3CO 2  Fe 2 O 3 + 3CO  H  = kJ CO + ½ O 2  CO 2  H  = – kJ  H  = – kJ