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InnovativeCadmium

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Badr University in Assiut

Prof. Dr. Hossieny Ibrahim

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physical chemistry enthalpy thermochemistry chemical reactions

Summary

This document is lecture 3 on physical chemistry, specifically covering concepts related to enthalpy and different types of heat (enthalpy) of reactions. It includes examples and calculations.

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Prof. Dr. Hossieny Ibrahim Badr University in Assiut School of Biotechnology [email protected] Office number: Bio-326 1 Unit 1: Enthalpy of A reaction  Thermochemical measurements are made either at: (a) constant volume or (b) constant pres...

Prof. Dr. Hossieny Ibrahim Badr University in Assiut School of Biotechnology [email protected] Office number: Bio-326 1 Unit 1: Enthalpy of A reaction  Thermochemical measurements are made either at: (a) constant volume or (b) constant pressure.  The magnitudes of changes observed under the two conditions are different.  The change in internal energy (ΔE) is the heat change accompanying a chemical reaction at constant volume because no external work is performed.  In the laboratory most of the chemical reactions are carried out at constant pressure.  In order to study the heat changes for reactions taking place at constant pressure and constant temperature, chemists have introduced a new term called Enthalpy. 2 Unit 1: Enthalpy of A reaction  The Enthalpy of a system is defined as the sum of the internal energy and the product of its pressure and volume. H = E + PV where E, is the internal energy, P , is the pressure, V , is the volume of the system  like E, enthalpy is also a state function and it is not possible to measure its absolute value.  However a change in enthalpy (ΔH) accompanying a process can be measured accurately and is given by the expression: H = HProducts – HReactants= HP – HR 3 Unit 1:Enthalpy of A reaction  Thus if ΔV be the change in volume in case of a reaction at constant temperature and pressure, the H observed will be the sum of the change in internal energy (ΔE) and the work done in expansion or contraction. H = E + P V  For reactions involving Solids and Liquids only:  (ΔV) is very small and  The term P ΔV is Negligible.  Thus H = E  In case of Gases:  Most of reactions are studied at constant pressure  The term PΔV is Appreciable.  Thus H = E + P V 4 Unit 1: Exothermic and Endothermic Reactions  Let us consider a general reaction at constant pressure: A+B C+D  If HA& HB, the enthalpies of reactants (A & B)  HC & HD the enthalpies products (C & D) H = HProducts – HReactants= (HC + HD) – (HA + HB)  The value of H may be either:  H = Zero : HProducts = Hreactants No heat is evolved or absorbed  H < Zero (Negative) : T Surroundings increases Tsystem decreases HProducts < Hreactants  Heat is evolved  Exothermic Reactions  Heat transfers from System to Surroundings 5 Unit 1: Exothermic and Endothermic Reactions T Surroundings decreases  H > Zero (Positive) : Tsystem increases HProducts > Hreactants  Heat is absorbed  Endothermic Reactions  Heat transfers from Surroundings to System Examples of Exothermic and Endothermic Processes Exothermic Processes Endothermic Processes Mixing water and strong acids Melting solid salts Mixing water with calcium chloride Evaporation of water Rusting iron Sublimation Freezing Mixing water and ammonium nitrate Condensation 6 Unit 1: Calculation of ΔH from ΔE and vice versa H = E + P V (i)  Let us consider a general reaction : aA + bB cC + dD  Change in number of moles = No. of moles of products – No. of moles of reactants = (c + d) – (a + b) = n  Let the volume occupied by one mole of the gas be V.  Then, ΔV = n × V  P V = P (n × V) or P V = PV n (ii) but PV = RT for one mole  Putting RT in place of PV in equation (ii) we get: P V = RT n  Substituting the value of P V in equation (i) we get: H = E + n RT 7 Unit 1: Solved Problem  Solved Problem : The heat of combustion of ethylene at 17ºC and at constant volume is – 332.19 kcals. Calculate the heat of combustion at constant pressure considering water to be in liquid state. (R = 2 cal degree–1 mol–1) Given that: Answer: 8 Unit 1: Solved Problem  Solved Problem : The heat of combustion of carbon monoxide at constant volume and at 17ºC is – 283.3 kJ. Calculate its heat of combustion at constant pressure (R = 8.314 J degree–1 mol–1). Given that: Answer: Quiz : The enthalpy of formation of methane at constant pressure and 300 K is −75.83 kJ. What will be the heat of formation at constant volume? (R= 8.3 JK−1mol−1). The equation is C (s) + 2H2 (g) = CH4(g). (Ans. −73.34 kJ) 9 Unit 1: Thermochemical Equations  An equation which indicates the amount of heat change (evolved or absorbed) in the reaction or process is called a Thermochemical equation.  It must essentially :  Be balanced  Give the value of ΔE or ΔH corresponding to the quantities of substances given by the equation.  Mention the physical states of the reactants and products.  The physical states are represented by the symbols:  ( s ) for solid,  ( l ) for liquid),  ( g ) for gas,  ( aq ) for aqueous states. 10 Unit 1: Thermochemical Equations Is H negative or positive? System gives off heat Exothermic H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = - 890.4 kJ 11 Unit 1: Thermochemical Equations  The stoichiometric coefficients always refer to the number of moles of a substance. H2O (s) H2O (l ) H = 6.01 kJ  If you Reverse a reaction, the sign of H changes: H2O (l ) H2O (s) H = - 6.01 kJ  If you multiply (or divided) both sides of the equation by a factor n, then H must change by the same factor n. 2H2O (s) 2H2O (l ) H = 2 x 6.01 kJ  The physical states of all reactants and products must be specified in thermochemical equations. H2O (l ) H2O (g) H = 44.0 kJ 12 Unit 1: Different Types of Heat (Enthalpy) of Reaction  The heat or enthalpy changes accompanying chemical reactions are expressed in different ways, depending on the nature of the reaction: (1) Heat of Formation ( HF ): The change in enthalpy that takes place when ONE mole of the compound is formed from its elements.  Example, the heat of formation of ferrous sulphide (FeS) and acetylene (C2H2) may be expressed as : The heat of formation of hydrogen chloride (HCl), therefore, would be – 44.0/2 = – 22.0 kcal 13 Unit 1: Different Types of Heat (Enthalpy) of Reaction (2) Heat of Combustion ( 𝚫𝑯𝒄 ): the change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen. Examples: Methane: 𝚫𝑯𝒄 = Octane: 𝚫𝑯𝒄 = It should be noted clearly that the heat of combustion of a substance (𝚫𝑯𝒄 ) is always negative. (3) Heat of Solution (∆Hsolution): the change in enthalpy when one mole of a substance is dissolved in a specified quantity of solvent at a given temperature. 𝚫𝑯𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝚫𝑯𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 14

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