Chapter-2-The first law of thermodynamics and thermochemistry

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1 Chapter-2-The first law of thermodynamics and thermochemistry Dr. El Hassane ANOUAR Chemistry Department, College of Sciences and Humanities, Prince Sattam bin Abdulaziz University, P.O. Box 83, Al-Kharij 11942, Saudi Arabia. 10/24/2016 College Of Science and humanities, PSAU 1

2 II-Thermochemistry Thermochemistry = The study of the energy transferred as heat during the course of chemical reactions is called thermochemistry Exothermic process = release of heat ΔH < 0. Endothermic process = Absorption of heat ΔH > 0 10/24/2016 College Of Science and humanities, PSAU 2

3 The standard enthalpy change, ΔH is a change in enthalpy for a process in which the initial and final substances are in their standard states. The standard state of a substance at a specified temperature is its pure form at 1 bar. Examples (i) The standard state of liquid ethanol at 298 K is pure liquid ethanol at 298 K and 1 bar. (ii) The standard state of solid iron at 500 K is pure iron at 500 K and 1 bar. 10/24/2016 College Of Science and humanities, PSAU 3

4 ΔH for a reaction (or a physical process) at specified T: rea H = H Prducts H Reactants Example: The standard enthalpy of vaporization, Δ vap H, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar, as in: H 2 O(l) H 2 O(g) Δ vap H (373 K) = kj mol 1 Note The conventional temperature for reporting thermodynamic data is K (corresponding to C) 10/24/2016 College Of Science and humanities, PSAU 4

2.7.1. Enthalpies of physical change The standard enthalpy change that accompanies a change of physical state is called the standard enthalpy of transition and is denoted Δ trs H.

5 Enthalpies of physical change The standard enthalpy change that accompanies a change of physical state is called the standard enthalpy of transition and is denoted Δ trs H. Standard enthalpies of fusion and vaporization at the transition temperature, Δ trs H (kj mol 1 ) H 2 O(s) H 2 O(l) Δ fus H (273 K) = kj mol 1 H 2 O(l) H 2 O(g) Δ vap H (373 K) = kj mol 1 10/24/2016 College Of Science and humanities, PSAU 5

6 Enthalpies of physical change Different types of enthalpies of transitions encountered in thermochemistry 10/24/2016 College Of Science and humanities, PSAU 6

7 Enthalpies of physical change H is a state function => ΔH is independent of the path between the two states. Example: conversion of a solid to a vapour (i) Direct path (one step) H 2 O(s) H 2 O(g) Δ sub H (ii) Indirect path (two steps) H 2 O(s) H 2 O(l) Δ fus H H 2 O(l) H 2 O(g) Δ vap H Overall: H 2 O(s) H 2 O(g) Δ sub H = Δ fus H + Δ vap H As Δ fus H > 0 => Δ sub H > Δ vap H at a given T ΔH (A B) = - ΔH (B A) 10/24/2016 College Of Science and humanities, PSAU 7

8 Enthalpies of physical change The lattice enthalpy (ΔH L ) is the change in standard molar enthalpy for this process. MX(s) M + (g) + X (g) Experimental values of ΔH L are obtained by using a Born Haber cycle. It is a closed path of transformations starting and ending at the same point, one step of which is the formation of the solid compound from a gas of widely separated ions. At T = 0 => ΔH L = ΔU L At normal temperatures, ΔH L and ΔU differ by only a few kilojoules per mole, and the difference is normally neglected. 10/24/2016 College Of Science and humanities, PSAU 8

9 Enthalpies of physical change Example: A typical Born Haber cycle, for potassium chloride Because the sum of these enthalpy changes is equal to zero, we can infer from ΔHL/(kJ mol 1) = 0 Thus, ΔH L = +717 kj mol 1 10/24/2016 College Of Science and humanities, PSAU 9

10 Enthalpies of chemical change There are two ways of reporting the change in enthalpy that accompanies a chemical reaction: (i) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) (ii) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) ΔH = 890 kj Δ r H = 890 kj mol 1 In general, for a chemical reaction: Reactants Products r H = νh m Products νh m Reactants where H m molar enthalpies of the species are multiplied by their (dimensionless and positive) stoichiometric coefficients, ν 10/24/2016 College Of Science and humanities, PSAU 10

11 Hess s law Hess s law: The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. Example: Using Hess s law The standard reaction enthalpy for the hydrogenation of propene is 124 kj mol 1. CH 2 =CHCH 3 (g) + H 2 (g) CH 3 CH 2 CH 3 (g) ΔrH = 124 kj mol 1 The standard reaction enthalpy for the combustion of propane is 2220 kj mol 1. CH 3 CH 2 CH 3 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) ΔrH = 124 kj mol 1 Calculate the standard enthalpy of combustion of propene? 10/24/2016 College Of Science and humanities, PSAU 11

12 Hess s law Answer The skill to develop is the ability to assemble a given thermochemical equation from others. Add or subtract the reactions given, tog ether with any others needed, so as to reproduce the reaction required. Then add or subtract the reaction enthalpies in the same way. Add additional data. The combustion reaction we require is C 3 H 6 (g) + 9/2O 2 (g) 3 CO 2 (g) + 3 H 2 O(l) This reaction can be recreated from the following sum: C 3 H 6 (g) + H 2 (g) C 3 H 8 (g) 124 C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) 2220 H 2 O(l) H 2 (g) + 1/2O 2 (g) +286 C 3 H 6 (g) + O2(g) 3 CO 2 (g) + 3 H 2 O(l) 2058 Δ r H /(kj mol 1 ) 10/24/2016 College Of Science and humanities, PSAU 12

13 2.8 Standard enthalpies of formation The standard enthalpy of formation, Δ f H, of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states, i.e., the most stable state at the specified temperature and 1 bar. Example Δ f H of liquid benzene at 298 K, for example, refers to the reaction below is kj mol 1 6 C(s, graphite) + H 2 (g) C 6 H 6 (l) Notes Δ f H of elements in their reference states are zero at all temperatures because they are the enthalpies of such null reactions as N2(g) N2(g) => Δ f H = 0. Δ f H (H +, aq) = 0 10/24/2016 College Of Science and humanities, PSAU 13

14 II-Thermochemistry 2.8 Standard enthalpies of formation Reaction enthalpy in terms of enthalpies of formation Δ r H = νδ f H Products Example Consider the reaction : νδ f H Reactants Or Δ r H = ν J Δ f H 2 HN 3 (l) + 2 NO(g) H 2 O 2 (l) + 4 N 2 (g) The standard enthalpy of the reaction is calculated as follows: Δ r H = {Δ f H (H 2 O 2,l) + 4Δ f H (N 2,g)} {2Δ f H (HN3,l) + 2Δ f H (NO,g)} J (J) = { (0)} kj mol 1 {2(264.0) + 2(90.25)} kj mol 1 = kj mol 1 10/24/2016 College Of Science and humanities, PSAU 14

15 2.8 Standard enthalpies of formation Enthalpies of formation and molecular modelling Computer-aided molecular modelling software packages use to calculate the standard enthalpy of formation of a molecule drawn on the computer screen. These techniques can be applied to different conformations of the same molecule. Example: Methylcyclohexane, the calculated conformational energy difference ranges from 5.9 to 7.9 kj mol 1, with the equatorial conformer having the lower standard enthalpy of formation. These estimates compare favourably with the experimental value of 7.5 kj mol 1. Equatorial conformer Axial conformer 10/24/2016 College Of Science and humanities, PSAU 15

16 2.9 The temperature dependence of reaction enthalpies The standard enthalpies of many important reactions have been measured at different temperatures. However, in the absence of this information, standard reaction enthalpies at different temperatures may be calculated from heat capacities and the reaction enthalpy at some other temperature. where Δ r H T 2 = Δ r H T 1 + T 2 Δ r C p dt T 1 Kirchhoff s law Δ r C p = νc p,m Products νc p,m Reactants 10/24/2016 College Of Science and humanities, PSAU 16

17 2.9 The temperature dependence of reaction enthalpies Example: Using Kirchhoff s law The standard enthalpy of formation of H 2 O(g) at 298 K is kj mol 1. Estimate its value at 100 C given the following values of the molar heat capacities at constant pressure: H 2 O(g): J K 1 mol 1 ; H 2 (g): J K 1 mol 1 ; O2(g): J K 1 mol 1. Assume that the heat capacities are independent of temperature. Answer When ΔC p is independent of temperature in the range T1 to T2, the integral in eqn 2.36a evaluates to (T 2 T 1 )Δ r C p. Therefore, Δ r H T 2 = Δ r H T 1 + (T 2 T 1 )Δ r C p 10/24/2016 College Of Science and humanities, PSAU 17

18 2.9 The temperature dependence of reaction enthalpies Example: Using Kirchhoff s law To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate Δ r C p from the data. The reaction is so Δ r C p = C p,m H O, g C p,m J K 1 mol 1 It then follows that H 2 (g) O 2(g) H 2 O(g) H, g + 1 C 2 2 p,m O, g = 2 Δ r H 373 K = kj mol K 9.92 J K 1 mol 1 = kj mol 1 10/24/2016 College Of Science and humanities, PSAU 18

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