Thermochemistry: Chapter 2 - The First Law of Thermodynamics PDF

Thermochemistry Thermochemistry = The study of the energy transferred as heat during the course of chemical reactions is called thermochemistry Exothermic process =...

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 43  Standard enthalpy changes 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. 44  Standard enthalpy changes ΔH° for a reaction (or a physical process) at specified T: ∆𝒓𝒆𝒂 𝑯° = 𝑯° 𝑷𝒓𝒅𝒖𝒄𝒕𝒔 − 𝑯° 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 Example: The standard enthalpy of vaporization, ΔvapH°, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar, as in: H2O(l) → H2O(g), ΔvapH° (373 K) = +40.66 kJ mol−1 45  Standard enthalpy changes 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 ΔtrsH°. Standard enthalpies of fusion and vaporization at the transition temperature, ΔtrsH° (kJ mol−1) H2O(s) → H2O(l) ΔfusH°(273 K) = +6.01 kJ mol−1 H2O(l) → H2O(g) ΔvapH°(373 K) = +40.66 kJ mol−1 46  Standard enthalpy changes Enthalpies of physical change Different types of enthalpies of transitions encountered in thermochemistry 47  Standard enthalpy changes 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) H2O(s) → H2O(g) ΔsubH° (ii) Indirect path (two steps) H2O(s) → H2O(l) ΔfusH° H2O(l) → H2O(g) ΔvapH° Overall: H2O(s) → H2O(g) ΔsubH° = ΔfusH° + ΔvapH° ΔH° (A→B) = - ΔH° (B→A) 48  Standard enthalpy changes Enthalpies of physical change The lattice enthalpy (ΔHL) is the change in standard molar enthalpy for this process. MX(s) → M+(g) + X−(g) Experimental values of ΔHL 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 => ΔHL = ΔUL At normal temperatures, ΔHL and ΔU differ by only a few kilojoules per mole, and the difference is normally neglected. 49  Standard enthalpy changes 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 89 + 122 + 418 − 349 − ΔHL/(kJ mol−1) + 437 = 0 Thus, ΔHL = +717 kJ mol−1 50  Standard enthalpy changes Enthalpies of chemical change There are two ways of reporting the change in enthalpy that accompanies a chemical reaction: (i) CH4(g) + 2 O2(g)→CO2(g) + 2 H2O(l) ΔH° = −890 kJ (ii) CH4(g) + 2 O2(g)→CO2(g) + 2 H2O(l) ΔrH° = −890 kJ mol−1 In general, for a chemical reaction: Reactants → Products ∆𝒓 𝑯° = ෍ 𝝂𝑯°𝒎 − ෍ 𝝂𝑯°𝒎 𝑷𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 where 𝑯°𝒎 molar enthalpies of the species are multiplied by their (dimensionless and positive) stoichiometric coefficients, 𝝂 51  Standard enthalpy changes 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. CH2=CHCH3(g) + H2(g) → CH3CH2CH3(g) ΔrH° = −124 kJ mol−1 The standard reaction enthalpy for the combustion of propane is −2220 kJ mol−1. CH3CH2CH3(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔrH° = −2220 kJ mol−1 Calculate the standard enthalpy of combustion of propene ? 52  Standard enthalpy changes Hess’s law Answer Add additional data. The combustion reaction we require is C3H6(g) + 9/2O2(g)→3 CO2(g) + 3 H2O(l) This reaction can be recreated from the following sum: ΔrH°/(kJ mol−1) C3H6(g) + H2(g) → C3H8(g) −124 C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) −2220 H2O(l) → H2(g) + 1/2O2(g) +286 C3H6(g) + 9/2 O2(g) → 3 CO2(g) + 3 H2O(l) −2058 53  Standard enthalpies of formation The standard enthalpy of formation, ΔfH°, 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 ΔfH° of liquid benzene at 298 K, for example, refers to the reaction below is +49.0 kJ mol−1 6 C(s, graphite) + H2(g)→C6H6(l) Notes ΔfH° of elements in their reference states are zero at all temperatures. ΔfH°(H+, aq) = 0 54  Thermochemistry: Standard enthalpies of formation Reaction enthalpy in terms of enthalpies of formation 𝜟𝒓 𝑯° = ෍ 𝝂𝜟𝒇 𝑯° − ෍ 𝝂𝜟𝒇 𝑯° 𝑷𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 Example Consider the reaction : 2 HN3(l) + 2 NO(g) → H2O2(l) + 4 N2(g) The standard enthalpy of the reaction is calculated as follows: ΔrH° = {ΔfH°(H2O2,l) + 4ΔfH°(N2,g)} − {2ΔfH°(HN3,l) + 2ΔfH°(NO,g)} = {−187.78 + 4(0)} kJ mol−1 − {2(264.0) + 2(90.25)} kJ mol−1 = −896.3 kJ mol−1 55  The temperature dependence of reaction enthalpies The standard enthalpies may be calculated from heat capacities and the reaction enthalpy at some other temperature. Kirchhoff’s law 𝑻𝟐 𝜟𝒓 𝑯° 𝑻𝟐 = 𝜟𝒓 𝑯° 𝑻𝟏 + න 𝜟𝒓 𝑪°𝒑 𝒅𝑻 𝑻𝟏 𝚫𝐫 𝐇 ° 𝐓𝟐 = 𝚫𝐫 𝐇 ° 𝐓𝟏 + (𝐓𝟐 − 𝐓𝟏 )𝚫𝐫 𝐂𝐩° where Kirchhoff’s law Δ𝑟 𝐶𝑝° = ෍ ° 𝜈𝐶𝑝,𝑚 − ෍ ° 𝜈𝐶𝑝,𝑚 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 56  The temperature dependence of reaction enthalpies Example: Using Kirchhoff’s law The standard enthalpy of formation of H2O(g) at 298 K is −241.82 kJ mol−1. Estimate its value at 100°C given the following values of the molar heat capacities at constant pressure: H2O(g): 33.58 J K−1 mol−1; H2(g): 28.82 J K−1 mol−1; O2(g): 29.36 J K−1 mol−1. Assume that the heat capacities are independent of temperature. Answer Δ𝑟 𝐻 ° 𝑇2 = Δ𝑟 𝐻 ° 𝑇1 + (T2 −T1 )𝜟𝒓 𝑪°𝒑 57  The temperature dependence of reaction enthalpies Example: Using Kirchhoff’s law To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate Δ𝑟 𝐶𝑝° from the data. The reaction is 1 H2 (g) + 2 O2 (g) → H2 O(g) so ° ° ° 𝟏 ° 𝜟𝒓 𝑪𝒑 = 𝑪𝒑,𝒎 𝑯 𝐎, 𝒈 − 𝑪𝒑,𝒎 𝑯 , 𝒈 + 𝑪𝒑,𝒎 𝑶 , 𝒈 𝟐 𝟐 𝟐 𝟐 = −9.92 J K −1 mol−1 It then follows that 𝜟𝒓 𝑯° 𝟑𝟕𝟑 𝑲 = −241.82 kJ mol−1 + 75 K × −9.92 J K −1 mol−1 = −242.6 kJ mol−1 58