Chapter 9 Energy, Enthalpy and Thermochemistry The study of energy and its interconversions is called thermodynamics. Kinetic Energy: energy due to the motion of the object (1/2 mv 2 ) Potential Energy: energy due to position or composition Heat: the transfer of energy between two objects due to a temperature difference Work: a force acting over a distance

Figure 9.1 (a): Initial position of balls

Figure 9.1 (b): Final position of balls

Energy of Matters

State Function A property of the system depends only on its present state. A state function does not depend in any way on the system’s past. Energy is a state function, but work and heat are not state function

Combustion of methane

Nitrogen/oxygen

First Law of Thermodynamics The energy of the universe is constant In closed system ∆E= ∆U=q+w ( 內能 = 熱能 + 功 ) q: the heat added to the system during the process w: the work done on the system during the process q>0 heat flows into the system from the surroundings q<0 an outflow of heat from the system to the surroundings w>0 work is done on the system by the surroundings w<0 the system does work on the surroundings

P-V Work

Enthalpy ( 焓 ) The heat q p absorbed in a constant-pressure process equals the system’s enthalpy change.

For a chemical reaction ΔH=ΔH products -ΔH reactants If ΔH reactants <ΔH products (endothermic) If ΔH reactants >ΔH products (exothermic) Consider a constant-volume process dw=-PdV=0 ( 體積固定不做功 ) ∆U=q+w=q v ∆U=q v

Heat Capacity ( 熱容量 )

Thermodynamics of Ideal Gases

Heat Capacity of Heating an Ideal Monatomic Gas Under constant volume, the energy flowing into the gas is used to increase the translational energy of the gas molecules.

Heat Capacity of Ideal Monatomic Gases GasC v (J/K mol) C p (J/K mol) He Ne Ar

Heat Capacity of Diatomic Gases GasC v (J/K mol) C p (J/K mol) H2H N2N

Heat Capacity of Polyatomic Gases GasC v (J/K mol) C p (J/K mol) N2ON2O CO C2H6C2H

Heat Capacity of Heating a Polyatomic Gas Polyatomic gases have observed values for C v that are significantly greater than 3/2 R. This larger value for C v results because polyatomic molecules absorb energy to excite rotational and vibrational motions in addition to translational motions.

C v and C p of molecules

C v and C p of Monatomic Gas

C v and C p of H 2 O at 373K

2 mol of monatomic ideal gas Calculate q, w, ∆U and ∆H for both pathway

T A =122K, T C =366K, T B =183K, T D =61K C v =3/2R, C p =5/2R Path 1(A→C) w 1 =-P∆V=-2atm×(30-10)L×101.3J/Latm=-4.05×10 3 J q 1 =q p =nC p ∆T=2×5/2(R)×( )=1.01×10 4 J=∆H 1 ∆U 1 =nC v ∆T= 2×3/2(R)×( )=6.08×10 3 J Path 2(C→B) q 2 =q v =nC v ∆T=2×3/2(R)×( )=-4.56×10 3 J=∆U 2 ∆H 2 =nC p ∆T= 2×5/2(R)×( )=-7.6×10 3 J ∆V=0 w 2 =-P∆V =0

Path 3(A→D) q 3 =q v =nC v ∆T=2×3/2(R)×(61-122)=-1.52×10 3 J=∆U 3 ∆H 3 =nC p ∆T= 2×5/2(R)×(61-122)=-2.53×10 3 J ∆V=0 w 2 =-P∆V =0 Path 4(D→B) w 1 =-P∆V=-1atm×(30-10)L×101.3J/Latm=-2.03×10 3 J q 4 =q p =nC p ∆T=2×5/2(R)×(183-61)=5.08×10 4 J=∆H 4 ∆U 4 =nC v ∆T= 2×3/2(R)×(183-61)=3.05×10 3 J

Summary Path 1 q path1 =q 1 +q 2 =5.5 ×10 3 J w path1 =w 1 +w 2 = -4.05×10 3 J ∆H path1 = ∆H 1 +∆H 2 = 2.55×10 3 J ∆U path1 = ∆U 1 +∆U 2 = 1.52×10 3 J Path 2 q path2 =q 3 +q 4 =3.56×10 3 J w path2 =w 3 +w 4 = -2.03×10 3 J ∆H path2 = ∆H 3 +∆H 4 = 2.55×10 3 J ∆U path2 = ∆U 3 +∆U 4 = 1.52×10 3 J

Calorimetry Specific heat capacity with unit JK -1 g -1 Molar heat capacity with unit JK -1 mol -1

Coffee Cup Calorimeter A constant-pressure calorimetry is used in determining the change in enthalpy equals the heat. ∆H=q p =nC p ∆T

∆V=0 ∆U=q v +0=q v ∆U=q v =nC v ∆ T Bomb Calorimeter

2SO 2 (g)+O 2 (g)→2SO 3 (g) ∆H=-198 KJ 2 mol. 1 mol. 2 mol. Calculate ∆H and ∆U

P is constant, ∆H=q p =-198 KJ (energy flow out of system) ∆U= q p +w w=-P∆V and ∆V=∆n(RT/P) T and P are constant, ∆n=n final -n initial =-1 mol So w=-P∆V=-P × ∆n × (RT/P) =- ∆nRT=-(-1)(8.314)(298)=2.48 kJ ∆U= q p +w=-198 kJ+2.48 kJ=-196 kJ

Hess’s Law If a reaction is carried out in a series of steps, ∆H for the reaction will be equal to the sum of the enthalpy changes for the individual steps The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.

Consider the combustion reaction of methane to form CO 2 and liquid H 2 O CH 4 (g) + 2O 2 (g)→ CO 2 (g) + 2H 2 O(l) ∆H 1 =-890KJ/mol This reaction can be thought of as occurring in two steps: CH 4 (g) + 2O 2 (g)→ CO 2 (g) + 2H 2 O(g) ∆H 2 = -802 kJ/mol 2H 2 O(g)→2H 2 O(l) ∆H 3 =-88KJ/mol ∆H 1 = ∆H 2 + ∆H 3

Standard Enthalpies of Formation The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states. The superscript zero indicates that the corresponding process has been carried out under standard conditions.

Petroleum and Natural Gas Coal Present Sources of Energy New Energy Sources Coal Conversion Hydrogen as a fuel

CO 2 capture