First Law of Thermodynamics Doba Jackson, Ph.D. Dept of Chemistry & Biochemistry Huntingdon College

Outline of Chapter 2 Basic Concepts: Heat, Work, Energy First Law of Thermodynamics Expansion work Measurement of Heat Enthalpy changes Adiabatic changes (Cooling) Thermochemistry Inexact Differntials

System, Surroundings, Universe System- Part of the world of interest Surroundings- Region outside the system Open system- Allows matter and energy to pass Closed system- Cannot allow matter to pass Isolated system- Cannot allow matter or energy to pass

Work Work- the amount of energy transferred by motion against an opposing force ▪

Heat, and Heat Transfer Heat- is the transfer of energy from one body to another by thermal contact Heat can be transferred by Convection Conduction Radiation

Internal Energy: Energy that exist internal to the system First Law: The internal energy of an isolated system is a constant; Energy cannot be created or destroyed. ΔU= change in internal energy q = heat (released or absorbed) w = work done on or by the system Internal energy can have different forms: Translational Kinetic Energy Rotational Kinetic energy Vibrational Kinetic energy *Energy within the atom

Internal Energy of a monoatomic gas RT Monoatomic gas RT Linear molecule RT Non-Linear molecule RT RT

Expression for Work Change equation to relate pressure and volume True expression for Work

Types of Expansion Work Free Expansion- work done by expansion against zero opposing force or zero pressure

Types of Expansion Work Expansion against a constant external pressure (irreversible) If external pressure does, not vary, this expression can be integrated

Types of Expansion Work Expansion is reversible In this case, the external pressure Pex exactly equals the internal pressure P at each stage of the expansion. The integral cannot be evaluated because temperature is not a constant throughout the integration.

Types of Expansion Work Expansion is Isothermal, reversible The system is kept at a constant temperature. The external pressure Pex equals the internal pressure P at each stage of the expansion. Moles is always assumed constant unless otherwise noted.

Problems: Calculate the work needed for a 65 kg person to climb up 4 Problems: Calculate the work needed for a 65 kg person to climb up 4.0 meters on (a) the earth, (b) the moon (g=1.60 m/s2)

Problem 1: A chemical reaction takes place in a container of cross sectional area of 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an constant external pressure of 121 kPa. Calculate the work (in Joules) done by the system

Problem 2: A sample consisting of 2 Problem 2: A sample consisting of 2.00 mol of He is expanded at 22ºC from 22.8 dm3 to 31.7 dm3. Calculate q, w and ΔU (in Joules) when the expansion occurs under the following conditions:   (a) Freely   (b) Isothermal, and constant external pressure equal to the final pressure of the gas   (c) Isothermal, Reversibly  

Reversible, Nonspontaneous Reversible Process: The thermodynamic process in which a system can be changed from its initial state to its final state then back to its initial state leaving all thermodynamic variables for the universe (system + surroundings) unchanged. A truly reversible change will: - Occur in an infinite amount of time - All variables must be in equilibrium with each other at every stage of the change Reversible Irreversible

Irreversible, Spontaneous Irreversible Process: The thermodynamic process in which a system that is changed from its initial state to its final state then back to its initial state will change some thermodynamic variables of the universe. A truly irreversible change will: - Occur in an finite amount of time - All variables will not be in equilibrium with each other at every stage of the change Reversible Irreversible

Heat exchange can be measured in a Bomb Calorimeter First Law: The internal energy of an isolated system is a constant. Constant Volume: no work done

Calorimetry is the study of heat transfer during a chemical or physical process Heat is measured by the change in temperature of the water surrounding the calorimeter Heat The calorimeter constant is the heat capacity of the system

Heat Capacity at a constant volume Heat Capacity (Cv) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant volume. We need partial derivatives because the change holds the volume constant

Heat Capacity at a constant volume We need partial derivatives because the change holds the volume constant

How we can evaluate the Heat capacity equation If the heat capacity (Cv) does not vary with temperature during the range of temperatures used, then the equation can be broken down to: Because at constant volume, change in internal energy is equal to the heat (qv).

Problem 2.4: A sample consisting of 1.00 mol of a perfect gas, for which Cv,m = (3/2)R, initially at P = 1.00 atm and T = 300 K, is heated reversibly to 400 K at a constant volume. Calculate the final pressure, ΔU, q, and w

We cannot measure internal energy when the volume is not constant Typical reactions occur a constant external pressure

Heat exchange can be measured in a Differential Scanning Calorimeter at a constant pressure First Law: The internal energy of an isolated system is a constant. Lets define Enthalpy as:

Heat Capacity at a constant pressure Heat Capacity (Cp) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant pressure. We need partial derivatives because the change holds the pressure constant

How we can evaluate the Heat capacity equation If the heat capacity (CP) does not vary with temperature during the range of temperatures used, then the equation can be broken down to: Because at constant volume, change in internal energy is equal to the heat (qv).

Problem 2.20: When 2.25 mg of anthracene, C14H10 (s) was burned in a bomb calorimeter, the temperature rose by 1.35 K. Calculate the calorimeter constant for the system. By how much will the temperature rise if 135 mg of phenol (C6H5OH) is burned in the calorimeter under the same conditions? (ΔcHΘ(C14H10) = -7061 kJ/mol)

Adiabatic Expansion: work is done but no heat enters the system When a gas expands adiabatically, work is done but no heat enters or leaves the system. The internal energy falls and the temperature also falls.

Adiabatic changes to an Ideal Gas can be related by the following equations C = CV/nR γ = Cp/CV The derivation of these equations is quite long and follows the next slide

Adiabatic Expansion of an Ideal Gas (Derivation) Assume the heat capacity does not vary during the temperature interval For an adiabatic change: q = 0

Problem 2.1- A 3.75 mole sample of an ideal gas with Cv,m = 3R/2 initially at a temperature Ti=298 K, and Pi= 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by placing a 725 kg mass on the piston of diameter 25.4 cm. Calculate the work done in this process and the distance the piston travels. Assume the mass of the piston is negligible.

Thermochemistry The study of energy transferred as heat during the course of chemical reactions. In thermochemistry, we rely on calorimetry to measure the internal energy (∆U) enthalpy (∆H). Changes in matter are: Physical Changes Solid  Liquid Melting (Fusion) Liquid  Gas Boiling (Vaporization) Solid  Gas Sublimation Chemical Changes Formation reactions Combustion reactions Other reactions

The Thermodynamic Standard State Chapter 8: Thermochemistry: Chemical Energy 4/25/2017 The Thermodynamic Standard State The standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. The International Union of Pure and Applied Chemistry (IUPAC) recommends using a standard pressure po = 1 bar (100 kilopascals). 3CO2(g) + 4H2O(g) C3H8(g) + 5O2(g) ∆H = -2044 kJ 3CO2(g) + 4H2O(l) C3H8(g) + 5O2(g) ∆H = -2220 kJ The conversion from liquid to gas phases requires energy. Therefore, it’s important that the states are specified. Putting the “°” superscripted next to the ∆H means standard state. The standard pressure is actually 1 bar. The enthalpy difference is usually small, however. Thermodynamic Standard State: Most stable form of a substance at 1 bar and at a specified temperature, usually 25 °C (for non-solutions). 3CO2(g) + 4H2O(g) C3H8(g) + 5O2(g) ∆H° = -2044 kJ Copyright © 2010 Pearson Prentice Hall, Inc.

Physical and Chemical Transitions Chapter 8: Thermochemistry: Chemical Energy Physical and Chemical Transitions 4/25/2017 Copyright © 2010 Pearson Prentice Hall, Inc.

State Functions Internal energy and enthalpy are is a state function. Work and Heat are path function. This means we can never write Δq or Δw

Enthalpies of Physical Change Chapter 8: Thermochemistry: Chemical Energy Enthalpies of Physical Change 4/25/2017 Hess’s Law Copyright © 2010 Pearson Prentice Hall, Inc.

Chapter 8: Thermochemistry: Chemical Energy Example of Hess’s Law Chapter 8: Thermochemistry: Chemical Energy 4/25/2017 Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Germain Hess Example 1: The two reactions below are known methods to produce ammonia by the Haber process. Use the information to solve for the enthalpy change in the reaction below. Hydrazine, N2H4, is a reactive intermediate (more on that in Ch. 12: Chemical Kinetics). Chemical reactions are far more complicated than discussed in Ch. 6 (stoichiometry). The overall reaction and step 2 can be measured, but step 1 cannot be measured easily. A fascinating book about Haber is: Charles, Daniel; Master Mind: The Rise and Fall of Fritz Haber, The Nobel Laureate Who Launched the Age of Chemical Warfare; HarperCollins Publishers Inc., 2005. 2NH3(g) 3H2(g) + N2(g) ∆H°1 = -92.2 kJ 2NH3(g) N2H4(g) + H2(g) ∆H°2 = -187.6 kJ Solve for the reaction below N2H4(g) 2H2(g) + N2(g) ∆H°3 = ? Copyright © 2010 Pearson Prentice Hall, Inc.

Example 1: 2NH3(g) 3H2(g) + N2(g) 2NH3(g) N2H4(g) + H2(g) 2NH3(g) Chapter 8: Thermochemistry: Chemical Energy Example 1: The two reactions below are known methods to produce ammonia by the Haber process. Use the information to solve for the enthalpy change in the reaction below. 4/25/2017 2NH3(g) 3H2(g) + N2(g) ∆H°1 = -92.2 kJ 2NH3(g) N2H4(g) + H2(g) ∆H°2 = -187.6 kJ Steps to solve the problem - Find known reactions that will add up to the desired reaction - Add up the reactions and add up the ΔH, ΔU, ΔG or other state functions 2 2NH3(g) 3H2(g) + N2(g) ∆H°1 = -92.2 kJ ∆H°2 = 187.6 kJ Solve for the reaction below ∆H°3 = -92.2 kJ + (187.6 kJ) = 95.4 kJ N2H4(g) 2H2(g) + N2(g) Copyright © 2010 Pearson Prentice Hall, Inc. 41

Chapter 8: Thermochemistry: Chemical Energy Hess’s Law Example 4/25/2017 Copyright © 2010 Pearson Prentice Hall, Inc.

Standard Heats of Formation Chapter 8: Thermochemistry: Chemical Energy 4/25/2017 Standard Heats of Formation Standard Heat of Formation (∆fH° ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states. Standard states CH4(g) C(s) + 2H2(g) ∆H°f = -74.8 kJ 1 mol of 1 substance Students need to remember solid, liquid, gases, and diatomics as discussed previously. Copyright © 2010 Pearson Prentice Hall, Inc.

Chapter 8: Thermochemistry: Chemical Energy 4/25/2017 Standard Heats of Formation are used to determine reaction enthalpies (ΔrHº) Reaction enthalpies (ΔrHº) can be determined by the difference of the product enthalpies of formation and the reactant enthalpies of formation. Using table values. cC + dD aA + bB Example: ∆rH° = {c ∆fH°(C) + d ∆fH°(D)} – {a ∆fHº(A) + b ∆fH°(B)} Products Reactants Copyright © 2010 Pearson Prentice Hall, Inc.

Standard Heats of Formation Chapter 8: Thermochemistry: Chemical Energy 4/25/2017 Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O. C6H12O6(s) + 6*O2(g) 6*CO2(g) + 6*H2O(l) ∆rH° = ? Standard State (ΔfHº)= 0 Answer ΔrH° = [∆H°f (C6H12O6)] - [6*∆H°f (CO2) + 6*∆H°f (H2O)] Using table values. The units for standard enthalpies of formation are kJ/mol while the unit for the standard enthalpy change is kJ. Result is endothermic. Products Reactants ∆rH° = [(1 mol)(-1273.3 kJ/mol)] - [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] = 2802.5 kJ Copyright © 2010 Pearson Prentice Hall, Inc.

Heats of Combustion (ΔcHº) is a special type of reaction enthalpy Chapter 8: Thermochemistry: Chemical Energy Heats of Combustion (ΔcHº) is a special type of reaction enthalpy 4/25/2017 Calculate the standard enthalpy of combustion for methane (CH4(g)). Answer CO2(g) + H2O(l) CH4(g) + O2(g) 2 2 ∆cH°= [∆fH°(CO2) + 2 ∆fH°(H2O)] - [∆fH°(CH4)] = [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] - ΔfHº (CH4) = -74.8 kJ/mol ΔfHº (CO2) = -393 kJ/mol [(1 mol)(-74.8 kJ/mol)] = -890.3 kJ ΔfHº (H2O) = -285 kJ/mol Combustion reactions are always balanced with 1 as the coefficient of the hydrocarbon Copyright © 2010 Pearson Prentice Hall, Inc.

Temperature dependence of reaction enthalpy (ΔrHº) Often standard reaction enthalpies need to be corrected for temperature differences. If Cp is independent of temperature during the temperature range, the integral can be evaluated. The equation doesn’t work if there is a phase transition through the temperature range.

Problem: Balance the reactions and determine ΔrHΘ and ΔrUΘ NO: 91.3 kJ/mol NH3: -45.9 kJ/mol H2O: -241.8 kJ/mol