Section 3.0. The 1 st Law of Thermodynamics. (CHANG text Chapter 4) 3.1. Revisiting Heat Capacities Definitions and Concepts

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1 Section 3.0. The 1 st Law of Thermodynamics (CHANG text Chapter 4) 3.1. Revisiting Heat Capacities 3.2. Definitions and Concepts 3.3. The First Law of THERMODYNAMICS 3.4. Enthalpy 3.5. Adiabatic Expansion of an Ideal Gas 3.6. Thermochemistry 3.7. Bond Energies and Bond Enthalpies

2 3.6. Thermochemistry Goals. (1) To define enthalpy of reaction (Δ R H ) (2) To understand how to calculate Δ R H from Δ f H for compounds (1) To use Hess s Law to calculate Δ f H

3 3.6. Thermochemistry is the study of energy changes in chemical reactions. For a constant pressure process the heat of reaction (q p ) is equal to the enthalpy of reaction (Δ R H ) Exothermic reaction (gives off heat) Δ R H = negative value Endothermic (absorbs heat) Δ R H = positive value at a pressure of 1 bar and T = 298 K we say Δ R H = Δ R H the standard enthalpy of reaction Δ R H has units of kj Δ R H is the enthalpy change when reactants in standard state are converted to products in standard state

4 Exothermic Endothermic

5 The standard enthalpy change for a chemical rxn. is defined as: Δ R H = v H ( products ) - v H ( reactants ) H is the standard molar enthalpy v is the stoichiometric coefficient (number of moles) Example: a A + b B c C + d D The standard enthalpy of reaction is Δ R H = + (c mol) H (C) + (d mol) H (D) (a mol) H (A) (b mol) H (B) Unfortunately we can t measure the absolute values of H for substances so we use standard molar enthalpy of formation instead Δ f H so Δ R H = v Δ f H ( products ) - v Δ f H ( reactants )

6 How can we calculate Δ R H? Elements Enthalpy Δ f H (reactants) Δ f H (products) We can calculate H simply by measuring the difference between the heat of formation (Δ f H ) of the products and the reactants. R H Reactants Products

7 Example: C (graphite) + O 2 (g) CO 2 (g) R H = kj Δ R H = v Δ f H ( products ) - v Δ f H ( reactants ) = (1 mol)δ f H (CO 2 ) - (1 mol)δ f H ( graphite) (1 mol)δ f H (O 2 ) = kj Note: we assign Δ f H = 0 for elements in their most stable allotropic forms at a particular temperature example at 298 K Δ f H (O 2 ) = 0 kj mol -1 Δ f H ( graphite) = 0 kj mol -1 so Δ R H for combustion of graphite may be expressed as: Δ R H = (1 mol) Δ f H (CO 2 )= kj

8 Allotropy is the ability of a chemical to exhibit in a number of different and physically distinct forms in its pure elemental state. Carbon, for instance can exist as graphite, diamond and fullerene. pure elemental substances that can exist with different crystalline structures

9 So..standard molar enthalpies of formation (Δ f H ) are important values..with them we can calculate the enthalpy of a reaction (Δ R H ) How do we obtain the values of Δ f H for chemicals? There are two methods: Direct Method Indirect Method The direct method only works for compounds that can be directly synthesized from their elements this is not usually the case so we have to use the indirect method.

10 Indirect Method for Determining Δ f H for Compounds based on Hess s Law Hess s Law is stated as follows: when reactants are converted to products the change in enthalpy is the same whether the reaction takes place in one step or a series of steps remember enthalpy is a state function so it s value is path independent let s look at an example to understand this.

11 Let s use Hess s Law to calculate Δ f H for CO C (graphite) + ½ O 2 (g) CO(g) We have the following two reactions to work with.. (1) C (graphite) + O 2 (g) CO 2 (g) Δ R H = kj (2) CO (g) + ½ O 2 (g) CO 2 (g) Δ R H = kj let s take these two reactions and rearrange them so they add up to give overall desired reaction shown above we need C(graphite) and ½ O 2 in the reactants and CO in the products let s reverse equation (2); and then add up eqn (1) and reverse of (2) Δ f H (CO) = ( kj kj)/(1 mol) = kj mol -1

12 Hess s Law continued so we add up the equations such that everything cancels out to leave only the desired reactants and products if we reverse an equation we must change the sign of Δ R H if we multiply the reaction by a coefficient we must also multiply Δ R H by the same factor Let s work through one more example together

13 Calculate the standard molar enthalpy of formation of acetylene (C 2 H 2 ) from its elements. 2C(graphite) + H 2 (g) C 2 H 2 (g) The equations for combustion and the corresponding enthalpy changes are: (1) C(graphite) + O 2 (g) CO 2 (g) Δ R H = kj (2) H 2 (g) + ½ O 2 (g) H 2 O (l) Δ R H = kj (3) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) Δ R H = kj Solution: Need to rearrange and add equations 1, 2 and 3 so that: 1 - C, H 2 are reactants and C 2 H 2 is product 2 - O 2, CO 2 and H 2 O are eliminated

14 Solution continued. Note: C and H 2 are already in reactants in eqns. 1 and 2 but we need to reverse eq. 3 to get C 2 H 2 into products. 1. Reverse equation 3 to get C 2 H 2 on the product side 2. Now multiply eqn 1 by 4 and eqn 2 by 2 in order to eliminate O 2, CO 2 and H 2 O (1) 4C(graphite) + 4O 2 (g) 4CO 2 (g) Δ R H = kj (2) 2H 2 (g) + O 2 (g) 2H 2 O (l) Δ R H = kj (3) 4CO 2 (g) + 2H 2 O (l) 2C 2 H 2 (g) + 5O 2 (g) Δ R H = kj 4 C(graphite) + 2 H 2 (g) 2 C 2 H 2 (g) Δ R H = kj Thus.2 C(graphite) + H 2 (g) C 2 H 2 (g) Δ R H = kj Δ f H (C 2 H 2 (g)) = Δ R H /(1 mol) = kj mol -1

15 most stable allotropic form of carbon

16 some compounds have lower values of Δ f H than elements they are formed from (e.g. H 2 O (l) ) some compounds have higher values of Δ f H than elements they are formed from (e.g. C 2 H 2(g) ) *** compounds that have negative values for Δ f H are usually more stable than those with positive values WHY? Because neg. value for Δ f H means heat is released during formation so must be supplied for decomposition.

17 Heat of Combustion: the heat involved in the oxidation of a compound..basically a compound is burned in the presence of O 2 to produce CO 2 and H 2 O i.e. C x H y + O 2 (g) CO 2 (g) + H 2 O (l) The heat of combustion for various foods gives us an indication of their ability to store energy Fuel Specific Enthalpy c H (kj/g) (kj/mol) Hydrogen Methane Gasoline (octane) Tristearin (beef fat) 38 Glucose 16 Phys. Chem. With Applications in Biology P.Atkins

18 A sample problem Metabolism is the stepwise breakdown of food we eat to provide energy for growth and function. A general overall equation for this complex process represents the degradation of glucose (C 6 H 12 O 6 ) to CO 2 and H 2 O: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) Calculate the standard enthalpy of reaction at 298K. Solution: Δ R H = v Δ f H ( products ) - v Δ f H ( reactants ) = (6 mol)(δ f H CO 2 ) + (6 mol)(δ f H H 2 O) (1 mol)(δ f H C 6 H 12 O 6 ) (6 mol)(δ f H O 2 ) = (6 mol)( kj/mol) + (6 mol)( kj/mol) (1 mol)( kj/mol) (6 mol)(0 kj/mol) = kj

19 3.6. Thermochemistry Goals. (1) To define enthalpy of reaction (Δ R H ) (2) To understand how to calculate Δ R H from Δ f H for compounds (3) To use Hess s Law to calculate Δ f H Progress (1) Δ R H is the enthalpy change when reactants in standard state are converted to products in standard state (2) Δ R H = v Δ f H ( products ) - v Δ f H ( reactants ) (3) Break reaction down into several steps, rearrange individual equations so can cancel out to leave required reactants and products

20 Section 3.0. The 1 st Law of Thermodynamics (CHANG text Chapter 4) 3.1. Revisiting Heat Capacities 3.2. Definitions and Concepts 3.3. The First Law of THERMODYNAMICS 3.4. Enthalpy 3.5. Adiabatic Expansion of an Ideal Gas 3.6. Thermochemistry 3.7. Bond Energies and Bond Enthalpies

21 3.7. Bond Enthalpies (Section 4.7 in Chang) For diatomic molecules the bond enthalpy..has special meaning since there is only one bond e.g. N 2 (g) 2N(g) Δ R H = kj so bond enthalpy may be assigned to that bond..thus we call it the bond dissociation enthalpy For polyatomics there is more than one bond so we refer to it as the average bond enthalpy e.g. for H 2 O the energy required to break first O-H bond is different from energy required to break second. H 2 O (g) H(g) + OH(g) Δ R H = 502 kj OH(g) O(g) + H(g) Δ R H = 427 kj

22 OH avg. bond enthalpy = 460 kj/mol Triple bond stronger than double which are stronger than single

23 Δ R H, the enthalpy of reaction in the gas phase may be given by: Δ R H = BE (reactants) - BE (products) = total energy bonds broken total energy bonds formed BE is the average bond enthalpy If Δ R H is positive than reaction is endothermic and if negative than exothermic Note: if all the reactants and products are diatomics then we will be using bond dissociation enthalpies (which are known and accurate) so we will get accurate value for Δ R H But if we are working with polyatomics.we will be using avg. bond enthalpies so value obtained for Δ R H is only reasonable approximation.

24 Endothermic Exothermic

25 Sample Problem Estimate the enthalpy of combustion for methane CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) At 298 K, 1 bar using bond enthalpies. Solution Calculate the no. of bonds broken and formed: Bonds No. bonds Bond enthalpy Enthalpy change broken (kj/mol) (kj) C-H O = O Bonds formed C=O O-H

26 Solution Continued Δ R H = BE (reactants) - BE (products) = total energy bonds broken total energy bonds formed = [(1656 kj kj) (1598 kj kj)] = kj We can also calculate Δ R H using Δ f H values: Δ R H = (1 mol) Δ f H CO 2 + (2 mol) Δ R H (H 2 O) (1 mol)δ R H (CH 4 ) (2 mol) Δ R H (O 2 ) = (1 mol)( kj/mol) + (2 mol)( kj/mol) (1 mol)( kj/mol) (2 mol)(0 kj/mol) = kj

Chapter 5 Thermochemistry

Chapter 5 Thermochemistry Learning Outcomes: Interconvert energy units Distinguish between the system and the surroundings in thermodynamics Calculate internal energy from heat and work and state sign