Thermochemical Equations

Transcription

1 Assignment There are two correct answers for #9 on the exam choice b and c. If you have one of these and it is marked wrong--submit your exam and get 6 points.

2 Thermochemical Equations 1. The stoichiometric coefficients always refer to the number of moles of a substance H 2 O (s) H 2 O (l)!h = 6.01 kj 2. If you reverse a reaction, the sign of!h changes H 2 O (l) H 2 O (s)!h = kj 3. If you multiply both sides of the equation by a factor n, then!h must change by the same factor n. 2H 2 O (s) 2H 2 O (l)!h = 2 x 6.01 = 12.0 kj 4. The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (s) H 2 O (l)!h = 6.01 kj H 2 O (l) H 2 O (g)!h = 44.0 kj

3 !H Is The Standard Thermodynamic State In thermodynamics we also define a standard state of where conditions are rigorously defined. They are different than a gas at STP (1 atm and 0 C). THERMODYNAMIC STANDARD STATE --1 atmosphere pressure --25 C = K --1 Molar concentration for solutions containing solutes WE ADD A SUPERSCRIPT!H TO DENOTE STANDARD STATE CONDITIONS OR MEASUREMENT

4 !H Is The Standard Thermodynamic State Example of Reaction N2 (g) + 3H2 (g) > 2NH3 (g)!h = kj 1 mol of N 2 reacts with 3 mol H 2 to produce 3NH 3 and evolves kj at 1 atm, 25 C 2N2 (g) + 6H2 (g) > 4NH3 (g)!h = 2 X kj 1/3N2 (g) + H2 (g) ----> 2/3NH3 (g)!h = 1/3 X kj

5 Types of Heats of Reaction,!H rxn WHEN MEASURED UNDER STANDARD CONDITIONS THEY ALL BECOME!H rxn

6 Thermochemical Equations How much heat is evolved when 266 g of white phosphorus (P 4 ) is combusted in air? Molar mass of P4 is g/mol. P 4 (s) + 5O 2 (g) P 4 O 10 (s)!hcomb = kj!h = H (products) H (reactants)!h = 1 mol P g P 4 x g P 4 x kj 1 mol P 4 = kj

7 Using the Heat of Reaction (!H rxn ) to Find Amounts The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by Al 2 O 3 (s) 2Al(s) + 3/2O 2 (g)!h rxn = 1676 kj If aluminum is produced this way, how many grams of aluminum can form when x 10 3 kj of heat is transferred? SOLUTION: heat produced = kj 2 mol Al 1676 kj g Al 1 mol Al = g Al

8 Why Do We Care About State Functions? A change in a functions of state,! has a unique value between any two arbitrary states. State 2 mgh2 Hf!h!H State 1 h = 1 Hi

9 Enthalpy Energy Level Diagrams CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H 2 O(l) H 2 O(g) CH 4 + 2O 2 H initial H 2 O(g) H final Enthalpy, H Enthalpy, H!H < 0 heat out!h > 0 heat in CO 2 + 2H 2 O H final H 2 O(l) H initial A Exothermic process B Endothermic process

10 Why Do We Care About State Functions? 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. Property of a State Function! Suppose we want to calculate the enthalpy for the following reaction and can not measure it directly. Thermodynamics let s us calculate it. CH4 (g) + 2O2 ==> CO2 + 2H2O!H1 =? kj Data From A Table CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O!H2 = kj!h3 = kj

11 Hess Law Example 2 CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O CH4 (g) + 2O2 ==> CO2 + 2H2O!H2 = kj!h3 = kj!h1 =? kj

12 Tabulated Standard Heats of Reactions Are Used To Predict Any!H Using Hess s law. Using the characteristics of Hess s Law just described, calculate the enthalpy of the following reaction, using the reactions and their associated enthalpies from below. FeCl 2 (s) + 1/2 Cl 2 (g) FeCl 3 (s)!h =? 1. Fe(s) + Cl 2 (g) FeCl 2 (s)!h = kj/mol 2. Fe(s) + 3/2 Cl 2 (g) FeCl 3 (s)!h = kj/mol

13 Standard Enthalpy of Formation!Hf Standard Enthalpy of Formation,!H f, is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states k (25 C) and 1 atmosphere pressure. Examples: H 2 (g) + O 2 (g) "# H 2 O(l )!H f = kj/mol 3C(s) + 4H 2 (g) "# C 3 H 8 (g) Ag(s) + 1/2Cl 2 (g) > AgCl(s)!H f = kj/mol Always For One Mole of Product!---thus, we see fractional moles as Reactants on Occasion. Oxygen exists as O 2 gas at 25 C Carbon exists as solid graphite at 25 C. Sulfur exits as S8 as a solid at 25 C Water is H 2 O(l ) in its standard state (not ice or water vapor).

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15 Writing Formation Equations PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include!h 0 f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: Use the table of heats of formation for values.

16 Writing Formation Equations PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include!h 0 f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: SOLUTION: Use the table of heats of formation for values. (a) Ag(s) + 1/2Cl 2 (g) AgCl(s)!H 0 f = kj (b) Ca(s) + C(graphite) + 3/2O 2 (g) CaCO 3 (s)!h 0 f = kj (c) 1/2H 2 (g) + C(graphite) + 1/2N 2 (g) HCN(g)!H 0 f = 135 kj

17 Calculating!H f from tabulated data Applications of Enthalpies of Formation. Example: calculate the!h comb for the combustion of benzene, C 6 H 6 (l ) from enthalpies of formation taken from a table. 2 C 6 H 6 (l ) + 15 O 2 (g) "# 12 CO 2 (g) + 6 H 2 O(l )!H comb =? 1. From a Table C(s) + O 2 (g) "# CO 2 (g)!h f = kj H 2 (g) + O 2 (g) "# H 2 O(l )!H f = kj 1/2 6C(s) + 3H 2 (g) "# C 6 H 6 (l )!H f = kj 2. Rearrange thermochemical equations to satisfy desired equation C 6 H 6 (l ) + O 2 (g) "# 6CO 2 (g) + 3H 2 O(l)!H comb =?

18 Heats of Reaction From!H f 6C(s) + 6O 2 (g) "# 6CO 2 (g) 3H 2 (g) + O 2 (g) "# 3H 2 O(l ) C 6 H 6 (l ) "# 6C(s) + 3H 2 (g)!h f = 6( kj)!h f = 3( kj)!h f = ( kj) C 6 H 6 (l ) + O 2 (g) "# 6CO 2 (g) + 3H 2 O(l)!H comb =?!H comb = 6( kj) + 3( kj) + ( kj) = -3,267.4 kj that s for one mole of benzene only! 2 C 6 H 6 (l ) +15 O 2 (g) # 12 CO 2 (g) + 6 H 2 O(l )!H comb = 2 X kj/mol

19 Major Riff... We can obtain the same answer by using a balanced equation for one mole of product (combustion, neutralization, dissolution) a table of standard enthalpies and the following easy formula:!h rxn = $ ni!hi f (products) - $ mi!hi f (reactants) where $ means the sum of ni is the respective molar coefficient for i th product mi is the respective molar coefficient for each i th reactant

20 Heats of Reaction From!H f aa + bb cc + dd!hrx for any chemical reaction:!h =!H f (Products)!H f (Reactants!H = [c!h f (C) + d!h f (D)] [a!h f (A) + b!h f (B)]

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22 Calculating the Heat of Reaction from Heats of Formation Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate!H 0 rxn from!h 0 f values. PLAN: Look up the!h 0 f values and use Hess s Law to find!h rxn.!h 0 f NH 3 (g) = kj/mol!h 0 f O 2 (g) = 0!H 0 f H 2 O = kj/mol!h 0 f NO(g) = 90.4 kj/mol

23 Calculating the Heat of Reaction from Heats of Formation Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate!H 0 rxn from!h 0 f values. SOLUTION:!H rxn = $ m!h 0 f (products) - $ n!h 0 f (reactants)!h rxn = [4(!H 0 f NO(g) + 6(!H 0 f H 2 O(g)] - [4(!H 0 f NH 3 (g) + 5(!H 0 f O 2 (g)] = (4 mol)(90.4 kj/mol) + (6 mol)( kj/mol) - [(4 mol)(-46.3 kj/mol) + (5 mol)(0 kj/mol)]!h rxn = -904 kj

24 Heats of Reaction From!H f

25 Energy Diagram Depiction!H f The general process for determining!h 0 rxn from!h0 f values. Elements Enthalpy, H Reactants decomposition -!H 0 f formation!h 0 f H initial!h 0 rxn Products H final!h rxn = $ m!h f(products) - $ n!h f(reactants)