Learning Check q = c * m * ΔT How much heat, q, is required to raise the temperature of 1000 kg of iron and 1000 kg of water from 25 C to 75 C? (c water =4.184 J/ C g, c iron =0.450 J/ C g) q Fe = 0.450 q water = 4.184 J g 1.00x103 kg 1000g 1kg J g 1.00x103 kg 1000g 1kg 75 C 25 C 75 C 25 C = 2.25x10 7 J = 2.09x10 8 J 1
Practice Problem A 22.05 g solid is heated in a testtube to 100.00 C and added to 50.00 g of water in a coffee-cup calorimeter. The water temperature changes from 25.10 C to 28.49 C. Find the specific heat capacity of the solid. q = c * m * ΔT C water = 4.184 J/g K 2
Since the water and the solid are in contact, heat is transferred from the solid to the water until they reach the same T final. In addition, the heat given out by the solid (-q solid ) is equal to the heat absorbed by the water (q water ). DT water = T final T initial = (28.49 C 25.10 C) = 3.39 C = 3.39 K DT solid = T final T initial = (28.49 C 100.00 C) = -71.51 C = -71.51 K c solid = c x mass x DT H 2 O H 2 O H 2 O mass solid x DT solid 4.184 J/g K x 50.00 g x 3.39 K = = 0.450 J/g K 22.05 g x (-71.51 K) 3
Homework 50.0 ml of 0.500 M NaOH is placed in a coffee-cup calorimeter at 25.00 o C and 25.0 ml of 0.500 M HCl is carefully added, also at 25.00 o C. After stirring, the final temperature is 27.21 o C. Calculate q soln (in J) and the change in enthalpy, DH, (in kj/mol of H 2 O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/ml and c = 4.184 J/g K 4
Thermochemical Equation A balanced equation that includes ΔH rxn C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) ΔH = -2043 kj 1. ΔH < 0: reaction is exothermic. 2. 2043 kj is released and can be written as product. C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) + 2043 kj 3. Useful stoichiometric factors for calculations 1 mol C 3 H 8 (g) = -2043 kj 5 mol O 2 (g) = -2043 kj 3 mol CO 2 (g) = -2043 kj 5
Energy Transfer in a Reaction The relationship between amount (mol) of substance and the energy (kj) transferred as heat during a reaction. 6
Learning Check The major source of aluminum in the world is bauxite which is mostly aluminum oxide. Its thermal decomposition can be represented by: 2 Al 2 O 3 (s) 4Al(s) + 3O 2 (g) ΔHrxn = 3352 kj If aluminum is produced this way, how many grams of Al can form when 1.000 x 10 3 kj of heat is used to decompose the oxide? g Al = 1000kJ 4molAl 3352 kj 26.98 g Al 1 mol Al = 32.20 g Al 7
Learning Check Using the thermochemical equation, calculate how much heat is evolved when 266 g of white phosphorus (P 4, MW = 123.9 g/mol) is combusted in air. P 4 (s) + 5O 2 (g) P 4 O 10 (s) ΔHrxn = -3013 kj H = 266 g P 4 1 mol P 4 123.9 g P 4 3013 kj 1 mol P 4 = 6470 kj 8
Calculating Heat of Reaction 2 Ways or Problem Types for Enthalpy Calculations Rearrange Multiple Thermochemical Equations Use Standard Enthalpy of Formation, ΔH f 9
Hess s Law The enthalpy change for any reaction is equal to the sum of the enthalpy changes for any individual step in the reaction. CH 4 (g) + 2O 2 (g) CO + 2H 2 O + 1/2 O 2 ΔH 1 = -607 kj CO(g) + 2H 2 O + 1/2 O 2 (g) CO 2 + 2H 2 O ΔH 2 = -283 kj CH 4 (g) + 2O 2 (g) CO 2 + 2H 2 O ΔH 3 =? kj Because enthalpy is a state function, its value only depends on the final and initial state and not the path. This allows us to predict the enthalpy of many chemical reactions. 10
Hess s Law CH 4 (g) + 2O 2 (g) CO + 2H 2 O + 1/2 O 2 ΔH 1 = -607 kj CO(g) + 2H 2 O + 1/2 O 2 (g) CO 2 + 2H 2 O ΔH 2 = -283 kj CH 4 (g) + 2O 2 (g) CO 2 + 2H 2 O ΔH 3 =? kj 11
Enthalpy, H Energy Level Diagrams Useful pictorials of the thermodynamic transitions that take place during a chemical reaction. CH 4 (g) + 2O 2 (g) ΔH 1 = -607 kj ΔH 3 = -890 kj CO + 2H 2 O + 1/2 O 2 CO 2 + 2H 2 O ΔH 2 = -283 kj 12
Manipulating Thermochemical Equations 1. The stoichiometric coefficients always refer to the number of moles of a substance (it s extensive). H 2 O(s) H 2 O(l) ΔH = 6.01 kj 2. If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n. 2H 2 O(s) 2H 2 O(l) ΔH = 2(6.01) = 12.0 kj 3. If you reverse a reaction, the sign of ΔH changes. H 2 O(l) H 2 O(s) ΔH = -6.01 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 2 O(l) H 2 O(g) ΔH = 6.01 kj ΔH = 44.0 kj 13
Practice Problem Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) CO 2 (g) + 1/2N 2 (g) ΔH =? Given the following information, calculate the unknown ΔH. CO(g) + 1/2O 2 (g) CO 2 (g) N 2 (g) + O 2 (g) 2NO(g) ΔH = -283.0 kj ΔH = 180.6 kj 14
Practice Problem Calculate the enthalpy of the following reaction and draw an energy level diagram that summarize the thermodynamic enthalpies involved in this reaction. NO 2 (g) NO(g) + 1/2 O 2 (g) ΔH = 57.07 kj 1/2 N 2 (g) + O 2 (g) NO 2 (g) ΔH = 33.18 kj 1/2 N 2 (g) + O 2 (g) NO(g) + 1/2 O 2 (g) ΔH =? 15
Calculating an Unknown ΔH 1. Identify the target equation with the unknown ΔH. 2. Manipulate equations with known ΔH values so that products and reactants are on the correct sides. a) Change sign of ΔH if equation is reversed. b) Multiply n and ΔH by the same factor. 3. Add manipulated equations and their ΔH values. Double-check. 16