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Thermal Chemistry. V.B.3 a.Explain the law of conservation of energy in chemical reactions b.Describe the concept of heat and explain the difference between.

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Presentation on theme: "Thermal Chemistry. V.B.3 a.Explain the law of conservation of energy in chemical reactions b.Describe the concept of heat and explain the difference between."— Presentation transcript:

1 Thermal Chemistry

2 V.B.3 a.Explain the law of conservation of energy in chemical reactions b.Describe the concept of heat and explain the difference between heat energy and temperature c.Explain physical and chemical changes and endothermic or exothermic energy changes

3 Energy Def: the ability to do work Classified as either: Potential Energy Kinetic Energy

4 1 st Law of Thermodynamics The energy of the universe is constant Energy can neither be created or destroyed Can be converted from one form to another

5 Energy Energy is a state function Property of the system that changes independently of the pathway

6 Temperature & Heat Temperature is a measure of the motion of the particles of a substance Heat is the flow of energy due to a temperature difference

7 System & Surroundings

8 Internal Energy Always from the viewpoint of the system ΔE = q + w q = heat w = work

9 Special Conditions If the container is rigid and cannot expand, then no work is done by or on the system All energy must be in the form of heat (q) and therefore ΔE = q

10 V.B.3. d. Solve heat capacity and heat transfer problems involving specific heat, heat of fusion, and vaporization e. Calculate the heat of reaction for a given chemical reaction when given calorimetric data

11 Measuring Energy calorie (c) – the amount of energy needed to raise the temperature of 1g of water by 1 o C Joule (J) – 4.184 J = 1 calorie (SI unit)

12 Heating Depends On.. 1.The amount of substance being heated (in grams) 2.The temperature change 3.Specific heat capacity – the amount of energy needed to raise the temperature of one gram of a substance by 1 o C

13 Calculating Energy Change q = m c ΔT q = heat m = mass in grams c = specific heat capacity ΔT = change in temperature

14 V.B.4. a.Define enthalpy and explain how changes in enthalpy determine whether a reaction is endothermic or exothermic b.Compute ΔH rxn from ΔH f o and explain why the ΔH f o values for elements are zero

15 ENTHALPY

16 What is Enthalpy? Consider a process at constant pressure where the only work is PV work (w = -PΔV) ΔE = q p + w ΔE = q p – PΔV q p = ΔE + PΔV H = E + PV or ΔH = ΔE + Δ(PV) ΔH = ΔE + PΔV (pressure is constant) ΔH = q p

17 In Other Words… the terms heat of reaction and change in enthalpy are the same so ΔH = H products - H reactants

18 Endothermic – absorbs heat during reaction (feels cold) Exothermic – gives off heat during reaction (feels hot)

19 Calculate the Standard Enthalpy Change for the combustion of Methane CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) 1.CH 4 (g)  C(s) + 2H 2 (g) 2H 2 (g) + C(s)  CH 4 (g) -75 kJ 2.O 2 (g) 0 kJ 3.C(s) + O 2 (g)  CO 2 (g)-394 kJ 4.H 2 (g) + ½ O 2 (g)  H 2 O(l)-286 kJ

20 Example When 1 mole of CH 4 is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which 5.8 g of CH 4 is burned at constant pressure.

21 Calorimetry Calorimeter – used to determine the heat energy change during a reaction Carried out under constant pressure measures enthalpy (ΔH) Carried out under constant volume measures energy (ΔE)

22 Example A 110. g sample of copper (specific heat capacity = 0.20 J/C o ∙g) is heated to 82.4 o C and then placed in a container of water at 22.3 o C. The final temperature of the water and the copper is 24.9 o C. What was the mass of the water in the original container, assuming complete transfer of heat from the copper to the water?

23 Heat lost by copper = -(heat lost by copper) = (heat gained by water)

24 Heating Curve

25 Heat of Fusion ΔH fus = enthalpy change that occurs in melting a solid at its melting point Example: What quantity of heat is needed to melt 1.0 kg of ice at its melting point? ΔH fus =6.0 kJ/mol

26 Heat of Vaporization ΔH vap = the energy needed to vaporize one mole of a liquid at a pressure of 1 atm Example: What quantity of heat is required to vaporize 130. g of water?

27 Example Substance X has the following properties: ΔH vap = 20. kJ/mol ΔH fus = 5.0 kJ/mol Boiling point = 75 o C Melting point = -15 o C Specific heat Solid = 3.0 J/g o C Liquid = 2.5 J/g o C Gas = 1.0 J/g o C

28 Calculate the energy required to convert 250. g of substance X from a solid at - 50 o C to a gas at 100 o C. Assume that X has a molar mass of 75.00 g/mol. 5 Step Process 1.Heating solid 2.Melting solid 3.Heating liquid 4.Boiling liquid 5.Heating gas

29 Solution 1.q = m x c x ΔT = 250.g x (3.0 J/g o C) x 35 o C = 26 kJ 2.mol x ΔH fus = 3.33 mol x 5.0 kJ/mol =17 kJ 3.q = m x c x ΔT = 250.g x (2.5 J/g o C) x 90 o C =56 kJ 4.mol x ΔH vap = 3.33 mol x 20. kJ/mol =67 kJ 5.q = m x c x ΔT = 250.g x (1.0 J/g o C) x 25 o C =6.2 kJ 172 kJ

30 Heat of Formation ΔH f o The change in enthalpy of the formation of one mole of a compound from it elements in their standard states ΔH f ° 25 o C at 1 atm and 1 M

31 ΔH f o = 0 By definition, the standard heat of formation for elements in their standard states equals zero. Example: Which of the following will have standard heats of formation equal to zero? H 2 (g), Hg(s), CO 2 (g), H 2 O(l), Br 2 (l)

32 Example Write the balanced molecular equation representing the ΔH f ° for ethanol. Answer: 2C(s) + 3H 2 (g) + ½ O 2  C 2 H 5 OH

33 Hess ’ s Law

34 Based on… 1.State function 2.Enthalpy change is same for a reaction whether the reaction takes place in one or many steps

35 How to use Hess’s Law Manipulate equations to reach the desired reaction If the reaction given is reversed, so is ΔH If multiplying the equation to balance the coefficients also multiply ΔH by the same number

36 Example Calculate the enthapy for the following reaction: N 2 (g) + 2O 2 (g)  2NO 2 (g) ΔH° = ??? kJ N 2 (g) + O 2 (g)  2NO(g) ΔH° = +180 kJ 2NO 2 (g)  2NO(g) + O 2 (g) ΔH° = +112 kJ

37 For any reaction… ΔH° reaction = Σ nΔH° f(products) - Σ mΔH° f(reactants)

38 Calculate the Standard Enthalpy Change for the Combustion of Methane 1.2H 2 (g) + C(s)  CH 4 (g)-75 kJ 2.O 2 (g) 0 kJ 3.C(s) + O 2 (g)  CO 2 (g)-394 kJ 4.H 2 (g) + ½ O 2 (g)  H 2 O(l)-286 kJ

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