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**The study of the heat flow of a chemical reaction or physical change**

Thermochemistry The study of the heat flow of a chemical reaction or physical change

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**1st Law of Thermodynamics**

Energy cannot be created or destroyed but only converted from one form to another The energy of the universe is constant

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**Ex - Elevation vs distance**

State Function A state function is independent of the pathway (does not rely on the past or future of the system A state function depends only on the present state of the system Ex - Elevation vs distance

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**Energy The capacity to do work or provide heat**

Potential Energy (PE) = mhv Kinetic Energy (KE) = 1/2mv2 The total energy of the system is equal to the sum of the potential and kinetic energy Can be changed by a flow of heat/work ΔE = q + w

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Chemical Energy Energy will flow either from the system to its surroundings (exothermic) or from the surroundings to the system (endothermic) The system will be the reactants and products ΔE < 0 energy flows out of the system ΔE > 0 energy flows into the system

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**From the system’s point of view…**

q + Heat flows into the system - Heat flows out of the system w Work done on the system Work done by the system

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Internal Energy Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system ΔE = q + w ΔE = 15.6 kJ kJ = 17.0 kJ Work vs Energy flow

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Work Done by or to Gases

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**Work is done by the system so w is negative**

PV Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm w = -PΔV ΔV = 64 L – 46 L = 18 L w = -(15 atm x 18 L) w = -270 L∙atm Work is done by the system so w is negative

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**Internal Energy, Heat & Work**

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. Use 1 L∙atm = J

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**w = -(1.0 atm x (5.0 x 105 L)) = -5.0 x 105 L∙atm**

ΔE = q + w q = +1.3 x 108 J w = -PΔV ΔV = 5.0 x 105 L w = -(1.0 atm x (5.0 x 105 L)) = -5.0 x 105 L∙atm ΔE = (+1.3 x 108 J) + (-5.1 x 107 J) ΔE = 8 x 107 J

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**Enthalpy (H) H = E + PV E = internal energy P = pressure of the system**

V = volume of the system Enthalpy is a state function

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**ΔH = ΔE + PΔV (pressure is constant)**

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

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**ΔH = Hproducts - Hreactants**

In other words… the terms heat of reaction and change in enthalpy are the same so ΔH = Hproducts - Hreactants

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**2H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ**

Enthalpy How much heat is released when 4.03 g of hydrogen is reacted with excess oxygen? 2H2(g) + O2(g) 2H2O(l) ΔH = -572 kJ

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Enthalpy #2 When 1 mole of CH4 is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which 5.8 g of CH4 is burned at constant pressure. qp = ΔH = -890 kJ/mole CH4

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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)

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Heat Capacity Specific heat capacity – the amount of heat needed to raise one gram on a substance by 1oC (J/Co∙g or J/K∙g) Molar heat capacity – the amount of heat needed to raise one mole of a substance by 1oC (J/Co∙mol or J/K∙mol)

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**Heat Gained or Lost Depends on…**

The change in temperature during the reaction The amount of substance present The heat capacity of the substance

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

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**-(heat lost by copper) = (heat gained by water)**

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Calorimetry #2 When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0oC is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0oC in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1oC. Assuming that the calorimeter absorbs only a negligible amount of heat, that the specific heat capacity of the solution is 4.18 J/oC∙g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

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**Solution (no pun intended)**

Species present: Ba2+, NO3-, Na+, and SO42- Spectator ions: Na+ and NO3- Net Ionic equation: Ba2+(aq) + SO42-(aq) BaSO4(s)

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**q = qp = (2.0 x 103 g)(4.18J/oCg)(3.1oC)**

Heat evolved by reaction = heat absorbed by the solution = m x c x ΔT Since 1.00 L of each solution is used, total volume of the mixture is 2.00 L (2.0 x 103 g) Temperature increase = 28.1oC – 25.0oC = 3.1oC q = qp = (2.0 x 103 g)(4.18J/oCg)(3.1oC) 2.6 x 104 J = ΔH

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Hess’s Law

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**Based on… State function**

Enthalpy change is same for a reaction whether the reaction takes place in one or many steps

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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

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**Calculate the enthapy for the following reaction:**

N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = ??? kJ N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ 2NO2(g) ---> 2NO(g) + O2(g) ΔH° = +112 kJ

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**Solution N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ**

2NO(g) + O2(g) ---> 2NO2(g) ΔH° = -112 kJ + 68 kJ The second equation has been reversed including ΔH

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**Calculate ΔH° for this reaction:**

2N2(g) + 5O2(g) ---> 2N2O5(g) H2(g) + 1/2O2(g) H2O(l) ΔH° = kJ N2O5(g) + H2O(l) 2HNO3(l) ΔH° = kJ 1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l) ΔH° = kJ

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Solution

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**Standard Enthalpies of Formation**

The change in enthalpy of the formation of one mole of a compound from it elements in their standard states

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ΔH° 25oC at 1 atm and 1 M 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? H2(g), Hg(s), CO2(g), H2O(l), Br2(l)

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Example Write the balanced molecular equation representing the ΔHf° for ethanol. Answer: 2C(s) + 3H2(g) + 1/2O2(g) CH5OH(l)

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**ΔH°reaction = Σ ΔH°f(products) - Σ ΔH°f(reactants)**

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

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**Calculate the Standard Enthalpy Change for the combustion of Methane**

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) CH4(g) C(s) + 2H2(g) 2H2(g) + C(s) CH4(g) kJ O2(g) kJ C(s) + O2(g) CO2(g) kJ H2(g) + ½ O2(g) H2O(l) kJ

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Example #1 Calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water Use table 6.2 on page on Page 262

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**Heat of Fusion & Heat of Vaporization**

Heating Curves Heat of Fusion & Heat of Vaporization

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Heating Curves

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Heat of Fusion ΔHfus = 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? ΔHfus =6.0 kJ/mol

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**For liquid water ΔHvap = 43.9 kJ/mol**

Heat of Vaporization ΔHvap = 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? For liquid water ΔHvap = 43.9 kJ/mol

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**Example #2 Substance X has the following properties:**

ΔHvap = 20. kJ/mol ΔHfus = 5.0 kJ/mol Boiling point = 75oC Melting point = -15oC Specific heat Solid = 3.0 J/goC Liquid = 2.5 J/goC Gas = 1.0 J/goC

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Example #2 continued… Calculate the energy required to convert 250. g of substance X from a solid at -50oC to a gas at 100oC. Assume that X has a molar mass of g/mol. 5 Step Process Heating solid Melting solid Heating liquid Boiling liquid Heating gas

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**Example #2 continued… q = m x c x ΔT**

= 250.g x (3.0 J/goC) x 35oC = 26 kJ mol x ΔHfus = 3.33 mol x 5.0 kJ/mol = 17 kJ = 250.g x (2.5 J/goC) x 90oC = 56 kJ mol x ΔHvap = 3.33 mol x 20. kJ/mol =67 kJ = 250.g x (1.0 J/goC) x 25oC = 6.2 kJ 172 kJ

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