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**Energy Thermodynamics**

Chapter 6 Energy Thermodynamics

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**Energy is... …the ability to do work or produce heat. …conserved.**

…defined as kinetic or potential. …a state function (independent of the path, or how you get from point A to B...depends only on its present state). Work is a force acting over a distance. Heat is energy transferred between objects because of temperature difference.

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**The universe… …is divided into two halves.**

…the system and the surroundings. The system is the part you are concerned with (as a chemist). The surroundings are the rest. Exothermic reactions release energy to the surroundings. Endothermic reactions absorb energy from the surroundings.

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Heat Potential energy

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Heat Potential energy

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**Direction Every energy measurement has three parts:**

A unit ( Joules, kJ or Calories). A number. A sign to tell direction of energy flow. negative - exothermic Positive - endothermic

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Surroundings System Energy DE <0 Energy is leaving the system!

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Surroundings System Energy DE >0 Energy is entering the system!

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**Same rules for heat and work**

Heat given off is negative. Heat absorbed is positive. Work done by system on surroundings is negative. Work done on system by surroundings is positive. Thermodynamics- The study of energy and the changes it undergoes.

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**First Law of Thermodynamics**

The energy of the universe is constant. Law of conservation of energy. q = heat and w = work DE = q + w (Internal Energy = DE) DE is the sum of the KE and PE of all particles in a system…which can be changed by a flow of work and/or heat! Take the systems point of view to decide signs.

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**What is work? Work is a force acting over a distance. w= F x Dd**

P = F/ area Typically, we will let the physicist worry about the work!

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Work needs a sign… If the volume of a gas increases, the system has done work on the surroundings (it pushed against the surroundings). work is negative (or lost to the surr.) w = - PDV Expanding work is negative. Contracting, surroundings do work on the system…w is positive.

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**Enthalpy (for nerds) abbreviated H H = E + PV (that’s the definition)**

Calculated at a constant pressure. DH = DE + PDV the heat at constant pressure qp can be calculated from DE = qp + w = qp - PDV qp = DE + P DV = DH

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Enthalpy for Chemists Since qp = DE + P DV = DH…then we can say qp = DH In a “nutshell”: at constant pressure, the change in enthalpy of the system is equal to the energy flow as heat. Practical applications: the change in enthalpy is given by the equation DH = H products - H reactants

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**Calorimetry Science of measuring heat. Use a calorimeter: Two types…**

#1: Constant pressure calorimeter (called a coffee cup calorimeter) heat capacity for a material, C is calculated C= heat absorbed/DT = DH/DT specific heat capacity = C/mass (units J/ºC•g or J/K•g)

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**Calorimetry molar heat capacity = C/moles**

heat = specific heat x m x DT heat = molar heat x moles x DT Make the units work and you’ve done the problem right. A coffee cup calorimeter measures DH. An insulated cup, full of water. The specific heat of water is 1 cal/gºC Heat of reaction= DH = Cp x mass x DT

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Examples The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

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Calorimetry #2 Constant volume calorimeter is called a bomb calorimeter. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. The heat capacity of the calorimeter is known and tested. Since DV = 0, PDV = 0, DE = q

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**Bomb Calorimeter thermometer stirrer full of water ignition wire**

Steel bomb sample

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Properties Intensive property - not related to the amount of substance. density, specific heat, temperature. Extensive property - does depend on the amount of substance. Heat capacity, mass, heat from a reaction.

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**Hess’s Law Enthalpy is a state function (independent of the path).**

We can “add equations” to come up with the desired final product, and then add the DH values to arrive at the final enthalpy change! Two rules #1 - If the reaction is reversed the sign of DH is changed #2 - If the reaction is multiplied, so is DH

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**State Function One two-step pathway… N2 + O2 2NO ΔH = 180 kJ**

2NO + O2 2NO2 ΔH = -112 kJ A different single-step pathway… N2 + 2O2 2NO2 ΔH = 68 kJ Regardless of the pathway…68 kJ are absorbed in this endothermic reaction!

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**O2 +2NO -112 kJ H (kJ) 180 kJ 2NO2 68 kJ N2 +2O2 Two-step One-step N2**

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Standard Enthalpy The enthalpy change for a reaction under standard conditions (25ºC, 1 atm, 1 M solutions) Symbol DHº When using Hess’s Law, work by adding the equations up to obtain the “target equation”. The other species will cancel out…and you will sum the enthalpy values!

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**Procedure for solving…**

Goal: Establish a pathway to the Target Equation by rearranging the “given equations” Hints: Work backwards from the target equation Look for unique formulas (which only appear once) Avoid formulas which appear multiple times until you get started

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**Hess’s Law Example Given DHº= -1300. kJ**

calculate DHº for this reaction (target equation) DHº= kJ DHº= -394 kJ DHº= -286 kJ

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**Standard Enthalpies of Formation**

Hess’s Law is useful when you are given multiple sets of reactions…. However, the standard enthalpy of formation (DHfº) for a compound is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements in their standard states.

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**Standard Enthalpies of Formation**

Standard states are 1 atm, 1M and 25ºC See p. 246 of our text. For an element it is 0 (zero)! Why? It comes to us already formed! There is a table of values listed in Appendix 4 (pg A19-22).

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**Standard Enthalpies of Formation**

First, you need to be able to write the balanced equation from the elements. What is the equation for the formation of NO2 ? ½N2 (g) + O2 (g) ® NO2 (g) Have to make one mole to meet the definition. Write the equation for the formation of methanol CH3OH.

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**Since we can manipulate the equations…**

We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C2H5OH +3O2(g) ® 2CO2 + 3H2O which leads us to this rule:

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**Since we can manipulate the equations…**

We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C2H5OH +3O2(g) ® 2CO2 + 3H2O which leads us to this rule.

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**C2H5OH(l) +3O2(g) ® 2CO2(g) + 3H2O(l)**

Solve the problem: C2H5OH(l) +3O2(g) ® 2CO2(g) + 3H2O(l) ΔHf: (kJ/mol) ΔH = [(2•-393.5kJ)+(3•-286kJ)]-[-278kJ + 3•0kJ] ΔH = -1367kJ of energy (released) per mole of ethanol burned.

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Chapter 6 problems:

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