Thermodynamics PDF

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Summary

study guide for specifically thermodynamics...

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Study of heat released or absorbed in chemical reactions and physical change - Glucose is our major energy giving carbohydrate - Series of enzymatic steps in cellular respiration allows for C6H12O6 + 6O2 ← → 6H2O + 6CO2 - The heat/energy released is available to do work like the beat of the heart - The amount of heat released or absorbed at a constant pressure is called enthalpy qp= △ H (triangle means difference of enthalpy from products and reactants) - If the reaction gives a net release of heat (-△H), the reaction is exothermic - If the reaction requires heat (+△H), the reaction is endothermic State property - Enthalpy is a state property, the value is determined by current state - A state property is not dependent on path taken to obtain its current state - Hence state properties can be added or subtracted → △ H= Hp -HR - State properties: Energy, pressure, volume, temperature, heat capacity - Work and heat depend on the path taken so they are not state properties Physical or phase changes - Vaporization: liquid → vapor, endothermic - △Hvaporization = Hvapor -Hliquid, Hvapor > Hliquid - Melting (fusion): solid → liquid, endothermic - Enthalpy of fusion: Hliquid - Hsolid , Hliquid> Hsolid - Sublimation: solid → gas, endothermic - Enthalpy of sublimation: Hgas - Hsolid, Hgas > Hsolid - Enthalpy of sublimation= enthalpy change of melting + enthalpy change of vaporization Diagram of phase changes - Why is there no change in temperature at the melting or boiling point - This is because all the additional heat supplied is being used to break the bonds of the solid or liquid - Why does steam cause a more severe burn than hot water at the same temperature of 100 degrees and 100 grams of each

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Look at notebook for numbers

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Horizontal lines are the phase transitions The numbers at the top is the heat/energy needed to convert 1 mole of water to vapor or the heat/energy needed to heat 1 mole of ice, or the heat/energy to convert 1 mole of water to vapor When liquid touches someone it cools from 100 degrees lets say to 25 degrees which is 75 percent of 7.5 KJ per mole When steam touches someone it has to go through a phase transition which water doesn’t have to do and then it cools down and its enthalpy is much larger so releases greater amount of heat hence: because water has a larger enthalpy of condensation, it will burn more

Enthalpy changes are additive (hess’s law) - A state property is a value that is determined by the current state and is independent of the path taken to obtain that state - State properties can be added or subtracted - Enthalpy is a state function and hence it must be additive - Therefore enthalpy change at each step of a multistep reaction can be added to give the total enthalpy change 1st method (Hess’s law) → since enthalpy is a state function, enthalpy change for a reaction can be calculated easily - Ex: N2 (g) + O2 (g) → 2NO2, △ Hrxn = 180 kJ  2NO(g) + O2 (g) → 2NO2 △ Hrxn  = -112 kJ - Calculate the enthalpy change for the reaction: N2 (g) +2O2 → 2NO2 - Answer is 180 + -122 = 68 kJ Method 2: use bond enthalpies to calculate enthalpy change for reactions - All bond enthalpies are positive since breaking bonds requires energy → endothermic - Reactants, bonds broken, positive → products, bonds formed, negative - Ex: CH2=CH2 + H-Br→ CH3-CH2Br - Bond energies of the reactants minus bond energies of the products Additional comments - Tables of bond enthalpies refer to breaking bonds in gases - Therefore need to add enthalpy of phase changes for liquids and solids - Ex: Br2 (l) → Br2 (g) & Br2 (g) → 2Br (g) △ Hrxn= △Hvapor + △Hbond - Bond enthalpies of diatomic molecules are accurate (measured for the specific molecule however other bond enthalpies are averages from many different molecules - Thus this leads to the most inaccurate method Standard reaction enthalpies - Since reactants and products can be in different states, and reactions can occur at different pressures, the enthalpy of reaction can be different

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For example water in ice form and water in liquid form have the same reaction but different enthalpy changes - When all reactants and products are in their standard state at 1 atm, reaction enthalpy is called the standard reaction enthalpy (△H0) - Standard state for a gas is 1 atm while the standard state for a solution is 1 M at 1 Atm - For a pure liquid or solid it is the pure L or S; for an element it is the most stable phase at 1 atm and the temperature of interest (usually 25 degrees celsius) - Most reactions are given in their standard state △H0 - The standard reaction enthalpy for the formation of one mole of a substance from its elements in their stable state is called the standard enthalpy of formation △ Hf0 (kJmol-1) - Br2 is a liquid, mercury is a liquid, carbon is a solid (graphite), in their standard states - By definition, the standard enthalpy of formation of an element in its most stable form is zero - Ex: O2 (g) - Ex of a calculation: 4C (s) + 6H2 (g) → 2CH3 CH2Oh (l) △ Hrxn  = -555 kJ - △Hf0 = -227 3rd method: if bond enthalpies are not available, one can calculate the standard reaction enthalpy by using the standard enthalpy of formation of the products minus the standard enthalpy of formation of the reactants - Make sure it is in the standard form and it is balanced!!

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Lecture 3: - Measuring heat transfer - Heat required to raise the temperature of an object by 1 degree celsius is the heat capacity - Ex: 98 kJ of heat are required to raise the temperature of some ethanol by 2 degrees celsius - 98/2 is the heat capacity in kJ/C - Since the heat required depends on the amount of substance the heat capacity is an extensive property which is not desirable - If we divide heat capacity by the amount of substance present then we have an intensive property, called specific heat capacity - Units of kJ C-1  g-1  or replace C with K-1  - This is the heat required to raise the temperature of 1 gram of a substance by 1 degree celsius - If we divide heat capacity by the moles of the substance we get the molar heat capacity which is also an intensive property - Can be quoted in J not always kJ

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Molar heat capacity of a gas at constant volume: Cv Molar heat capacity of a gas at constant pressure: Cp - The Cp and Cv are equal for a liquid or solid - However, for a gas Cp > Cv because since system is doing work of expansion volume increases as pressure stays the same - If T increases and everything else stays the same then volume must also increase - Csp is the specific heat capacity Molar heat capacity of a gas at a constant pressure formula - N Cp triangle T = qp = triangle H Specific heat capacity is often called specific heat and is measured using a calorimeter Two main types of calorimetry: constant pressure calorimetry (gives enthalpy values) - Not scaled, no pressure buildup - Constant volume calorimetry: uses bomb calorimeter Equations - Remember to write out neutralization reactions and to indicate the enthalpy change is negative since neutralization reactions are exothermic - Qsystem +  qsurroundings = 0 - If system gains 10 kJ then the surroundings lost 10 kJ - Remember to put a negative on the answer if combustion or neutralization reaction Hess’ law - The enthalpy change of a chemical reaction is independent of the pathway taken between the initial and final states - Depends on state properties Thermodynamics “heat transfer” - Temperature indicates random motion of particles - Heat: transfer of energy due to temperature difference - We can also transfer energy through work (force acting over a distance) - Work is a form of energy In a thermodynamic approach one must: identify the system (object of interest) and surroundings (everything else) - System + surroundings = universe - We usually define surroundings as an entity that is so much bigger than the system so changes in the surroundings are insignificant Open system (matter and energy can exchange with surroundings) - Ex: beaker of water, water can evaporate and beaker isn’t insulated Closed system (energy can exchange with surroundings) - Ex: sealed beaker of water; beaker isn’t insulated Isolated system (nothing exchanges with surroundings) - Ex: combustion of glucose in a bomb calorimeter - Whatever is going on in the system says in the system

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3 ways of changing the energy of a system: - Add or remove the amount of substance in a system (only for open system) - Add 1 gram of glucose to 9 grams of beaker - Heat or cool the system (closed system) - Do work on the system (only for gas phases and in a closed system) - If the surroundings do work on the system, piston compresses and volume decreases, causing work to be positive - If we let the system do work on the surroundings the piston moves out, causing volume to increase, and work to be negative Work - work= - Pexternal △ V If the piston moves out then the energy of the system has decreased by an amount equal to energy lost as work done by the system - System does work expansion when change in volume increases - External pressure is the pressure that the system pushes against - Assume weight of the piston and friction dont matter Systems at equilibrium: we refer to all changes (ex, change in volume) as occurring in small steps - Slower it is → more efficient - Bad that its slow - If system is at equilibrium Pinternal = Pexternal - Then change in volume happens in infinitely small fluctuations - △ V = dV and Pinternal ≅Pexternal - To sum an infiinte number of steps use the integral - w= - ⌠ Pex dV - Yet Pex is a constant so we can move it out and we get the equation w= -Pex △ V - Look at book Equations - When you have a irreversible process: work= - Pexternal △ V - When you have a reversible process: w= -nRTln(final volume/initial volume) - For both the units are moles, L, 8.3145, and kelvin - 1 L atm= 101.325 which you multiply with irreversible processes not reversible

Cv = △ U/ △ T Cp = △ H/ △ T Ideal gases

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- Have internal energy change zero Molar heat capacity - Increases with molecular complexity → as more atoms are present in the molecule, there are more possible bond vibrations that can absorb added energy Remember - Volume change is final - initial - Always have to find the limiting reactant → consider standard reactions in terms of the limiting reactant - Gases take up volume - Even excess reactant gases take up volume Internal energy - Is a state property and so it is additive - △ u = final u - initial u - If change in internal energy is positive then final internal energy> initial internal energy - If change in internal energy is negative then final internal energy < initial internal energy Solid and liquid - They have barely if at all any volume change - This means that the -P △ V is equal to zero and is insignificant so U= qv Gases - When there is a change in the net moles of gas then the △V is significant and hence -P △V is significant so u= qp + - Pext △V Remember - For certain reactions if the pressure is constant → PV= nRT, pressure is constant, R is constant, T is constant, so change in volume is proportional to change in moles - Hence u= q + -△nRT 1st law of thermodynamics - The internal energy of an isolated system never changes, is a constant - The universe is an isolated system, internal energy never changes - Internal energy of a closed system → u= q + -Pext△V Phase changes...