Energy (E) Energy is the capacity for doing work or supplying

heat.

Two types of energy:

Kinetic

Potential

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Energy (E)

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Heat

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Exothermic Exothermic describes a process that releases

energy in the form of heat.

Forming a chemical bond releases energy and therefore is an exothermic process.

Exothermic reactions usually feel hot because it is giving heat to you.

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Endothermic Endothermic is a process or reaction that absorbs

energy in the form of heat.

Breaking a chemical bond requires energy and therefore is endothermic.

Endothermic reactions usually feel cold because it is taking heat away from you.

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Heat some physical changes involve an exchange of heat

Note: Dissolving does as well

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Exothermic vs. Endothermic a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it.

c) Water boils in a kettle being heated on a stove.

d) Water vapor condenses on a cold pipe.

e) Ice cream melts.

Exothermic

Endothermic

Endothermic

Exothermic

Endothermic

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Exothermic vs. Endothermic Exothermic Processes Endothermic Processes

freezing water

solidifying solid salts

condensing water vapor

making a hydrate from an anhydrous salt

forming an anion from an atom in the gas phase

Annihilation of matter E=mc2

splitting of an atom

melting ice cubes

melting solid salts

evaporating liquid water

making an anhydrous salt from a hydrate

forming a cation from an atom in the gas phase

splitting a gas molecule

separating ion pairs

cooking an egg

baking bread

Exothermic Reactions Endothermic Reactions

Combustion of hydrogen

dissolving lithium chloride in water

Burning of propane

dehydration of sugar with sulfuring acid

thermite

decomposition of hydrogen peroxide

decomposition od ammonium dichromate

halogenation of acetylene

Reaction of barium hydroxide octahydrate crystals with dry

ammonium chloride

dissolving ammonium chloride in water

reaction of thionyl chloride (SOCl2) with cobalt(II) sulfate

heptahydrate

mixing water and ammonium nitrate

mixing water with potassium chloride

reacting ethanoic acid with sodium carbonate

photosynthesis (chlorophyll is used to react carbon dioxide plus

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Involvement of Energy vs Heat

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Law of Conservation of Energy The law of conservation of energy states that energy

cannot be created or destroyed. Energy can only be transferred or transformed.

Therefore since you cannot gain or lose energy, heat lost must equal heat gained.

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Quantity of Heat (Q)

Q = Quantity of heat energy, in joules - [J]

m = mass of the substance, in grams [g]

c = specific heat capacity of the substance [J/(g °C)]

∆T = temperature change, in degrees Celsius [°C]

∆T = Tf - Ti

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Specific Heat Capacity (C)

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Quantity of Heat Example 1 Calculate the quantity of thermal energy absorbed by a 5.00 kg block of concrete to raise its temperature from 17.1°C to 35.5°C.

Q = ? m = 5.00kg = 5000g c = 2.10 J/(g °C) ∆T = 35.5-17.1°C = 18.4°C

Q = mc∆T = (5000g)(2.10)(18.4°C) Q = 193 200 J = 193.2 kJ

The concrete block must absorb 193 kJ.

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Quantity of Heat Example 2 In one experiment, a student placed 50.0 mL of 1.00 M HCl in a coffee-cup calorimeter and carefully measured its temperature to be 25.5oC. To this was added 50.0 mL of 1.00 M NaOH solution whose temperature was also 25.5oC. The mixture was quickly stirred, and the student noticed that the temperature of the mixture rose to 32.4oC. What was the heat of reaction?

These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.184 J/goC. The density of water is 1.00 g/mL and even though these are solutions we can assume that they are close enough to water to have the same density.

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Quantity of Heat Example 2 In one experiment, a student placed 50.0 mL of 1.00 M HCl in a coffee-cup calorimeter and carefully measured its temperature to be 25.5oC. To this was added 50.0 mL of 1.00 M NaOH solution whose temperature was also 25.5oC. The mixture was quickly stirred, and the student noticed that the temperature of the mixture rose to 32.4oC. What was the heat of reaction?

Q = ? m = 50 + 50 mL = 100 mL = 100 g c = 4.184 J/(g °C) ∆T = 32.4-25.5°C = 6.9°C

Q = mc∆T = (100g)(4.184)(6.9°C) Q = 2886.96 J

The heat of the reaction is 2886.96 J. This is the heat gained by the water, but in fact it is the heat lost by the reacting HCl and NaOH, therefore Q = -2.9 x 103 J. I.E. it is an exothermic reaction, heat was lost to the water and it got warmer.

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Calorimetry The purpose of calorimetry is to use an instrument

known as a calorimeter to determine the enthalpy of a substance undergoing a chemical change.

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Calorimetry When a reaction occurs at constant pressure inside a

calorimeter, the enthalpy change involves heat, and little heat is lost to the lab (environment) or gained from it.

If the reaction gives off heat, very nearly all of it stays inside the calorimeter, and it is this amount of heat absorbed or released by the reaction that can be calculated.

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Enthalpy (H) Enthalpy (H) can also be defined as the total heat

energy stored in chemical bonds.

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Exothermic Reactions heat is liberated

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Endothermic Reactions heat is absorbed

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Enthalpy Change (ΔH) Enthalpy change (ΔH) is the energy exchanged

between a system and its environment during a physical change or chemical reaction at constant pressure – also referred to as the heat of reaction.

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Enthalpy Change (ΔH) To calculate Enthalpy change (ΔH) we would:

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Enthalpy Change (ΔH) Heat of Formation:

NOTE: the enthalpy of formation for all elements (Zn, Fe...) and

diatomic elements (N2, O2...) = 0 (zero)

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Enthalpy (ΔH) Example 1

From the reference table: NO(g) ΔHf

o = +90.3 kJ mol-1 O2(g) ΔHf

o = 0 kJ mol-1 NO2(g) ΔHf

o = +34 kJ mol-1

ΔH = [2 x ΔHf(NO2(g)] - [2 x ΔHf(NO(g) + ΔHf(O2(g))]

ΔH = [(2 x +34)] - [(2 x +90.3) + (2 x 0)] = [68] – [180.6] ΔH = -112.6

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Enthalpy (ΔH) Example 2 Calculate the standard enthalpy (heat) of reaction for the oxidation of ammonia gas to produce nitrogen dioxide gas and water vapour given the following standard enthalpies (heats) of formation:

NH3(g) + 7/4O2(g) → NO2(g) + 3/2H2O(g) NH3(g) ΔHfo = -46 kJ mol-1

NO2(g) ΔHfo = +34 kJ mol-1

H2O(g) ΔHfo = -242 kJ mol-1

O2(g) ΔHfo = 0 kJ mol-1

ΔH = [ΔHf(NO2(g)) + 3/2 x ΔHf(H2O(g))] - [ΔHf(NH3(g)) + 7/4 x ΔHf(O2(g))]

ΔH = [+34 + (3/2 x -242)] - [-46 + (7/4 x 0)]

ΔH = [+34 + -363] - [-46 + 0] = -329 + 46 ΔH = -283 kJ

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Molar Enthalpy

m: mass [g] MM: Molar Mass [g/mol]

C: Molar concentration [mol/L] V: volume [L]

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Enthalpy (ΔH) Example 3

Hcl

Q = ? m = 100 + 50 g = 150 g C = 4.184 J/(g °C) ∆T = 35.5-20.0°C = 15.5°C

Q = (150 g)(4.184)(15.5°C) Q = 9727.8 J Q = 9.73 kJ

HCl: C = 2 mol/L V = 100 mL = 0.1L

nHCl = (C)(V) = (2mol/L)(0.1L)

nHCl = 0.2 mol

∆H = -Q/n = -(9.73kJ)/(0.2mol) = -48.6 kJ/mol

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Hess’ Law If two or more thermochemical equations are added to

give a final equation, then the enthalpies (ΔH) can be added together to give the enthalpy (ΔH) for the final equation.

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Hess’ Law To carry out the sum of the thermochemical equations:

The identical terms situated on the same side of the equation are added.

The identical terms on either side of the equation are subtracted.

If an equation is reversed, the ΔH sign is also reversed.

If the coefficients of a chemical reaction are modified by multiplying or dividing them by a common factor, then the value ΔH must also be multiplied or divided by this same common factor.

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Hess’ Law Example 1 Consider the following two thermochemical equations:

and use them to find the ΔH for the reaction.

ΔH = (- 110.5 kJ) + (- 283.0 kJ) = -393.5 kJ

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Hess’ Law Example 2

ΔH = (+ 26.74 kJ) + (3 x (- 283.0 kJ)) = - 822.26 kJ

Must be reversed

Multiply by a factor of 3

3 1.5 3 x 3

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Bond Energies Enthalpy from bond energies can be calculated by:

Check out this video for a quick lesson: https://www.youtube.com/watch?v=o8HF-kdQuys

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Bond Energies

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Bond Energies

H2

O2

N2

CH4

Here, there are four C-H bonds.

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Bond Energies

C4H10

Here, there are three C-C bonds and ten C-H bonds.

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Enthalpy from Bond Energies Example #1

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Enthalpy from Bond Energies Example #2

Calculate the enthalpy of combustion for propane:

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Thermodynamics & Stoichiometry Practice Problem #1

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