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Transcript of Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the...
![Page 1: Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the energy stored in stressed objects, like a bent bow.](https://reader036.fdocuments.net/reader036/viewer/2022062422/56649e925503460f94b98807/html5/thumbnails/1.jpg)
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Energy comes in many forms:
1. kinetic energy is the energy of motion.2. potential energy is the energy stored in
stressed objects, like a bent bow or compressed spring, or water at the top of a waterfall.
3. chemical energy is the energy stored in chemical bonds, which is released when a chemical reaction, like burning, occurs.
4. nuclear energy is the energy stored within the nucleus of an atom:• Fission• Fusion
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Heat
• Heat is thermal energy. • Heat is the product of motion. • All matter has heat in it, since all
matter is made up of particles which move.
• Something has no heat only if all motion has stopped inside it. This point is called absolute zero.
• Absolute zero is the lowest temperature that can be reached:– it is equal to - 273 ° Celsius– it is also equal to 0 Kelvins (K)
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Heat
• It is not possible to determine the amount of heat an object has, since it is tied up with the various types of energy.
• Heat always flows from a warmer to a colder object. It continues to flow until both objects are at the same temperature.
• Different masses or volumes of a substance require different amounts of heat to produce a similar rise in temperature.
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Heat
• Different substances of the same mass require different amounts of heat to produce a similar rise in temperature: – Given equal inputs of heat energy, antifreeze will
rise much more rapidly in temperature than water will.
– Put another way, if water and antifreeze are heated, water requires more heat energy per unit mass to produce a similar temperature rise.
– Put a third way, in equal masses of water and antifreeze at the same temperature, the water contains more heat.
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Specific Heat Capacity• Also called Specific Heat, or
Heat Capacity.• Is the amount of heat required
to heat 1 g or 1 kg of a substance by 1°C (or 1 K)
• Measured in kiloJoules per kilogram degree Celsius (kJ/kg·C).
• All substances have their own specific heat capacity; it is a characteristic physical property.
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Measuring Changes in Heat• Involves 3 things:
1. Heat Capacity – is a constant for each substance or phase of a substance.
2. Mass – the more mass, the more heat.
3. Temperature Change – can be positive or negative.
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Measuring Changes in HeatAmount of = Specific x Mass x
Change in Heat Heat
Temperature
Q = C x M x ΔT
Q = C x M x (Tf – Ti)
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Q = C x M x ΔT
• This is an equation of four variables. If you have three of the variables you should be able to calculate the fourth. Remember a few things, though: 1. Make sure the units are correct. Include metric
conversion. 2. Heat can go into or come out of a substance.
– If heat goes in temperature goes up the change is positive.
– If heat comes out temperature goes down the change is negative.
3. Use the specific heat table for the C values (see handout).
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Q = C x M x ΔT
1. Calculate the heat required to change the temperature of 1.5 kg water from 40.0°C to 80.0°C.
• Since Q = C x M x (Tf – Ti) and the substance is water, therefore
• C = 4.18 kJ/kgC• M = 1.5 kg• Tf = 80.0 C• Ti = 40.0 C• Q = (4.18 kJ/kgC)(1.5 kg)(80.0 C - 40.0 C)• = 250 kJ (2 significant digits)
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Q = C x M x ΔT
2.Calculate the heat required to change the temperature of 0.300 kg sucrose from 80.0°C to 19.0°C.
• C = 1.25 kJ/kgC• M = 0.300 kg• Tf = 19.0 C• Ti = 80.0 C• Q = (1.25 kJ/kgC)(0.300 kg)(19.0 C - 80.0 C)• = -22.9 kJ
(a negative value, since the temperature dropped)
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Q = C x M x ΔT
3. Calculate the heat required to increase the temperature of 650 g copper by 12.6°C.
•
• C = 0.385 kJ/kgC• M = 650 g x (1 kg / 1000 g) = 0.65 kg • ΔT = 12.6 C (only temp change is given)
• Q = (0.385 kJ/kgC)(0.65 kg)(12.6 C)• = 3.2 kJ (2 significant digits)
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ΔT = Q C x M
4.Calculate the temperature change that occurs when 2500 kJ of heat is added to 4.45 kg of asbestos.
• C = 0.820 kJ/kgC• M = 4.45 kg• Q = 2500 kJ
• ΔT = 2500 kJ (0.820 kJ/kgC)(4.45 kg)
• = 690 C
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ΔT = Q C x M
5. Calculate the temperature change that occurs when -652 J of heat is added to 241 g of mercury.
• C = 0.140 kJ/kgC• M = 241 g x (1kg / 1000 g) = 0.241 kg• Q = -652 J x (1 kJ / 1000 J) = -0.652 kJ
• ΔT = -0.652 kJ(0.140 kJ/kgC)(0.241 kg)
• = -19.3 C (the temperature dropped)
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M = Q C x ΔT
6.Calculate the mass of wood that would undergo a temperature change of 26.5°C when 1360 kJ of heat is added.
• C = 1.76 kJ/kgC• ΔT = 26.5 C• Q = 1360 kJ• M = 1360 kJ
(1.760 kJ/kgC)(26.5 C)• = 29.2 kg
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Phase Changes
0
2 0
4 0
6 0
8 0
1 0 0
1 2 0
1 4 0
-2 0
-4 0
-6 0
H ea tin g C u rv e o f W a te r
L IQU ID
GAS
MELT ING
FREEZ ING
CONDENSING
BO IL ING
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Phase Changes
• Heat goes into the breaking of physical bonds between the molecules of the substance, and putting distance between the molecules:
– When a substance melts it goes from a rigid, hard solid to a liquid. Bonds are broken so that molecules have the freedom of movement.
– When a liquid becomes a gas the molecules get much further apart.
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Phase Changes
• Heat of Fusion (Hf) - This is the heat required to melt or freeze a substance. Expressed in kilojoules per kilogram or kilojoules per mole. If a substance is freezing the Hf value is expressed as a negative.
• Heat of Vapourization (Hv) - This is the heat required to vapourize or boil a substance. Expressed in the same units as heat of fusion. If a substance is condensing the Hv value is expressed as a negative.
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Calculating the Energy of Phase Changes
• Heat Required to Melt/Freeze = Heat of Fusion x Mass of a Substance Substance
• Q = HfM (for water Hf is 334 kJ/kg)
• Heat Required to Boil/Condense = Heat of x Mass ofa Substance Vapourization Substance
• Q = HvM (for water Hv is 2260 kJ/kg)
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1. Calculate the heat required to melt 1.5 kg ice
• Since the phase change is melting the formula is
Q = Hf M
• where Hf = 334 kJ/kg
• and M = 1.5 kg
• Q = (336 kJ/kg)(1.5 kg)
= 5.0 x 102 kJ
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2. Calculate the heat required to freeze 4.50 x 102 kg water.
• Since the phase change is freezing the formula is
Q = Hf M
• where Hf = - 334 kJ/kg (heat is taken out)
• and M = 4.50 x 102 kg
• Q = (- 336 kJ/kg)(4.50 x 102 kg)
= - 1.50 x 105 kJ
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3. Calculate the heat required to boil 245 g water.
• Since the phase change is boiling the formula is
Q = Hv M
• where Hv = 2260 kJ/kg and M = 245 g x (1 kg/1000 g) = 0.245 kg
• Q = (2260 kJ/kg)(0.245 kg)
= 554 kJ
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4. Calculate the heat required to condense 1.2 t steam.
• Since the phase change is condensing the formula is
Q = Hv M
• where Hv = - 2260 kJ/kg and M = 1.2 t x (1000 kg/ 1 t) = 1200 kg
• Q = (- 2260 kJ/kg)(1200 kg)
= 2.7 x 106 kJ
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5. Calculate the heat required to melt 162 mol ice.
• Since the phase change is melting the formula is
Q = Hf M• where Hf = 334 kJ/kg • and M = is not given, but is 162 mol• The molar mass of water is 18.02 g/mol (from H2O) and• Mass = Molar mass x Number of moles
= (18.02 g/mol)(162 mol) = 2920 g x (1 kg / 1000 g) = 2.92 kg
• Q = (334 kJ/kg)(2.92 kg) = 975 kJ
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Combined Heat Problems
0
2 0
4 0
6 0
8 0
1 0 0
1 2 0
1 4 0
-2 0
-4 0
-6 0
H ea tin g C u rv e o f W a te r
L IQU ID
GAS
MELT ING
FREEZ ING
CONDENSING
BO IL ING
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how much heat is required to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ?
• 5 steps are required:– heat ice from -20.00°C to 0°C– melt ice– heat water from 0°C to 100°C– boil water– heat steam from 100°C to 110.0°C
• any change in temperature uses the formula Q = C M ΔT
• phase changes use the formula Q = HfM or HvM
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how much heat is required to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ?
• heat ice from -20.00°C to 0°CQ = C M ΔT
= (2.06 kJ/kg·°C)(2.00 kg)(0°C – (-20.0°C))
= 82.4 kJ• melt ice
Q = Hf M= (334 kJ/kg)(2.00 kg)= 668 kJ
• heat water from 0°C to 100°CQ = C M ΔT
= (4.18 kJ/kg·°C)(2.00 kg)(100°C – 0°C)= 836 kJ
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• boil waterQ = Hv M= (2260 kJ/kg)(2.00 kg)= 4520 kJ
• heat water from 100°C to 110°CQ = C M ΔT= (1.86 kJ/kg·°C)(2.00 kg)(110°C – 100°C)= 37.2 kJ
• Total = 84.2 + 668 + 836 +4520 kJ + 37.2 kJ
= 6150 kJ
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Enthalpy
• refers to the energy change in chemical reactions.
• the symbol for enthalpy is H.• it is impossible to determine the heat
or energy contained in an object or chemical, but we can determine the change in energy in a chemical reaction.
• change in enthalpy is ΔH
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Enthalpy Change
• reactions that give off energy are exothermic; their ΔH is negative.
• reactions that take in energy are endothermic; their ΔH is positive.
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Heat of Formation
• is the enthalpy change when a compound is formed from its elements.
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Heat of FormationElements Heat of reaction
(kJ/mol of product) H2 (g) + 1/2 O2 (g) H2O(g) - 241.8H2 (g) + 1/2 O2 (g) H2O(l) - 285.8S(s) + O2 (g) SO2 (g) - 296.8H2 (g) + S(s) + 2 O2 (g) H2SO4 (l) - 812S(s) + 3/2 O2 (g) SO3 (g) - 395.71/2 N2 (g) + 1/2 O2 (g) NO(g) + 90.371/2 N2 (g) + O2 (g) NO2 (g) + 33.851/2 N2 (g) + 3/2 H2 (g) NH3 (g) - 46.19C(s) + 1/2 O2 (g) CO(g) - 110.5C(s) + O2 (g) CO2 (g) - 393.5C(s) + 2 H2 (g) CH4 (g) - 74.862 C(s) + 3 H2 (g) C2H6 (g) - 83.8 3 C(s) + 4 H2 (g) C3H8 (g) - 104
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Using Heats of Formation
1. C (s) + 1/2 O2 (g) CO(g)
Δ H = - 110.5 kJ
2. CO(g) + 1/2 O2 (g) CO2 (g) ΔH = - 283.0 kJ
3. C (s) + O2 (g) CO2 (g) Δ H = - 393.5 kJ
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Using Heats of Formation
• C (s) + 1/2 O2 (g) CO(g) Δ H = - 110.5 kJ
• CO(g) + 1/2 O2 (g) CO2 (g)
ΔH = - 284.0 kJ
• C (s) + O2 (g) CO2 (g) Δ H = - 393.5 kJ
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Hess’ Law• Hess's law of constant heat
summation:
• The enthalpy change for any reaction depends only on the products and reactants and is independent of the pathway or the number of steps between the reactant and product.
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2 Al (s) + 3 CuO (s) 3 Cu(s) + Al2O3
(s) • Identify the compounds in the equation; find heat of
formation equations from the table that contain them (don’t worry about the elements. They take care of themselves):
• 2 Al (s) + 1½ O2 (g) Al2O3 (s) ΔH = - 1676.0 kJ/mol • Cu (s) + ½ O2 (g) CuO (s) ΔH = - 155.0 kJ/mol
• Flip the CuO equation to make the compound a reactant as well:
• Cu (s) + ½ O2 (g) CuO (s) ΔH = - 155.0 kJ/mol
• Becomes
• CuO (s) Cu (s) + ½ O2 (g) ΔH = + 155.0 kJ/mol
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2 Al (s) + 3 CuO (s) 3 Cu(s) + Al2O3
(s)
• Multiply the CuO equation by 3 to have the same number of molecules as in the original equation. The enthalpy change is multiplied by the same factor:
• 2 Al (s) + 1½ O2 (g) Al2O3 (s) ΔH = - 1676.0 kJ/mol
• 3 x CuO (s) Cu (s) + ½ O2 (g) ΔH = 3 x (+155.0
kJ/mol)
• This gives• • 2 Al (s) + 1½ O2 (g) Al2O3 (s) ΔH = - 1676.0 kJ/mol • 3 CuO (s) 3 Cu (s) + 1½ O2 (g) ΔH = + 465.0 kJ/mol
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2 Al (s) + 3 CuO (s) 3 Cu(s) + Al2O3
(s)• Add the two equations together. You treat the equations
just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added:
• 2 Al (s) + 1½ O2 (g) Al2O3 (s) ΔH = - 1676.0 kJ/mol • 3 CuO (s) 3 Cu (s) + 1½ O2 (g) ΔH = + 465.0 kJ/mol• 2 Al (s) + 1½ O2 (g) + 3 CuO (s) Al2O3 (s) + 3 Cu (s) + 1½ O2(g)
• ΔH = - 1211 kJ
• Cancel out like terms:• • 2 Al (s) + 3 CuO (s) Al2O3 (s) + 3 Cu (s) ΔH = - 1211 kJ
• You know you are right if the net equation you end up with is the same as the original equation you started with.
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2 Al (s) + 3 CuO (s) 3 Cu(s) + Al2O3
(s)• WHAT YOU ACTUALLY NEED TO SHOW:
• 2 Al (s) + 1½ O2 (g) Al2O3 (s) ΔH = - 1676.0 kJ/mol • 3 x 3 CuO (s) 3 Cu (s) + (3)½ O2 (g) ΔH = 3 x (+155.0
kJ/mol)
• 2 Al (s) + 3 CuO (s) Al2O3 (s) + 3 Cu (s) ΔH = - 1211 kJ
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2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O
(l)
• 2 x C2H6 (g) 2 C (s) + 3 H2 (g)
ΔH = 2 x(+ 84.0 kJ/mol)
• 4 x C (s) + O2 (g) CO2 (g)
ΔH = 4 x (- 394.0 kJ/mol)
• 6 x H2 (g) + ½ O2 (g) H2O (l)
ΔH = 6 x (- 286.0 kJ/mol)
• 2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l)
ΔH = - 3124 kJ
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SiO2 (s) + C (s) CO2 (g) + Si (s) Where the heat of formation of
SiO2 (s) = - 861 kJ/mol
• Heat of formation is formation of a compound from its elements. SiO2 is made of silicon (Si (s)) and oxygen (O2
(g)):
• 1x SiO2 (s) Si (s) + O2 (g)
ΔH = +861 kJ/mol
• 1x C (s) + O2 (g) CO2 (g)
Δ H = - 394 kJ/mol
• SiO2 (s) + C (s) CO2 (g) + Si (s)
Δ H = + 467 kJ
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Enthalpy and Entropy
• Enthalpy is one of the driving forces in the universe; reactions that release energy (negative ΔH) are favoured.
• Endothermic reactions do occur spontaneously; that is because of a second force, Entropy.
• Entropy is disorder• The more disorder, the more entropy.
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Examples of Increasing Entropy• tossed salad
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Examples of Increasing Entropy• Broken glass
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Examples of Increasing Entropy• Bedroom: Before
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Examples of Increasing Entropy• Bedroom: After
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Recognizing Entropy in Reactions:• Phase changes:
– solids have lowest entropy– gases have highest entropy
– C12H22O11 (s) C12H22O11 (l)
– increasing entropy
– C2H5OH (g) C2H5OH (l)
– decreasing entropy
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Recognizing Entropy in Reactions:
• Mixtures have higher entropy than pure substances:– C12H22O11 (s) C12H22O11 (aq)
– increasing entropy
– Na1+(aq) + Cl1-
(aq) NaCl (s)
– decreasing entropy
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Recognizing Entropy in Reactions:
• Side of reaction equation with more particles has higher entropy:
– 2 SO2 (g) + O2(g) 2 SO3 (g)
– decreasing entropy
– C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
– increasing entropy
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Spontaneous Processes
• Spontaneous processes are those that can proceed without any outside intervention.
• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously
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Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
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Spontaneous Processes• Processes that are spontaneous at one
temperature may be nonspontaneous at other temperatures.
• Above 0C it is spontaneous for ice to melt.• Below 0C the reverse process is spontaneous.
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Combining Enthalpy and Entropy• If enthalpy and entropy are both
forces that affect the spontaneity of reactions how can we determine whether a reaction will be spontaneous or not ?
• GIBB’S FREE ENERGY
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GIBB’S FREE ENERGY
• Gibb’s free energy (ΔG) combines enthalpy and entropy terms to tell if there is extra energy left over to cause a reaction to happen.
• ΔG = ΔH - T ΔS
• where T is the temperature in Kelvin (always positive)
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ΔG = ΔH - T ΔS• a reaction is spontaneous if the ΔG function is
negative. That means there are 4 possibilities of ΔH and ΔS:
• ΔH ΔS Effect on the Reaction
• positive positive ΔG will be negative only if TΔS is very large; this reaction will be spontaneous only at high temperature.
• positive negative ΔG cannot be negative under any
conditions. This reaction will not proceed spontaneously.
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ΔG = ΔH - T ΔS
• ΔH ΔS Effect on the Reaction
• negative positive ΔG will be negative under all conditions. This reaction will proceed spontaneously at all temperatures.
• negative negative ΔG will be negative only if TΔS is very small; this reaction will be spontaneous
only at low temperature.
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1. H2O (l) H2O (s) ΔH = - 6 kJ
• ΔH is negative. The reaction is exothermic.
• ΔS is negative. This is a phase change and the solid is more ordered than the liquid, so entropy is decreasing.
• ΔG = ΔH - TΔS
• this reaction will be spontaneous only if TΔS is as small as possible, so as to be smaller than the enthalpy change.
• This reaction will only proceed at low temperatures.
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2. CaCO3 (s) + 181 kJ CaO (s) + CO2 (g)
• ΔH is positive. The energy term is on the reactant side, which means it is going into the reaction, so it is endothermic.
• ΔS is positive. This is a decomposition reaction producing more molecules and one product is a gas.
• ΔG = ΔH - TΔS
• this reaction will be spontaneous only if TΔS is as large as possible, so as to be bigger than the enthalpy change.
• This reaction will only proceed at high temperatures.
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3. CS2 (g) + 3 O2 (g) CO2 (g) + 2 SO2 (g)
ΔH = + 1110 kJ• ΔH is positive. The reaction is endothermic.
• ΔS is negative. All the substances are gases, but the reactants have more particles than the products.
• ΔG = ΔH - TΔS
• this reaction is not spontaneous at any temperature because Gibbs Free Energy cannot be made to be positive.
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4. NaCl (s) Na1+ (aq) + Cl1- (aq)
+ energy • ΔH is negative. The reaction is exothermic,
since it appears on the product side of the equation.
• ΔS is positive. Since the salt is dissolving the entropy is increasing.
• ΔG = ΔH - TΔS • since both factors are favourable for a
spontaneous reaction this reaction will be spontaneous at any temperature.
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Kinetics
• Free energy tells us if a reaction is likely to be spontaneous, but tells us nothing about how fast it will be.
• Kinetics studies the rate at which a chemical process occurs.
• Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).
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The Collision Model
• In a chemical reaction, bonds are broken and new bonds are formed.
• Molecules can only react if they collide with each other.
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The Collision Model
Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.
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Activation Energy• In other words, there is a minimum amount of
energy required for reaction: the activation energy, Ea.
• Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.
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Factors That Affect Reaction Rates
• Physical State of the Reactants– In order to react, molecules must come in
contact with each other.– The more homogeneous the mixture of
reactants, the faster the molecules can react.
– nature of the reactants - the more complex the reaction, the slower the rate. • Reactions that only involve a change in charge
will proceed more rapidly than a reaction that involves breaking and reforming chemical bonds
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Factors That Affect Reaction Rates
• Concentration of Reactants– As the concentration of reactants
increases, so does the likelihood that reactant molecules will collide.
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Factors that Affect Reaction Rates• Surface Area
– the more surface area the more molecules exposed to react (more collisions).
– Gases react more quickly than solids
– powders react more quickly than larger lumps.
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Factors That Affect Reaction Rates
• Temperature– At higher temperatures, reactant
molecules have more kinetic energy, move faster, and collide more often and with greater energy.
– The minimum energy needed for the reaction to occur is threshold energy.
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Effect of Temperature on Rate
• Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.
• At any temperature there is a wide distribution of kinetic energies.
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Effect of Temperature on Rate
• As the temperature increases, the curve flattens and broadens.
• Thus at higher temperatures, a larger population of molecules has higher energy.
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Effect of Temperature on Rate• If the dotted line represents the activation
energy, as the temperature increases, so does the fraction of molecules that can overcome the threshold energy barrier.
• As a result, the reaction rate increases.
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Factors That Affect Reaction Rates
• Presence of a Catalyst– Catalysts speed up reactions by
changing the mechanism of the reaction.
– Catalysts are not consumed during the course of the reaction.
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Catalysts• Catalysts increase the rate of a reaction by
decreasing the activation energy of the reaction.
• Catalysts change the mechanism by which the process occurs.
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Catalysts
One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.
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Catalysts
• A second mechanism is to become part of the activated complex.
• Enzymes perform this function.
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Enzymes• Enzymes are catalysts in
biological systems.• The substrate fits into the
active site of the enzyme much like a key fits into a lock.
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Inhibitors
• are the opposite of catalysts; they increase the activation energy and reduce the rate of a chemical reaction.
• they often act by forming a complex with one of the reactants, preventing it from having successful collisions with other reactants.
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Potential Energy DiagramsIt is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.
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Potential Energy Diagrams
• It shows the energy of the reactants and products (and, therefore, E).
• The high point on the diagram is the transition state.
• The species present at the transition state is called the activated complex.
• The energy gap between the reactants and the activated complex is the activation energy barrier.
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Drawing Potential Energy Diagrams• reactant energy = 150 kJ• threshold energy = 400 kJ• product energy = 250 kJ• activation energy =• ΔH = • exo- or endo-themic ?
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Drawing Potential Energy Diagrams• reactant energy = 450
kJ• threshold energy = • product energy = • activation energy = 50
kJ• ΔH = - 200 kJ• exo- or endo-themic ?
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Drawing Potential Energy Diagrams• reactant energy = • threshold energy = • product energy = 350
kJ• activation energy =
250 kJ• ΔH = +150 kJ• exo- or endo-themic ?
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Drawing Potential Energy Diagrams
• reactant energy = • threshold energy = 300 kJ• product energy = • activation energy = 200
kJ• ΔH = -50 kJ• exo- or endo-themic ?
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Reaction Rates
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.
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Reaction Rates
One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.
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Concentration and Rate
Comparing Experiments 1 and 2, when [NH41+]
doubles, the initial rate doubles.
NH41+
(aq) + NO21-
(aq) N2 (g) + 2 H2O (l)
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Concentration and Rate
Likewise, comparing Experiments 5 and 6, when [NO2
1-] doubles, the initial rate doubles.
NH41+
(aq) + NO21-
(aq) N2 (g) + 2 H2O (l)
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Reaction Order
• is a description of the relationship between the reactants and their effect on the rate of a reaction
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Reaction Order
• Zeroth Order – a change in concentration of a reactant has no effect on the rate of the reaction.
• In the rate equation the concentration is raised to the zero power.
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Reaction Order
• First Order – a doubling in concentration of a reactant doubles the rate of the reaction.
• In the rate equation the concentration is raised to the first power.
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Reaction Order
• Second Order – a doubling in concentration of a reactant causes the rate of the reaction to increase 4 times.
• In the rate equation the concentration is raised to the second power.
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Rate Laws
• The overall reaction order can be found by adding the exponents on the reactants in the rate law.
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Concentration and Rate
When [NH41+] doubles, the rate doubles.
When [NO21-] doubles, the rate doubles.
NH41+
(aq) + NO21-
(aq) N2 (g) + 2 H2O (l)
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Writing Rate Law Equations• This means
Rate [NH41+]
Rate [NO21-]
Rate [NH1+] [NO21-]
or
Rate = k [NH41+] [NO2
1-]• This equation is called the rate law,
and k is the rate constant.• This reaction is second order overall.
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Writing Rate Law Equations
• 2 ICl + H2 I2 + 2 HCl
• if [ICl] doubles, the rate doubles (first order)
• if [H2] quadruples, rate quadruples (first order)
• Rate = k [ICl]1 [H2]1
• Rate is second order overall
Experiment
[ICl] (mol/L)
[H2] (mol/L)
Initial Rate(mol/(L·s))
1 0.10 0.01 0.002
2 0.20 0.01 0.004
3 0.10 0.04 0.008
![Page 103: Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the energy stored in stressed objects, like a bent bow.](https://reader036.fdocuments.net/reader036/viewer/2022062422/56649e925503460f94b98807/html5/thumbnails/103.jpg)
Calculating Rate Law Constant
• Rate = k [ICl]1 [H2]1
• k = Rate . [ICl][H2]
• k = (0.002 mol/(L·s)) (0.10 mol/L)(0.01 mol/L)
• k = 2
• Rate = 2 [ICl][H2]
Experiment [ICl] (mol/L)
[H2] (mol/L) Initial Rate(mol/(L·s))
1 0.10 0.01 0.002
2 0.20 0.01 0.004
3 0.10 0.04 0.008
![Page 104: Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the energy stored in stressed objects, like a bent bow.](https://reader036.fdocuments.net/reader036/viewer/2022062422/56649e925503460f94b98807/html5/thumbnails/104.jpg)
Writing Rate Law Equations
• A + B C
• if [A] doubles, the rate quadruples (second order)
• if [B] doubles, rate unchanged (zeroth order)• Rate = k [A]2 [B]0 • Rate is second order overall
Experiment
[A] (mol/L) [B] (mol/L) Initial Rate(mol/(L·s))
1 0.002 0.05 2.0 x 10-5
2 0.004 0.05 8.0 x 10-5
3 0.002 0.10 2.0 x 10-5
![Page 105: Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the energy stored in stressed objects, like a bent bow.](https://reader036.fdocuments.net/reader036/viewer/2022062422/56649e925503460f94b98807/html5/thumbnails/105.jpg)
Calculating Rate Law Constant
• Rate = k [A]2 [B]0 • k = Rate .
[A]2
• k = (2.0 x 10-5 mol/(L·s)) (0.002 mol/L)2
• k = 5• Rate = 5 [A]2
Experiment
[A] (mol/L) [B] (mol/L) Initial Rate(mol/(L·s))
1 0.002 0.05 2.0 x 10-5
2 0.004 0.05 8.0 x 10-5
3 0.002 0.10 2.0 x 10-5
![Page 106: Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the energy stored in stressed objects, like a bent bow.](https://reader036.fdocuments.net/reader036/viewer/2022062422/56649e925503460f94b98807/html5/thumbnails/106.jpg)
Writing Rate Law Equations
• H2O2 (aq) + 2 HI (aq) 2 H2O (l) + I2 (aq)
• if [H2O2] doubles, the rate doubles (first order)
• if [HI] doubles, rate doubles (first order)
• Rate = k [H2O2]1 [HI]1
• Rate is second order overall
Experiment
[H2O2] (mol/L)
[HI] (mol/L)
Initial Rate(mol/(L·s))
1 0.05 0.05 0.002
2 0.05 0.10 0.004
3 0.10 0.05 0.004
![Page 107: Energy comes in many forms : 1.kinetic energy is the energy of motion. 2.potential energy is the energy stored in stressed objects, like a bent bow.](https://reader036.fdocuments.net/reader036/viewer/2022062422/56649e925503460f94b98807/html5/thumbnails/107.jpg)
Writing Rate Law Equations
• Rate = k [H2O2]1 [HI]1
• k = Rate . [H2O2][HI]
• k = (0.002 mol/(L·s)) (0.05 mol/L)(0.05 mol/L)
• k = 0.8• Rate = 0.8 [H2O2][HI]
Experiment [H2O2] (mol/L)
[HI] (mol/L) Initial Rate(mol/(L·s))
1 0.05 0.05 0.002
2 0.05 0.10 0.004
3 0.10 0.05 0.004