Exothermic and endothermic reactions. Chemical Reactions usually involve a temperature change (heat...
-
Upload
loraine-norman -
Category
Documents
-
view
226 -
download
1
Transcript of Exothermic and endothermic reactions. Chemical Reactions usually involve a temperature change (heat...
Exothermic and endothermic reactions
• Chemical Reactions usually involve a temperature change
(heat is given out or taken in)
Law of conservation of energy
• Energy cannot be created or destroyed, but only changed from one form into another
Exothermic Reactions
• Examples include:– Burning reactions including the
combustion of fuels.– Detonation of explosives.– Reaction of acids with metals.
Thermit reaction
Magnesium reacting with acid
Exothermic reactions increase in temperature.
• Magnesium + Hydrochloric acid
Gets hot
25o C 45o C
magnesium
Hydrochloricacid
Heatenergygivenout
Exothermic Reactions
Reactants convert chemical energy to heat energy.
The temperature rises.
25o C45o C
Exothermic Reactions
45o C
• Almost immediately the hot reaction products start to lose heat to the surroundings and eventually they return to room temperature.
25o C
Exothermic Reactions
Ene
rgy
/ kJ
)
Progress of reaction (time)
Energy Level Diagram for an Exothermic Reaction
reactants
Reactants have more chemical energy.
Some of this is lost as heat which spreads out into the room.
productsProducts now have less chemical energy than reactants.
Exothermic reactions give out energy. There is a temperature rise
and H is negative.
Exothermic Reaction - Definition
products
Ene
rgy
/ kJ
)
Progress of reaction
reactants
H is negative
Heat changes also happen when substances change state.
An exothermic reaction
• When hydrocarbons burn in oxygen they produce carbon dioxide and water vapour.
• The reaction also involves the loss of heat so it is an exothermic reaction.
Endothermic reaction
• An endothermic reaction is when heat is taken in in a reaction.
Endothermic Reactions
• Endothermic chemical reactions are relatively rare.
• A few reactions that give off gases are highly endothermic - get very cold.
Cools
Heatenergytakenin as the mixture returns back to room temp.
Starts 25°C Cools to 5°C
Ammoniumnitrate
Water
Endothermic reactions cause a decrease in temperature.
Returns to 25°C
Endothermic Reactions
25o C
• The cold reaction products start to gain heat from the surroundings and eventually return to room temperature.
5o C The reactants gain energy.25o C
This comes from the substances used in the reaction and the reaction gets cold.
Eventually heat is absorbed from the surroundings and the mixture returns to room temperature.
Overall the chemicals have gained energy.
Endothermic Reactions
Endothermic reactions take in energy. There is a temperature
drop and H is positive.
Endothermic Reaction Definition
H=+
products
Ene
rgy
/ kJ
Progress of reaction
reactants
Heat of reaction
• The heat of reaction is the heat change when the number of moles of reactants indicated in the balanced equation react completely.
For an exothermic reaction:
the heat of reaction is always negative e.g ∆H = -34kJ
For an endothermic reaction:
the heat of reaction is always positive e.g ∆H = +34kJ
Chemical hazard notes
Concentrated hydrochloric acid : Very corrosive to eyes and skin, and its vapour is very irritating to lungs.
Sodium hydroxide : Caustic, harmful to skin and especially to eyes. Always wear eye protection.
Measuring the heat of reaction of hydrochloric acid and sodium hydroxide
A polystyrene cup is used as it is an insulator of heat – it doesn’t let the heat escape ( has negligible heat capacity)
The equation for the reaction is
HCl + NaOH → NaCl + H2O
Other precautions to ensure an accurate result!
1. Make sure both solutions are at the same temperature before you start!
2. Wash the thermometer and dry it before switching solutions.
3. Stir the mixture slowly and make sure none of the mixture is splashed out of the cup.
Results
Temperature of HCl solution before mixing = oC Temperature of NaOH solution before mixing = oC Highest temperature reached after mixing = oC Temperature rise = oC Number of moles of acid used = Number of moles of base used =
Calculations –
What is the heat change in the reaction carried our:
Heat change = mc ∆ T
Mass in kilogramsSpecific heat capacity
Temperature rise
Q260. Calculations for the heat of reaction
• 261. Calculate the heat of reaction (using the formula ΔH = mcΔT) for the reaction between and nitric acid sodium hydroxide from the following experimental results:
• Volume of nitric acid = 100 cm3 of 1.0 M• Volume of sodium hydroxide = 100 cm3 of 1.0 M• Initial temperature of the solutions = 17.5 oC• Final temperature of the solutions = 24.4 oC• Specific heat capacity of the mixture = 4080 J Kg—1oC—1
Heat change=
(.2kg) x (4080Jkg-1K-1) x 6.9 oC
Heat change = 5630J
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Mass in kilograms Specific heat capacity
Temperature rise
2. How many moles of Nitric acid were reacted?:
• 100cm3 of a 1M solution of HNO3 was reacted
1 x 100 = 0.1 moles1000
Number of moles of HNO3 reacted = 0.1moles
Question: What is the heat of reaction?
Balanced equation:HNO3 + NaOH NaNO3 + H2O1 1 1 1
0.1 moles of HNO3 = 5630J of heat1 mole of HNO3 = 5630 X10 = 56300
The heat of reaction = - 56300J (or – 56.3kJ)The negative sign is because the reaction is
exothermic ( heat is given out – the temperature went up!)
Q261. Calculations for the heat of reaction
• A student carried out an experiment to measure the heat of reaction (neutralisation) of nitric acid by sodium hydroxide in a container made of plastic of negligible heat capacity. She used 100 cm3 of 1.0 M nitric acid and 100 cm3 of 1.0 M sodium hydroxide. The initial temperature of the solutions was 15.6 oC and the final temperature of the solution was 22.4 oC.
Given that specific heat capacity of the solution is 4080 J Kg—1 K—1, calculate the heat of reaction.
• (Assume that the density of the solution is 1 g cm—3)
Heat change=
(.2kg) x (4080Jkg-1K-1) x 6.8 oC
Heat change = 5548.8J
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Mass in kilograms Specific heat capacity
Temperature rise
2. How many moles of Nitric acid were reacted?:
• 100cm3 of a 1M solution of HNO3 was reacted
• 1000cm3 of solution = 1 mole in it.• 100cm3 of solution = x moles
(1) = x10
Number of moles of HNO3 reacted = 0.1moles
Question: What is the heat of reaction?
Balanced equation:HNO3 + NaOH NaNO3 + H2O1 1 1 1
0.1 moles of HNO3 = 5548J of heat1 mole of HNO3 = 5548 x 10 = 55480kJ
The heat of reaction = -55480kJ (or – 55.48kJ)The negative sign is because the reaction is
exothermic ( heat is given out – the temperature went up!)
Q262. Calculations for the heat of reaction
• A student mixed 250 cm3 of 0.5 M HCl with an equal volume of 0.5 M NaOH in a plastic container. The original temperature of both solutions was 14.8 oC and the final temperature was 18.2 oC. Calculate the heat of reaction (neutralisation) of hydrochloric acid and sodium hydroxide.
• Assume that the density of the final solution is 1 g cm—3 and its specific heat capacity is 4060 J Kg—1 K—1.
Heat change=
(.5kg) x (4060Jkg-1K-1) x 3.4 oC
Heat change = 6902J
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Mass in kilograms Specific heat capacity
Temperature rise
2. How many moles of Hydrochloric acid were reacted?:
• 250cm3 of a 0.5M solution of HCl was reacted
1000cm3 of solution = 0.5 mole in it.250cm3 of solution = (0.5/ 1000) x 250 = 0.125moles
Number of moles of HCl reacted = 0.125moles
Question: What is the heat of reaction?
Balanced equation:HCl + NaOH NaCl + H2O1 1 1 1
0.125 moles of HCl = 6902J of heat1 mole of HNO3 = 6902 x 8 = 55216
The heat of reaction = -55216J (or – 55.216kJ)The negative sign is because the reaction is
exothermic ( heat is given out – the temperature went up!)
Q266
(d) In the experiment 50 cm3 of 1 M hydrochloric acid (HCl) and 50 cm3 of 1M sodium hydroxide (NaOH) were mixed. The temperature rise was recorded as 6.8 K. Assuming the densities and heat capacities of both solutions are the same as that of water, calculate the heat produced by the reaction.
• [Density of water is 1g /cm3 specific heat capacity of water is 4.2 kJ kg―1 K—1.]
(e) How many moles of hydrochloric acid were used in the experiment?
Calculate the heat of reaction (ΔH) when 1 mole of each solution is used.
Heat change=
(.1kg) x (4.2kJkg-1K-1) x 6.8 K
Heat change = 2.856J
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Mass in kilograms Specific heat capacity
Temperature rise
Q266
(d) In the experiment 50 cm3 of 1 M hydrochloric acid (HCl) and 50 cm3 of 1M sodium hydroxide (NaOH) were mixed. The temperature rise was recorded as 6.8 K. Assuming the densities and heat capacities of both solutions are the same as that of water, calculate the heat produced by the reaction.
• [Density of water is 1g /cm3 specific heat capacity of water is 4.2 kJ kg―1 K—1.]
(e) How many moles of hydrochloric acid were used in the experiment?
Calculate the heat of reaction (ΔH) when 1 mole of each solution is used.
How many moles of Hydrochloric acid were reacted?:
• 50cm3 of a 1M solution of HCl was reacted
• 1000cm3 of solution = 1 mole in it.• 50cm3 of solution = x moles
(50)(1) = x(1000)
(1) x (50) = x1000
Number of moles of HCl reacted = 0.05moles
Q266
(d) In the experiment 50 cm3 of 1 M hydrochloric acid (HCl) and 50 cm3 of 1M sodium hydroxide (NaOH) were mixed. The temperature rise was recorded as 6.8 K. Assuming the densities and heat capacities of both solutions are the same as that of water, calculate the heat produced by the reaction.
• [Density of water is 1g /cm3 specific heat capacity of water is 4.2 kJ kg―1 K—1.]
(e) How many moles of hydrochloric acid were used in the experiment?
Calculate the heat of reaction (ΔH) when 1 mole of each solution is used.
Question: What is the heat of reaction?
Balanced equation:HCl + NaOH NaCl + H2O1 1 1 1
0.05 moles of HCl = 2.856kJ of heat1 mole of HNO3 = 2.856 /20 = 57.12 J of heat
The heat of reaction = -57.12kJ The negative sign is because the reaction is
exothermic ( heat is given out – the temperature went up!)
• Q267• (f) Calculate the number of moles of acid neutralised in this
experiment.• In an experiment to measure the heat of reaction for the reaction
between sodium hydroxide with hydrochloric acid, a student added 50 cm3 of 1.0 M HCl solution to the same volume of 1.0 M NaOH solution in a polystyrene foam cup.Taking the total heat capacity of the reaction mixture used in this experiment as 420 J K–1, calculate the heat released in the experiment if a temperature rise of 6.7 ºC was recorded.
• Hence calculate the heat of reaction for NaOH + HCl → NaCl + H2O
How many moles of Hydrochloric acid were reacted?:
• 50cm3 of a 1M solution of HCl was reacted
• 1000cm3 of solution = 1 mole in it.• 50cm3 of solution = (( 1/1000) x 50) = 0.05 moles
Number of moles of HCl reacted = 0.05moles
• Q267• (f) Calculate the number of moles of acid neutralised in this
experiment.• In an experiment to measure the heat of reaction for the reaction
between sodium hydroxide with hydrochloric acid, a student added 50 cm3 of 1.0 M HCl solution to the same volume of 1.0 M NaOH solution in a polystyrene foam cup.Taking the total heat capacity of the reaction mixture used in this experiment as 420 J K–1, calculate the heat released in the experiment if a temperature rise of 6.7 ºC was recorded.
• Hence calculate the heat of reaction for NaOH + HCl → NaCl + H2O
Heat change=
(.1kg) x (420Jkg-1K-1) x 6.7 oC
Heat change = 281.4J
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Mass in kilograms Specific heat capacity
Temperature rise
Question: What is the heat of reaction?
Balanced equation:HCl + NaOH NaCl + H2O1 1 1 1
0.05 moles of HCl = 281.4J of heat1 mole of HCl = (281.4 x 20) = 5628 J of heat
The heat of reaction = -5628J ( - 5.628kJ)The negative sign is because the reaction is
exothermic ( heat is given out – the temperature went up!)
Bond energy
Breaking chemical bonds
• Breaking chemical bonds requires energy – is an endothermic process.
Heat taken in
Energy needed to overcome the
bonding between the atoms
En
erg
y in
ch
emic
als
Energy needed
Activation Energy.
– Before new bonds can be formed we need to break some existing chemical bonds.
– This requires an energy input –known as the activation energy (Eactivation energy (Eaa or E Eactact))
– Activation energy is the energy input into a reaction to Activation energy is the energy input into a reaction to allow chemical bonds to be broken.allow chemical bonds to be broken.
Making chemical bonds
• Energy will be given out in an exothermic process when bonds are formed.
Heat given out
Energy given out as bonds form between
atoms
En
erg
y in
ch
emic
als
Energy given out
Bond energy
This is the energy needed to break 1 mole of covalent bonds
or
The energy released when 1 mole of covalent bonds are made
Changes to chemical bonds
• Again some existing bonds are broken (endothermic)
Energy taken in as old bonds break
• And new bonds are formed (exothermic)
En
erg
y in
ch
emic
als
reactants
products
Energy given out as new bonds form
H
Overall exothermic – in this case
• The formation of nitrogen (IV) oxide (formula NO2) from reaction of nitrogen with oxygen in car engines has a H value of +33.2kJ per mol of nitrogen oxide.
1. Write a word equation for the reaction.2. Write a chemical equation for the reaction.3. Is H positive or negative?4. Is the reaction exothermic or endothermic?5. Draw an simple energy diagram for the reaction
(showing bond breaking and forming.) 6. Which involves the biggest energy change: bond
breaking or bond forming?
Activity
1. Nitrogen + oxygen nitrogen(IV)oxide2. N2 + 2O2 2NO2.3. H positive (+33.2kJ/mol).4. The reaction is endothermic.5. Energy diagram 6. Bond breaking involves the biggest energy change.
Answer
Ene
rgy
/ kJ
)
Progress of reaction
reactants
products
H= -
Activation Energy and Exothermic Reactions
Ea= +Activation energy
Ene
rgy
/ kJ
)
Progress of reaction
reactants
products
H=+
Ea= +Activation energy
Activation Energy and Endothermic Reactions
Copy the energy diagram and use it to help you explain why garages can store petrol safely but always have notices about not smoking near the petrol pumps.
Ene
rgy
/ kJ
)
Progress of reaction
Petrol +
oxygen
Carbon dioxide + water
ActivationEnergy
Activity
The reaction is exothermic but requires the Activation energy to be provided before the reaction can get underway.This is necessary to break some of the bonds in the oxygen or petrol before new bonds can start forming.
Ene
rgy
/ kJ
)
Progress of reaction
Petrol +
oxygen
Carbon dioxide + water
ActivationEnergy
Answer
• This is an exothermic reaction
Bond Forming
BondBreaking
Progress of reaction
En
erg
y in
ch
emic
als
OO
OO
H
CH
HH
O OOO
C H H H H
O C OO
O
H H
H H
H
Burning Methane
Heat of combustion
• The heat change that occurs when 1 mole of substance is burned in excess oxygen
Uses of heat generated from burning substances
Transportation
Generating electricity
A bomb calorimeter
• Use : measuring the energy contents of fuels or foods
Kilogram Calorific value Amount of heat generated when 1 kilogram
of fuel when it is completely burned.
Fuels Gross calorific value/ MJ kg−1
Ethanol 30
General purpose coal (5–10% water) 32–42
Peat (20% water) 16
Diesel fuel 46
Gas oil 46
Heavy fuel oil 43
Kerosine 47
Petrol 44.8–46.9
Wood (15% water) 16
Natural gas `54
Hydrogen 141.9
Heat of formation • The heat change when 1 mole of a
substance is formed from its elements in their standard states
C(S) + 2H2(g) = CH4(g) ∆H = -74.9kJmol-1
H2(g) + S(S) +2O2(g)= H2SO4(g) ∆H = -811kJmol-1
O2(g)= O2(g) ∆H = 0kJmol-1 The heat of formation of any element is zero
Hess’s Law• The heat change for a
reaction is the same whether it takes place in one step or in a series of steps
• This rule can be used for calculations using Hess’s Law
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
An exampleWhat is the heat of reaction of
SO2 + 1/2O2 SO3?
The heat formation of SO3 and SO2 are -395 kJmol-1 and -297kJmol-1 respectively
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
∑ ∆Hf products = 1 ( -395) = - 395 kJmol-1
∑ ∆Hf reactants = 1 ( - 297 ) + 1/2 (0) = -297∆Hr = ∑ ∆Hp - ∑ ∆Hr
∆Hr = (-395) – (-297) = - 98 kJmol-1 heat of reaction for the given equation
Q270. Using Hess’s Law• The combustion of liquid benzene is described by the following equation
2C6H6 + 15O2 12CO2+ 6H20
• Given that the heats of formation of carbon dioxide gas, liquid water and liquid benzene are -394,-286 and 49kJmol-1 respectively, calculate the heat of combustion of liquid benzene
• ∑ ∆Hf products = 12 ( -394) + 6 (-286) = -6444kJmol-1
• ∑ ∆Hf reactants = 2 ( 49) + 14 (0) = 98
• ∆Hr = ∑ ∆Hp - ∑ ∆Hr
• ∆Hr = -6444 – = - 6542kJmol-1 heat of reaction for the given equation ( for 2 moles of benzene)
• Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of benzene is ( - 6542/ 2) = - 3271kJmol -1
∆Hr = ∑ ∆Hp - ∑ ∆Hr ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Q271. Using Hess’s Law• The combustion of butane is described by the following equation
2C4H10 + 13O2 8CO2+ 10H20
• Given that the heats of formation of butane, carbon dioxide gas, liquid water are -125,-394 and -286kJmol -1 respectively, calculate the heat of combustion of butane
• ∑ ∆Hf products = 8( -394) + 10 (-286) = -6012kJmol-1
• ∑ ∆Hf reactants = 2 ( -125) + 13 (0) = -250kJmol-1
• ∆Hr = ∑ ∆Hp - ∑ ∆Hr
• ∆Hr = - 6012 – (-250) = - 5762kJmol-1 heat of reaction for the given equation ( for 2 moles of butane)
• Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of butane is (5762 / 2) = - 2881kJmol -1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Q272. Using Hess’s Law• The combustion of cyclohexane is described by the following equation
C6H12 + 9O2 6CO2+ 6H20
• Given that the heats of formation of cyclohexane, carbon dioxide gas, liquid water are -156, -394,-286 kJmol -1 respectively, calculate the heat of combustion of cyclohexane
• ∑ ∆Hf products = 6( -394) + 6 (-286) = -4080kJmol-1
• ∑ ∆Hf reactants = 1( -156) + 9(0) = -156 kJmol-1
• ∆Hr = ∑ ∆Hp - ∑ ∆Hr
• ∆Hr = -4080 – (-156) = -3924 kJmol-1 heat of reaction for the given equation ( for 1 mole of cyclohexane)
• Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of cyclohexane is - 3924 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Q273. Using Hess’s LawPropane may be used in gas cyclinders for cooking appliances
Propane burns according to the following equatioon:
C3H8 + 5O2 3CO2+ 4H20
(i) Given that the heats of formation of propane, carbon dioxide gas, liquid water are -104, -394,-286 kJmol -1 respectively, calculate the heat of combustion of propane
• ∑ ∆Hp = 3 ( -394) + 4 (-286) = -2326 kJmol-1
• ∑ ∆Hr = 1( -104) + 5(0) = -104
• ∆Hr = ∑ ∆Hp - ∑ ∆Hr
• ∆Hr = - 2326 – (-104) = - 2222kJmol-1 heat of reaction for the given equation ( for 1 mole of propane)
• Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of propane is - 2222kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• If 500kJ of energy are needed to boil a kettle of water what mass of propane gas must be burned to generate this mass of heat? Give your answer to the nearest gram
x moles of propane = 500kJ of heat released when combusted
1 mole of propane = 2222Kj of heat released when combusted
500 x1 = x
2222
0.225022502 = x
0.225 = number of moles of propane needed to release 500 kJ of heat when combusted
Find number of moles needed Find the number of grams needed
0.225 moles of propane , how many grams?
(44)(0.225) = 9.901
Answer :9.901 g of propane would generate 5000Kj of heat
• If 500kJ of energy are needed to boil a kettle of water what mass of propane gas must be burned to generate this mass of heat? Give your answer to the nearest gram
Find number of moles needed Find the number of grams needed
X RMM
Q274. Using Hess’s LawI. Write a balanced equation for the combustion of ethanol C2H5OH.
II. Given that the heats of formation of ethanol, carbon dioxide and water are -278, -394,-286 kJmol-1 respectively, calculate the heat of combustion of ethanol.
i. C2H5OH + 3O2 2CO2+ 3H20
• ii)• ∑ ∆Hf products = 2( -394) + 3(-286) = -1646kJmol-1
• ∑ ∆Hf reactants = 1( -278) + 3(0) = -278 kJmol-1
• ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• ∆Hr = -1646 – (-278) = -1368 kJmol-1 heat of reaction for the given equation ( for 1 mole of ethanol)
• Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of ethanol is - 1368 kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Q275. Using Hess’s LawI. Write a balanced equation for the combustion of methanol CH3OH.
II. Given that the heats of formation of ethanol, carbon dioxide and water are -239, -394,-286 kJmol-1 respectively, calculate the heat of combustion of methanol.
i. CH3OH + 1½O2 1CO2+ 2H20
• ii)• ∑ ∆Hf products = 1( -394) + 2(-286) = -966 kJmol-1
• ∑ ∆Hf reactants = 1( -239) + 1.5(0) = -239 kJmol-1
• ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• ∆Hr = -966 – (-239) = - 727 kJmol-1 heat of reaction for the given equation ( for 1 mole of methanol)
• Heat of combustion is heat change when 1 mole of a substance completely reacts in excess oxygen so heat of combustion of methanol is - 727 kJmol -1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Q276b) Using Hess’s Law• The combustion of methane is described by the following balanced equation
CH4 + 2O2 CO2+ 2H20 ∆H = -890.4kJmol-1
• Given that the heats of formation of carbon dioxide gas, liquid water are -394,-286 kJmol -1 respectively, calculate the heat of formation of methane
• ∆Hr = -890.4kJmol-1
• ∑ ∆Hf products = 1( -394) + 2 (-286) = -966kJmol-1
• ∑ ∆Hf reactants = 1( x ) + 2(0) = x kJmol-1
• ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• -890.4kJmol-1 = -966kJmol-1 – (x kJmol-1 )
• - 75.6kJmol-1 = x kJmol-1 • Heat of formation of methane is - 75.6kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of methane
Q277) Using Hess’s Law• The combustion of ethyne is described by the following balanced equation
C2H2 + 2½O2 2CO2+ H20 ∆H = -1299kJmol-1
• Given that the heats of formation of carbon dioxide gas, liquid water are -394,-286 kJmol -1 respectively, calculate the heat of formation of ethyne
• ∆Hr = -1299kJmol-1
• ∑ ∆Hf products = 2( -394) + 1(-286) = -1074kJmol-1
• ∑ ∆Hf reactants = 1( x ) + 2½ (0) = x kJmol-1
• ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• -1299kJmol-1 = -1074kJmol-1 – (x kJmol-1 )
• - 225kJmol-1 =- x kJmol-1 • Heat of formation of ethyne is 225kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of ethyne
Q278) Using Hess’s Law• The combustion of propane is described by the following balanced equation
C3H8 + 5O2 3CO2+ 4H20 ∆H = -2222kJmol-1
• Given that the heats of formation of carbon dioxide gas, liquid water are -394,-286 kJmol -1 respectively, calculate the heat of formation of propane
• ∆Hr = -2222kJmol-1
• ∑ ∆Hf products = 3( -394) + 4(-286) = -2326kJmol-1
• ∑ ∆Hf reactants = 1( x ) + 5 (0) = x kJmol-1
• ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• -2222kJmol-1 = -2326kJmol-1 – (x kJmol-1 )
• - 104kJmol-1 = x kJmol-1 • Heat of formation of propane is - 104kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of propane
Q279) Using Hess’s Law• Write a balanced equation for the combustion of ethanol
i. C2H5OH + 3O2 2CO2+ 3H20 ∆H = -1300kJmol-1
• Given that the heats of formation of carbon dioxide gas, liquid water are -394,
-286 kJmol-1 respectively, calculate the heat of formation of ethanol
• ∆Hr = -1300kJmol-1
• ∑ ∆Hf products = 2( -394) + 3 (-286) = -1646kJmol-1
• ∑ ∆Hf reactants = 1( x ) + 3(0) = x kJmol-1
• ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
• -1300kJmol-1 = -1646kJmol-1 – (x kJmol-1 )
• - 346kJmol-1 = x kJmol-1 • Heat of formation of ethanol is - 346kJmol-1
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
Let x = heat of formation of ethanol