Transcript Exothermic and endothermic reactions

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
Exothermic reactions increase in temperature.
•
Examples include:
–
–
–
Burning reactions including the
combustion of fuels.
Detonation of explosives.
Reaction of acids with metals.
Magnesium
reacting with
acid
Thermit reaction
Exothermic Reactions
•
Magnesium + Hydrochloric acid
25o C
45o C
magnesium
Gets hot
Hydrochloric
acid
Heat
energy
given
out
Exothermic Reactions
45
25oo C
Reactants convert chemical energy to heat
energy.
The temperature rises.
Exothermic Reactions
•
Almost immediately the hot reaction products start
to lose heat to the surroundings and eventually
they return to room temperature.
25o C
45
Energy Level Diagram for an
Exothermic Reaction
reactants
Reactants have more chemical
energy.
Energy / kJ)
Some of this is lost as heat
which spreads out into the
room.
products
Progress of reaction (time)
Products now have less
chemical energy than
reactants.
Exothermic Reaction - Definition
Exothermic reactions give
out energy.
There is a temperature rise
is negative.
Energy / kJ)
and H
reactants
H is negative
products
Progress of reaction
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.
Endothermic Reactions
Endothermic reactions cause a decrease in temperature.
Heat
energy
taken
in as the
mixture
returns
back to
room
temp.
Ammonium
nitrate
Cools
Water
Starts 25°C
Cools to 5°C
Returns to 25°C
Endothermic Reactions
•
The cold reaction products start to gain heat
from the surroundings and eventually return
to room temperature.
o ooCC
25
5
The reactants gain energy.
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 Reaction Definition
Endothermic reactions take in
energy. There is a temperature
is positive.
products
Energy / kJ
drop and H
H=+
reactants
Progress of reaction
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
vapour is very irritating to lungs.
Sodium hydroxide
wear eye protection.
: Very corrosive to eyes and skin, and its
: Caustic, harmful to skin and especially to eyes. Always
Measuring the heat of reaction of
hydrochloric acid and sodium hydroxide
The equation for
the reaction is
HCl + NaOH →
NaCl + H2O
A polystyrene cup
is used as it is an insulator of heat – it doesn’t let the heat
escape ( has negligible heat capacity)
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
=
o
Temperature of NaOH solution before mixing
=
o
Highest temperature reached after mixing
=
o
Temperature rise
=
o
Number of moles of acid used
=
Number of moles of base used
=
C
C
C
C
Calculations –
What is the heat change in the reaction carried our:
Heat change = mc ∆ T
Temperature rise
Mass in kilograms
Specific heat
capacity
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
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Temperature rise
Mass in kilograms
Heat change=
(.2kg) x (4080Jkg-1K-1) x 6.9 oC
Heat change = 5630J
Specific heat
capacity
2. How many moles of Nitric acid were reacted?:
• 100cm3 of a 1M solution of HNO3 was reacted
1
x 100 = 0.1 moles
1000
Number of moles of HNO3 reacted = 0.1moles
Question: What is the heat of reaction?
Balanced equation:
HNO3 + NaOH
1
1
NaNO3 + H2O
1
1
0.1 moles of HNO3 = 5630J of heat
1 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)
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Temperature rise
Mass in kilograms
Heat change=
(.2kg) x (4080Jkg-1K-1) x 6.8 oC
Heat change = 5548.8J
Specific heat
capacity
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) = x
10
Number of moles of HNO3 reacted = 0.1moles
Question: What is the heat of reaction?
Balanced equation:
HNO3 + NaOH
1
1
NaNO3 + H2O
1
1
0.1 moles of HNO3 = 5548J of heat
1 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.
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Temperature rise
Mass in kilograms
Heat change=
(.5kg) x (4060Jkg-1K-1) x 3.4 oC
Heat change = 6902J
Specific heat
capacity
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
1
1
NaCl + H2O
1
1
0.125 moles of HCl = 6902J of heat
1 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.
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Temperature rise
Mass in kilograms
Heat change=
(.1kg) x (4.2kJkg-1K-1) x 6.8 K
Heat change = 2.856J
Specific heat
capacity
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) = x
1000
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
1
1
NaCl + H2O
1
1
0.05 moles of HCl = 2.856kJ of heat
1 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
1. What is the heat change in the reaction:
Heat change = mc ∆ T
Temperature rise
Mass in kilograms
Heat change=
(.1kg) x (420Jkg-1K-1) x 6.7 oC
Heat change = 281.4J
Specific heat
capacity
Question: What is the heat of reaction?
Balanced equation:
HCl + NaOH
1
1
NaCl + H2O
1
1
0.05 moles of HCl = 281.4J of heat
1 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.
Energy needed to
overcome the
bonding between
the atoms
Energy in chemicals
Heat taken in
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 (Ea or Eact)
–
Activation energy is the energy input into a reaction to
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
Energy in chemicals
•
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
reactants
Overall
exothermic –
in this case
Energy given out
as new bonds
form
H
Energy in chemicals
• And new bonds are formed (exothermic)
products
Activity
•
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.
2.
3.
4.
5.
Write a word equation for the reaction.
Write a chemical equation for the reaction.
Is H positive or negative?
Is the reaction exothermic or endothermic?
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?
Answer
1.
2.
3.
4.
5.
6.
Nitrogen + oxygen  nitrogen(IV)oxide
N2 + 2O2  2NO2.
H positive (+33.2kJ/mol).
The reaction is endothermic.
Energy diagram
Bond breaking involves the biggest energy change.
Activation Energy and Exothermic
Reactions
Ea= +
Energy / kJ)
reactants
Activation energy
H= products
Progress of reaction
Activation Energy and Endothermic
Reactions
Ea= +
Energy / kJ)
Activation
energy
products
H=+
reactants
Progress of reaction
Activity
ActivationE
nergy
Energy / kJ)
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.
Petrol
+
oxygen
Carbon dioxide +
water
Progress of reaction
Answer
ActivationE
nergy
Energy / kJ)
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.
Petrol
+
oxygen
Carbon dioxide +
water
Progress of reaction
Burning Methane
•
This is an exothermic reaction
H H H H
O
Bond
Breaking
H
C
H
H
H
O
O
O
Bond
Forming
O
O
O
O
O
H
Energy in chemicals
C
H
O
Progress of reaction
C
O
H
O
H
H
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
Hydrogen
`54
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 example
What 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
∆Hr = ∑ ∆H∆H
∆Hf rreactants
f products
r = ∑ ∆Hp- ∑
- ∑∆H
•
•
•
•
∑ ∆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
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
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
∑ ∆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
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
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
∑ ∆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
Q273. Using Hess’s Law
Propane 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
∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
∑ ∆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
• 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 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
• 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
0.225 moles of propane , how many grams?
(44)(0.225) = 9.901
Answer :9.901 g of propane would generate
5000Kj of heat
Q274. Using Hess’s Law
I. 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) ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
•
∑ ∆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
Q275. Using Hess’s Law
I. 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) ∆Hr = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
•
∑ ∆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
•
•
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 = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
•
•
•
∆Hr = -890.4kJmol-1
∑ ∆Hf products = 1( -394) + 2 (-286) = -966kJmol-1
Let x = heat of
∑ ∆Hf reactants = 1( x ) + 2(0) = x kJmol-1
formation of methane
∆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
•
•
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 = ∑ ∆Hf products - ∑ ∆Hf reactants
•
•
•
•
•
•
•
∆Hr = -1299kJmol-1
∑ ∆Hf products = 2( -394) + 1(-286) = -1074kJmol-1
∑ ∆Hf reactants = 1( x ) + 2½ (0) = x kJmol-1 Let x = heat of
formation of ethyne
∆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
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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 = ∑ ∆Hf products - ∑ ∆Hf reactants
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∆Hr = -2222kJmol-1
∑ ∆Hf products = 3( -394) + 4(-286) = -2326kJmol-1
Let x = heat of
∑ ∆Hf reactants = 1( x ) + 5 (0) = x kJmol-1
formation of propane
∆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
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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 = ∑ ∆Hf products - ∑ ∆Hf reactants
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∆Hr = -1300kJmol-1
∑ ∆Hf products = 2( -394) + 3 (-286) = -1646kJmol-1
Let x = heat of
∑ ∆Hf reactants = 1( x ) + 3(0) = x kJmol-1
formation of ethanol
∆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