Download Heat of reaction

Heat of reaction
• Chemical reactions involve breaking of bonds
(in reactants) and making of bonds (products).
• At the end of the reaction, energy may be
released to the surroundings or it may be
absorbed from the surroundings.
• A reaction that releases energy to its
surroundings is called an exothermic reaction.
• A reaction that absorbs energy from the
surroundings is called an endothermic
reaction.
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Effects of heats of reactions
Type of
reaction
Experimental
effect
What
happens to
the system
Sign of “q”
Endothermic
Reaction vessel
cools
Energy is
added
+
Exothermic
Reaction vessel
warms
Energy is
released
-
Ba(OH)2·8H2O(s) + 2NH4NO3(s)
C3H8(g) + O2(g)
2NH3(g) + 10H2O(l) +
Ba(NO3)2(aq) q = + 170.4 kJ
CO2(g) + H2O(l) q = - 890 kJ
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State functions
• Any property that does not depend on the history of
the system is called a state function.
• The state of a system is defined as the chemical
composition and its temperature, pressure and
volume.
• Pressure, temperature and volume are examples of
state functions.
• Changes in state functions are usually measured
and are denoted by Δ followed by whatever state
function is being measured.
• The change in temperature ΔT can be measured by:
ΔT = Tfinal - Tinitial
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Enthalpy (H) & Enthalpy change (ΔH)
• The heat absorbed or released by a system
usually depends on the conditions under which
the reaction is performed.
• Normally, reactions are performed in vessels
open to the atmosphere and hence at constant
atmospheric pressure.
• Enthalpy is an extensive property of a
substance that can be used to obtain the heat
evolved or absorbed in a chemical reaction.
• Enthalpy is a state function.
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Enthalpy of a reaction
• State functions are properties that are
independent of the previous history of the
system. They depend on the present state of
the system.
• The ΔH for a reaction at a given temperature
and pressure is obtained by:
ΔH = H(products) – H(reactants)
• The change in enthalpy in a chemical reaction
is = heat of reaction at constant pressure:
ΔH=qp
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Thermochemical Equations
• It is often useful to write the ΔH with the
balanced equation.
• A thermochemical equation is a balanced
chemical equation (including physical states)
with the ΔH written directly after the equation.
2H2(g) + O2(g)
2H2O(g)
ΔH = -483.7 kJ
2H2(g) + O2(g)
2H2O(l)
ΔH = -571.7 kJ
NaHCO3(aq) + HCl(aq)
NaCl(aq) + H2O(l) + CO2(g)
ΔH = +11.8 kJ
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Facts to Remember
• For changes that are reverse of each other the ΔH
values are numerically the same, but their signs are
opposite. For the evaporation of water ΔH = +44.0
kJ/mol, while for condensation of water vapor, ΔH = 44.0 kJ/mol.
• When a chemical equation is reversed the value of
ΔH stays the same but the sign is reversed.
N2(g) + 3H2(g)
2NH3(g) ΔH = -91.8 kJ
2NH3(g)
N2(g) + 3H2(g) ΔH = +91.8 kJ
• When a chemical equation is multiplied by any number
the value of ΔH is also multiplied by that number.
N2(g) + 3H2(g)
1/2N2(g) + 3/2H2(g)
2NH3(g) ΔH = -91.8 kJ
NH3(g) ΔH = -45.9kJ
Only when solving thermodynamics problems, can we have fractional
coefficients
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Stoichiometry and ΔH
• How much heat is produced by the combustion of
1.00 lb (454 g) of propane (C3H8)? The ΔH for the
combustion of propane is -2220 kJ mol-1.
• Convert the grams of propane to moles of propane
and then find the amount of heat produced:
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Hess’s Law
• The numerical value and the sign of the ΔH allows us
to make informed decisions.
• For several reactions a direct measurement can be
done with a calorimeter.
• Many times this is impossible or it is a time
consuming task which makes it very hard.
• Hess’s law allows us to manipulate equations for
calculating ΔH for a given reaction.
• If a reaction is the sum of two or more other reactions
then the ΔH fro the overall process must be the sum
of the constituent reactions.
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Applying Hess’s law
• What is the ΔH for the formation of methane
from solid C and hydrogen gas?
CH4(g) ΔH =?
C(s) + 2H2(g)
• Use the following information:
ΔH (kJ)
Reaction
1
2
3
C(s) + O2(g)
CO2(g)
H2(g) + 1/2O2(g)
CH4(g) + 2O2(g)
H2O(l)
CO2(g) + H2O(l)
-393.5
-285.8
-890.3
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• The 3 equations as they are written cannot be added for the
formation of CH4 from its elements.
• We need the CH4 to be a product, but it is a reactant in
equation 3. (Reverse equation 3, remember to change the
sign also)
• Equation (3’) needs 2 moles of H2O as reactants. Equation
(2) is written for only one mole of H2O. Multiply equation (2)
by 2 and multiply the ΔH value also by 2 to get:
• After these modifications we can add the 3 equations to get
the formation of CH4 from its elements.
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ΔH (kJ)
Reaction
1
C(s) + O2(g)
2’
2H2(g) + O2(g)
3’
CO2(g) + 2H2O(l)
C(s) + 2H2(g)
CO2(g)
2H2O(l)
CH4(g) + 2O2(g)
CH4(g)
-393.5
-571.6
+ 890.3
- 74.8
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Standard Enthalpies of Formation
• The enthalpy change in a reaction in which reactants in
their standard state produce products in their standard
state is called the standard enthalpy of reaction (ΔH°).
• The standard state of a substance is its pure and most
stable form at 1 atmosphere pressure and at 25°C.
• The enthalpy change for the formation of one mole of a
compound in its standard state from its elements in their
reference form and in their standard states is known as
the enthalpy of formation ΔH°f of that compound.
• The ΔH°f values (in appendix C in our book) can be used
for calculating ΔH°.
• The ΔH°f values for all elements are assigned a value =
zero.
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Using ΔH°f
• Consider the reaction between methane and chlorine:
CCl4(l) + 4HCl(g) ΔH° = ?
CH4(g) + 4Cl2(g)
• From Appendix C we can look up the values for the
ΔH°f of CH4(g), CCl4(l) and HCl(g) and then write:
ΔH°f
(kJ)
Reaction
1
C(graphite) + 2H2(g)
CH4(g)
-74.9
2
C(graphite) + 2Cl2(g)
CCl4(l)
-139
3
1/2H2(g) + 1/2Cl2(g)
HCl(g)
-92.3
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• Apply Hess’s law.
• Since the ____needs to be on the ____ and the ____
and _____ on the right, _______ equation _ and
multiply equation _ by _ to get:
ΔH°f
(kJ)
Reaction
1’
CH4(g)
C(graphite) + 2H2(g)
2
C(graphite) + 2Cl2(g)
3’
2H2(g) + 2Cl2(g)
CH4(g) + 4Cl2(g)
CCl4(l)
4HCl(g)
CCl4(l) + 4HCl(g)
The ΔH° for the reaction is ______
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Simplification
• The previous problem can be greatly
simplified by using the following formula:
ΔH o =
∑ n ΔH (products) −∑ m ΔH (reactants)
o
f
o
f
ΔH° = [ΔH°f (CCl4) + 4 ΔH°f (HCl)] – [ΔH°f (CH4)
+ 4 ΔH°f (Cl2)
ΔH° = [(-139 kJ) + 4(-92.3 kJ)] – [(-74.9 kJ) +
4(0 kJ)]
ΔH° = -433 kJ
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Another example
• Calculate the ΔH°vap of CS2 at 25°C. This is a
phase change process. The vaporization
process is given by:
CS2(l)
ΔH =
o
∑n
CS2(g)
ΔH of (products) −
∑ m ΔH (reactants)
o
f
• ΔH° =
• ΔH° =
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Another one
• NH3 is used to prepare HNO3. Oxidation of NH3 to
produce NO is the first step. What is the standard
enthalpy change for this reaction?
4NH3(g) + 5O2(g)
ΔH o =
4NO(g) + 6H2O(g)
∑ n ΔH (products) −∑ m ΔH (reactants)
o
f
o
f
• ΔH° =
• ΔH° =
• ΔH° =
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Calorimetry
• A technique for measuring the
heat evolved by a reaction.
• A weighed sample is placed in a
dish that is encased in a “bomb”.
• The bomb is then placed in a
well-insulated container.
• The bomb is filled with pure O2
and then the sample is ignited.
• The heat released by this
burning raises the temperature
of the water around it.
• System = Sample + O2
• Surroundings = Bomb + water
• Heat lost by system = heat
qreaction = -(qwater + qbomb)
gained by surroundings.
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ΔH by Calorimetry
• Octane is the primary component of gasoline.
Suppose a 1.00 g sample of Octane (C8H18) is
burned in a bomb calorimeter containing 1.20 kg of
water. The temperature of the water and the bomb
rise from 25.00°C to 33.50°C. If the heat capacity of
the bomb = 837 J/K, calculate the amount heat
transferred by the combustion of 1.00 g of octane.
Also calculate the amount of heat transferred per
mole.
C8H18(l) + 25/2 O2(g)
8CO2(g) + 9H2O(l)
qwater = (sp.heat of water) x (mass) x (ΔT)
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• The heat released by the reaction also heats up the
bomb and this is calculated by:
• The total heat released by the reaction is calculated
by:
• This shows that the amount of heat released by the
combustion of 1.00 g of Octane.
• The amount of heat released by the combustion of 1
mole of Octane can be calculated by:
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7
Coffee-cup Calorimetry
• A bomb calorimeter is sometimes inconvenient
to use, instead a coffee-cup calorimeter is
used.
• A coffee-cup calorimeter is simple,
inexpensive and operates at constant
(atmospheric) pressure.
• The heat of reaction measured in a coffee-cup
calorimeter is the measure of the reaction
enthalpy (qp = ΔH).
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Coffee-cup Calorimetry
• 0.500g of Magnesium chips are placed in a coffee
cup calorimeter containing 100.0 ml of 0.1M HCl.
The temperature of the solution increases from
22.2°C to 44.8°C. What is the ΔH for this reaction
per mole of Mg? Assume that the heat capacity of
the solution is 4.20 J/gK and the density of HCl is
1.00 g/ml
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
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