calorimetry
DESCRIPTION
Calorimetry. The enthalpy change associated with a chemical reaction or process and the specific heat of a substance can be measured experimentally using calorimetry . The experimental measurement of the amount of heat gained or lost Determined by measuring the temperature change that occurs. - PowerPoint PPT PresentationTRANSCRIPT
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Calorimetry
The enthalpy change associated with a chemical reaction or process and the specific heat of a substance can be measured experimentally using calorimetry.The experimental measurement of
the amount of heat gained or lost Determined by measuring the
temperature change that occurs.
Calorimeter:An instrument used to measure the
heat gained or lost
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CalorimetryTwo types of calorimetry are commonly
used:
Constant-pressure CalorimetrySpecific heatHrxn or Hsoln
Bomb Calorimetry (Constant Volume Calorimetry)Hcombustion
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Calorimetry In constant-pressure calorimetry,
Measurements are made in an “open” container. Reactions occur at constant pressure
Calorimeter is insulated Assume that the amount of heat
gained from or lost to the surroundings is negligible any heat gained or lost during a chemical reaction or process comes from or goes into the solution being studied.
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Calorimetry
In order to measure the specific heat of a substance, we need to know three values:qsubstance
massT
In order to measure a molar enthalpy change for a reaction, we need to know two values:qsubstance or qreactant
moles of substance or reactant
Cs = qsubst
mass x T
H = qsubst
moles subst.
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Calorimetry
In both cases, the “unknown” that must be determined experimentally is the amount of heat gained or lost by the substance being studied (i.e. qsubstance).
In constant-pressure calorimetry, qsubstance is determined indirectly by measuring the heat gained or lost by the liquid present in the calorimetry (whose specific heat is known).
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Calorimetry
We then apply the First Law of Thermodynamics: If heat is lost by the chemicals (or
object) during a reaction or process, then it must be gained by the solution (liquid) and vice versa.
The heat gained or lost by the reactants (or object) and the solution (liquid) are equal in magnitude but opposite in sign.
qobject = - qsoln or qrxn = -qsoln
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Calorimetry Calorimetry Simulation:
Measuring the specific heat of a metal
How is the experiment performed?
What kind of data is collected?
How are the data used to determine specific heat?
http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/animationsindex.htm
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Calorimetry Determining specific heat or molar
heat capacity experimentally:Heat an object with a known mass to
95-100oC.Measure initial temperature of
object.Place known mass of water into
calorimeter and measure its temperature.
Add hot object to the waterMeasure the equilibrium
temperature (i.e. final temperature) of the water and object.
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Calorimetry Types of data obtained from or used in a
calorimetry experiment often include:Sample Data
Mass of sample (object or compound)
Tinitial (sample) Tfinal (sample)
Solution Data Mass of solution Tinitial (solution) Tfinal (solution) Csoln
Often use specific heat of water if water is the solvent
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Calorimetry Using the data to find specific heat:
Calculate the amount of heat gained by the water: qwater = Cwater x mwater x Twater
Calculate the amount of heat lost by the object qobj = - qwater
Calculate specific heat of object Cs = qobj
mobj x Tobj
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CalorimetryExample: A 55.0 g piece of aluminum metal at 100.0oC was added to 51.3 g of water at 20.0oC. The equilibrium temperature of the system was 35.0oC. If the specific heat of water is 4.18 J/g.K, what is the specific heat of aluminum?
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Calorimetry
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Calorimetry This is the same procedure you will use
to calculate the specific heat of the metal sample you use in the lab.
Follow this procedure to do the calculations for Expt. 2.
You should expect a similar problem on your exam!
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Calorimetry Calorimetry Simulation:
Measuring the molar heat of solution (Hsoln)
How is the experiment performed?
What kind of data is collected?
How are the data used to determine molar heat of solution?
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Calorimetry Determining molar heat of solution
(Hsoln per mole of solute) from calorimetry data.Place known mass of water into
calorimeter.Measure the initial temperature of
the water.Add known mass of solute to water.Record equilibrium temperature of
the resulting solution.
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Calorimetry Types of data collected when
measuring molar heat of solution:Mass of water Initial temperature of waterMass of soluteFinal temperature of solution
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Calorimetry Determining Hsoln per mole of solute
from calorimetry data.
Assume Csoln = Cwater = 4.18 J/g-K unless told otherwise
Calculate the heat gained or lost by the entire solutionqsoln = Csoln x mass of solution x Tsoln
Mass of solution = mass of H2O + mass of solute
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Calorimetry
Calculate heat gained or lost by the solute (salt)qsolute = - qsoln
Calculate molar heat of solution
Hsoln per mole = qsolute
mole solute
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Calorimetry
Example: When a 2.125 g sample of solid ammonium nitrate dissolves in 30.000 g of water in a constant pressure calorimeter, the temperature drops from 22.0oC to 16.9oC. Calculate the molar Hsoln (in kJ/mole) for the dissolution process:
NH4NO3 (s) NH4+ (aq) + NO3
- (aq)
Assume that the specific heat of the solution is the same as pure water (4.18 J/g.K).
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Calorimetry
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Calorimetry You should expect a calorimetry
problem similar to this on your exam.
What would you actually observe when the dissolution process is exothermic?
What would you actually observe when the dissolution process is endothermic?
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Calorimetry
For an exothermic process, the heat produced causes the temperature of the solution to increase. (Tsoln >0)
Chemical Chemical particleparticle
Heat released by chemical particle
q
q
q
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Calorimetry
For an endothermic process, the heat gained comes from the reaction mixture and causes the temperature of the solution to decrease (Tsoln < 0)
Chemical Chemical particleparticle
Heat gained by chemical particle from the rxn mixture
q
q
q
q
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Calorimetry
Bomb CalorimetryConstant Volume
Calorimetry
Used to study combustion reactions
Measure Hcombustion
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Calorimetry
As combustion occurs,Heat is releasedHeat is absorbed by the calorimeter
and its contentsTemperature of calorimeter &
contents increases.
The change in temperature of the calorimeter and its contents can be used to determine the heat of combustion.
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Calorimetry To calculate the molar heat of combustion
from a calorimetry experiment: Calculate the heat absorbed by the
calorimeter and its contents using the heat capacity of the calorimeter:
qcal = Ccal x T (Notice that you do not need a mass term
because heat capacity has units of J/K)
Calculate the heat lost by the reactants:
qrxn = -qcal
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Calorimetry Calculate molar heat of combustion
(Hcomb per mole reactant)
Hcomb = qrxn
mole reactant
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Calorimetry
Example: When 2.00 g of methylhydrazine (CH6N2) is burned in a bomb calorimeter, the temperature of the calorimeter increases from 25.00oC to 32.25oC. If the heat capacity of the calorimeter is 7.794 kJ/oC, what is the molar heat of combustion for CH6N2?
2 CH6N2 (l) + 5 O2 (g) 2 N2 (g) + 2 CO2(g) + 6 H2O (g)
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Calorimetry
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Hess’s Law The heats of reaction (Hrxn) have been
measured and tabulated for many chemical reactions.
There are two approaches to determining the heat of reaction for a particular chemical reaction:CalorimetryUse tabulated Hrxn to calculate the
heat of reaction for another reaction of interest
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Hess’s Law The enthalpy change for a reaction or
process is a state function:Depends only on the amount of
reactants and products used/formed and on their physical state
Does not depend on how the reaction was done. one step vs. multiple steps
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Hess’s Law
Hess’s Law: If a reaction is
carried out in a series of steps, H for the overall (one-step) reaction is equal to the sum of the H’s for the individual steps.
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Hess’s Law
IMPORTANT:When applying Hess’s Law, if you need
to multiply or divide the coefficients of an equation by a number then the H must also be multiplied by the same number H depends on amount of material
H2O (g) H2O (l) H = - 44 kJ
2 H2O (g) 2 H2O (l) H = 2 (- 44 kJ) = - 88 kJ
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Hess’s Law IMPORTANT:
If a reaction has to be reversed, the magnitude of H stays the same but the sign must be reversed.
H2O (g) H2O (l) H = - 44 kJ
H2O (l) H2O (g) H = + 44 kJ
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Hess’s Law
Example: Calculate the Hrxn for the incomplete combustion of C forming CO:
C (s) + ½ O2 (g) CO (g) H = ???
given the following reactions:
C (s) + O2 (g) CO2 (g) H = - 393.5 kJ
2 CO (g) + O2 (g) 2 CO2 (g) H = - 566.0 kJ
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Hess’s Law
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Hess’s Law
Example: Calculate Hrxn for
NO (g) + O (g) NO2 (g) H = ???
using the following thermochemical equations.
NO (g) + O3 (g) NO2 (g) + O2 (g) H = -198.9 kJO3 (g) 3/2 O2 (g) H = -142.3 kJO2 (g) 2 O (g) H = 495.0
kJ
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Hess’s Law