Download Chapter 5: thermochemstry

5/14/2010
Chapter 5: thermochemstry
tonight’s goals
Energy and Enthalpy Review
Enthalpies of Reaction
Calorimetry
Hess’ Law
Enthalpies of Formation
Internal Energy: E
•E = The sum of all kinetic and potential energies of all
components of the system.
• E after a reaction/process is: E = Efinal − Einitial
– If E is positive, energy was gained by the system
– If E is negative, energy was lost to the surroundings
• When energy is exchanged between the system and
surroundings (E), it is exchanged as either heat (q) or
work (w): E = q + w
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Work = force x distance
• In an open container in the lab, the only work
done is by a gas pushing on the surroundings
or by the surroundings pushing on a gas
• In a closed system with a movable piston, that work
can be measured by the rising and falling of a piston
• Gas pushes piston up: work is done by
the system; energy goes to surroundings
• Piston is pushed down: work is done on
the system, energy is gained by system
• This is called PV work (Pressure and Volume are changing)
• When pressure is constant (our assumption); w = -PV
Enthalpy
• Enthalpy (H) is defined as internal energy + pressure x
volume (work): H = E + PV
• When the system changes at constant pressure, the
change in enthalpy is H = E + PV
• Since E = q + w and w = -PV, we can substitute these
into the enthalpy expression:
H = E + PV = (q+w) − w
H = q (at constant pressure)
• Thus, enthalpy is a measure of the heat flow between a
system and the surroundings
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Enthalpy of Reaction, H
The change in enthalpy, H, after a reaction is the enthalpy of
the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants
H is called the enthalpy or heat of reaction
H depends on the states of products and reactants.
H is numerically the same, but opposite sign for reverse reaction
Calorimetry: measurement of
heat flow
We can’t know exact enthalpy of reactants and products, but
we can measure H at constant pressure using calorimetry, the
measurement of heat flow.
If we perform a reaction in aqueous solution, the
molecules/atoms involved in the reaction are the
system and the water/container are the surroundings
We can measure T of the surroundings easily
T = Tfinal– Tinitial
Using T and information on the mass and specific
heat capacities of the water and container, we can
calculate qsurroundings and thus, H of the reaction
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Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a
substance by 1 K (1C) is its heat capacity.
Specific heat capacity (or specific heat) is the energy required
to raise the temperature of 1 g of a substance by 1 K.
Other
Substances
Sand
0.835
Soil
0.800
Wood
1.7 (1.2 to 2.3)
Specific Heat capacity
Look at the units for specific heat: J/g-K
Specific heat =
heat transferred (J)
Mass (m)  temp change (T)
Cs =
q
m  T
or
q = Cs m
T
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Back to the coffee cup
By carrying out a reaction in aqueous
solution in a calorimeter, we can estimate
the heat change for the system by
measuring ΔT for the contents &
calorimeter (the surroundings).
Δqsys = - Δqsurr = ΔHrxn
Δqsurr
= Δqcontents + Δqcup
Δqcontents = Cs (water) mcontents
Δqcup
= C (cup) ΔT
ΔT
For today’s lab; Cs (water) = 4.184 J/g-oC
C (cup) = 50.0 J/g-oC
Measuring ΔH Using a Coffee-Cup Calorimeter
A student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup
calorimeter . The temperature of the solution increases from 21.0 C to 27.5 C.
Calculate the ΔH for the reaction in kJ/mol HCl, assuming:
the calorimeter loses only a negligible quantity of heat,
the total volume of the solution is 100 mL
the solution density is 1.0 g/mL and its Cs is 4.18 J/g-K.
Solution
Analyze: Write reaction equation: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
We need to calculate the heat produced per mole of HCl, given the temperature
increase of the solution, the number of moles of HCl and NaOH involved, and the
density and specific heat of the solution.
Plan: The total heat produced can be calculated using qrxn = -Cs x m x Δ T. The #
moles of HCl can be calculated from the volume and M of the solution and used to
determine the heat produced per mol HCl.
Solve:
V of solution is 100 mL, mass is:
The temperature change is:
We can calculate qrxn
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Measuring ΔH Using a Coffee-Cup Calorimeter (con’t)
To find ΔH on a molar basis, we need to calculate # mol HCl:
Volume x M (mol/L) = # of mol HCl
0.050 L x 1 mol/L = 0.050 mol HCl
ΔHrxn = -2.7 kJ, so ΔHrxn/mol HCl = -2.7 kJ/0.050 M HCl
= -54 kJ/mol HCl
Reality check: The sign of ΔHrxn is negative, which it should be for
an exothermic reaction
Another cup of coffee?
When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are
mixed in a constant-pressure calorimeter, the temperature of the
mixture increases from 22.30 C to 23.11 C. The temperature
increase is caused by the following reaction:.
AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
Calculate ΔH for this reaction in kJ/mol AgNO3, assuming the
solution has a mass of 100.0 g and a specific heat of 4.18 J/g C.
ΔHrxn = qrxn = -Cs x m x ΔT
= -4.18 J/g C x 100.0 g x 0.81 C
= -339 J
# mol AgNO3 = 0.050 L x 0.100 M AgNO3 = 0.0050 mol AgNO3
ΔHrxn in kJ/mol = 339 J/0.0050 mol AgNO3 x 1kJ/1000 J = -68 kJ/mol
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hess’ law
Hess’s law states that “If
a reaction is carried out in
a series of steps, H for
the overall reaction will
be equal to the sum of the
enthalpy changes for the
individual steps.”
Intermediate step is formation
of carbon monoxide (incomplete
combustion), followed by further
combustion to carbon dioxide.
More on hess’ law
• Hess' law allows ΔH rxn to be calculated even when it can’t be
measured directly.
• To do this, we perform arithmetic operations on chemical
equations and known ΔH values.
– Chemical equations may be multiplied or divided by a
whole number.
– When an equation is multiplied by a constant, so is ΔH
– If an equation is reversed, ΔH for reaction is reversed (-ΔH)
• Addition of chemical equations leads to a net equation
– If ΔH for each equation is added, the result will be the ΔH
for the net equation.
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let’s try one
Use the thermochemical equations shown below to determine
the enthalpy for the reaction:
2NH3(g)→N2(g) + 3H2(g)
CH2O(g) + N2(g) + 3H2(g) → N2H4(l) + CH4O(l)
N2H4(l) + H2(g) → 2NH3(g)
CH2O(g) + H2(g) → CH4O(l)
ΔH=18.5 KJ
Δ H=-9.0 KJ
Δ H=32.5 KJ
N2H4(l) + CH4O (l) → CH2O(g) + N2(g) + 3H2(g)
2NH3(g) → N2H4(l) + H2(g)
CH2O(g) + H2(g) → CH4O(l)
2NH3(g)→N2(g) + 3H2(g)
Δ H= -18.5 KJ
Δ H=+9.0 KJ
Δ H= 32.5 KJ
Δ H= 23.0 KJ
For more fun like this: http://proton.csudh.edu/lecture_help/lechelp.html
Scroll down the menu for the Hess Law DP (drill and practice)
Try some more Hess Math
Given:
B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g)
H2O (l) → H2O (g)
H2 (g) + (1/2) O2 (g) → H2O (l)
2 B (s) + 3 H2 (g) → B2H6 (g)
ΔH = 2035 kJ/mol
ΔH = 44 kJ/mol
ΔH = -286 kJ/mol
ΔH = 36 kJ/mol
Find the ΔHf of:
2 B (s) + (3/2) O2 (g) → B2O3 (s)
First we have to adjust the equations:
(3/2) O2
B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g)
3 H2O (g) → 3 H2O (l)
3x
3 H2O (l) → 3 H2 (g) + (3/2) O2 (g) 3x
2 B (s) + 3 H2 (g) → B2H6 (g)
ΔH = -2035 J/mol (reversed)
ΔH = -132 kJ/mol (reversed)
ΔH = 858 kJ/mol (reversed)
ΔH = 36 kJ/mol
Adding these equations and canceling out common terms, we get:
2 B (s) + (3/2) O2 (g) → B2O3 (s)
ΔH = -1273 kJ/mol
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Enthalpies of Formation
• Enthalpy of formation, Hf, is defined as:
the enthalpy change for a reaction in which a compound is made
from its constituent elements in their element forms
• Standard enthalpies of formation, Hf , are measured
under standard conditions (25 C and 1.00 atm pressure)
with the elements in their standard states
– The Hf of the most stable form of an element is zero
• The standard enthalpy of formation is represented by a
reaction where:
– each reactant is an element in its standard state
– the product is one mole of the compound.
– Example: 2Na(s) + ½ O2 (g) → Na2O (s)
Which reaction represents the ΔHf reaction for NaNO3?
1.
2.
3.
4.
5.
Na+ (aq) + NO3- (aq)
NaNO3 (aq)
+
Na (g) + NO3 (g)
NaNO3 (s)
Na (s) + NO 3 (s)
NaNO3 (s)
2 Na (s) + N2 (g) + 3 O2 (g)
2 NaNO 3 (s)
2 Na (s) + 1/2 N2 (g) + 3/2 O2 (g)
NaNO3 (s)
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Standard enthalpies of
formation, ΔHfo
Calculation of H using Hf
We can use Hess’s law in this way:
H = nHf products – mHf reactants
where n and m are the stoichiometric coefficients.
To do this we look at an equation:
Start with the reactants.
Decompose them into elements, then rearrange the
elements to form products.
The overall H is the sum of the Hs for each step.
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Calculation of H using Hf
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
•
Decomposing into elements (note O2 is elemental, so ∆H˚f =0)
C3H8(g)  3C(s) + 4H2(g)
∆H1 = –∆H˚f [C3H8(g)]
• Form the products CO2 and H2O from their elements:
3C(s) + 3O2(g)  3CO2(g)
∆H2 = 3 ∆H˚f [CO2(g)]
4H2(g) + 2O2(g)  4H2O(l)
∆H3 = 4 ∆H˚f [H2O(l)]
• We look up the values and add:
H = [ΣnHf of Products] - [Σ mHf of Reactants]
= [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) – (-103.85 kJ)
= -2219.9 kJ
Foods and Fuels
• Foods and fuels both involve combustion-type reactions
to produce heat
• The term “fuel value” is used to express the quantity of
heat released when a fuel is combusted
• Our body’s fuels are carbohydrates, fats and proteins
• Fats yield more energy/gram than carbohydrates or
proteins (38 kJ/g vs. 17 kJ/g)
• Carbohydrates and fats produce CO2 and H2O when
metabolized; proteins produce N-containing compounds
• Consume more than burn → excess stored as fat
• Can use our knowledge of energy to calculate calories
burned, etc
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Energy in Fuels
Most of the energy consumed in the US is from fossil fuels.
The higher the % of H and C in a fuel, the more energy is
produced per gram
Energy consumption is increasing dramatically; research into
more efficient use of earth’s energy resources is critical
Creative energy conversions
• Hybrid cars – capture some of the
waste energy during braking to charge
electric battery, which can provide
energy to move the car at slow speeds
• Wind turbines
• Solar panels
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Upcoming events
• In lab:
•
•
•
•
Standardization of NaOH Lab Report due
Heats of reaction pre-lab due
Chapter 4-5 review on Thursday
Mastering Chemistry Chap 5 due Thur 5/20
Second Lecture Quiz (Chap 4-5) on 5/20
We’ll start Chapter 6 next Tuesday
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