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CHAPTER 3
|
THERMOCHEMISTRY: DEVELOPMENT OF THE FIRST LAW OF THERMODYNAMICS
The conceptual plan has three parts. In the
first part, use the temperature change and the heat
capacity of the calorimeter to find qcal.
In the second part, use qcal to get qrxn (which
just involves changing the sign). Since the bomb
calorimeter ensures constant volume, qrxn is
equivalent to ∆Uchem for the amount of sucrose
burned.
In the third part, divide qrxn by the number of
moles of sucrose to get ∆ Uchem per mole of sucrose.
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Ccal, ∆T → qcal
qcal = Ccal × ∆T
qcal → qrxn
qrxn = –qcal
∆Uchem =
qrxn
mol C12H 22O11
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Qcal = Ccal × ∆T = –qrxn
molar mass C12H22O11 = 342.3 g/mol
Gather the necessary quantities in the correct
units and substitute these into the equation to
compute qcal.
Find qrxn by taking the negative of qcal.
Find Erxn per mole of sucrose by dividing qrxn
by the number of moles of sucrose (calculated
from the given mass of sucrose and its molar
mass).
Solution
∆T = Tf Ti
= 28.33°C 24.92°C
= 3.41°C
q cal = Ccal ∆T
kJ
3.41°C
°C
= 16.7 kJ
q cal = 4.90
q rxn = q cal = 16.7 kJ
qrxn
∆Uchem =
mol C12H 22O11
16.7 kJ
=
1 mol C12H 22O11
1.010 g C12H 22O11
342.3 g C12H 22O11
= 5.66 103 kJ/mol C12H 22O11
Check The units of the answer (kJ) are correct for a
change in internal energy. The sign of ∆Erxn is negative, as it should be for a combustion reaction that
gives off energy.
3.22
Check Yourself 4
When 1.010 g of sucrose (C12H22O11) undergoes combustion in
a bomb calorimeter, the temperature rises from 24.92°C to 28.33°C.
Find ∆Uchem for the combustion of sucrose in kJ/mol sucrose. The
heat capacity of the bomb calorimeter, determined in a separate experiment, is 4.90 kJ/°C. (You can ignore the heat capacity of the small
sample of sucrose because it is negligible compared to the heat capacity
of the calorimeter.)
Given 1.010 g C12H22O11, Ti = 24.92ºC, Tf = 28.33°C, Ccal = 4.90kJ/°C
Find ∆Erxn
Enthalpy
To this point, we have dealt with the change in chemical internal
energy, ∆Uchem, only under conditions of constant volume in our bomb
calorimeter, for which
ΔU = ΔU therm + ΔU chem = q + w = qV
where w = –p∆V = 0 because the rigid walls of the bomb prevented any
change in volume, ∆V = 0. So the release of chemical energy upon combustion appeared as an increase in thermal energy at constant volume,
qV. However, a great many chemical reactions occur under conditions
of constant pressure and not constant volume.
This immediately raises the question: When we measure the heat of
reaction at constant volume, qV, how does that compare quantitatively
with the heat of reaction measured at constant pressure?
We know, because the system will do work on its surroundings
(w < 0) at constant pressure (because the volume will increase), that the
heat of reaction at constant pressure, qp, will be greater (less negative)
than qV because
∆U = qv = w + qp
and w < 0. But by how much?
The relation qV = qp + w is the key starting point from which we can
deduce the answer to our question. First, we know that, for a process at
constant volume, ∆U = qV and we know that w = –p∆V so we can write
∆U = qv= qp + w = qp − p∆V
Thus, qp = ∆U + p∆V. But now we recognize that U, p, and V are all state
variables because their values are each independent of the path taken to
reach that given state.
Suppose now we define a new state variable
H = U + pV
Then the change in that state variable
ΔH = H f − H i
(
= (U
) (
)
−U ) + ( p V − p V )
= U f + pf Vf − U i + piVi
f
i
f
f
i
i
= ΔU + Δ( pV )
If the process is carried out at constant temperature and pressure, then ∆(pV) =
p∆V and
ΔH = ΔU + pΔV = (qp − pΔV) + pΔV
=qp
where qp is the heat released at constant pressure.
This new state variable, H, is called the enthalpy and ∆H = qp is the enthalpy
change for a chemical process at constant pressure, which is the thermal energy
(heat) produced by the chemical reaction at constant pressure.
Because we live, by and large, in a constant pressure world, the state variable
enthalpy is a variable of great importance. In fact, ∆H released in a chemical reaction is routinely referred to as the energy release resulting from the change in bond
structure in going from reactants to products in a chemical reaction. The enthalpy
change, ∆H, is listed, as we will see, in the appendix of all chemistry texts and all
thermochemistry data sources as the quantitative measure of the relative energy
contained in the bonds of molecules. A table of ∆H constitutes Appendix B of this
text. If we lived in a constant volume world, those tables in the appendix of textbooks would list ΔU, not ΔH.
It is important to consider the magnitude of the enthalpy change of a chemical
reaction, ∆H, and the p∆V work term associated with a given reaction. What is
their relative magnitude?
Suppose we react two moles of carbon monoxide in the gas phase, CO(g),
with a mole of O2(g) to form two moles of carbon dioxide in the gas phase:
2CO(g) + O2 → 2CO2(g)
If the pressure is constant, we discover that 566.0 kJ of energy is released at constant temperature such that
qp = –566.0 kJ
Reactions in the Liquid Phase: An
Important Example of Reactions at
Constant Pressure
Adenosine triphosphate (ATP) is
used in the cell for the formation of proteins. In the reaction, ATP is hydrolyzed
to form adenosine diphosphate (ADP)
and phosphate (HPO42−) in the reaction
ATP4− + H2O → ADP3− + HPO42− + H+
Problem:
If 10 grams of ATP hydrolyzed to
ADP and HPO42− in 50 grams of water
at constant pressure in a calorimeter, it
is observed that the temperature of the
water increases by 2.1°C. What is q of the
reaction? What is ΔH for the reaction?
Step 1:
We know that the temperature increased so the reaction is exothermic and
qrxn is thus negative.
q = (50g)(4.18J/g-°C)(2.1K) = 439J
Thus qrxn = −439J as heat was released by
the reaction.
The molecular weight of ATP is
573g/mole
So: ΔHrxn = (−439J/10g)(573g/mole) =
−25kJ/mole
and since ∆Hrxn = qp, it follows that
∆Hrxn = –566.0 kJ
To evaluate the pressure-work term, p∆V, we write, from the perfect gas law pV
= nRT,
p∆V = ∆nRT
where ∆n is the change in the number of moles of gas and T is the temperature of
the gas mixture, which we keep constant.
But the change in the number of moles is just ∆n = nf – ni = 2 – 3 = –1 moles.
Thus,
p∆V = RT∆n = (8.3 × 10–3 kJ/mole-K)(298 K)(–1) = –2.5 kJ
This is an important and fairly general result that in most cases the –p∆V
term is small compared with the change in enthalpy, ∆H, of a reaction, and thus
the thermal energy (heat) produced in a chemical reaction at constant pressure is
approximately equal to that produced at constant volume.
To summarize
∆H = qp
3.23
CHAPTER 3
|
THERMOCHEMISTRY: DEVELOPMENT OF THE FIRST LAW OF THERMODYNAMICS
∆U = qV
∆U = ∆H – p∆V
The conceptual plan has three parts. In the first
part, use the temperature change and the other given
quantities, together with the equation q = m × Cs × ∆T,
to find qsoln.
In the second part, use qsoln to get qrxn (which simply involves changing the sign). Because the pressure is
constant, qrxn is equivalent to ∆Hrxn for the amount of
magnesium that reacted.
In the third part, divide qrxn by the number of moles
of magnesium to get ∆Hrxn per mole of magnesium.
qsoln → qrxn
qrxn = –qsoln
∆Hrxn =
qrxn
mol Mg
Notice that the sign of qsoln is positive, meaning that the
solution absorbed heat from the reaction.
Find qrxn by simply taking the negative of qsoln. Notice that qrxn is negative, as expected for an exothermic
reaction.
Solution
and because a large fraction of chemical reactions occur at constant
pressure, it is the enthalpy, ∆H, that appears most frequently.
Check Yourself 5 - Measuring Hrxn in a Coffee-Cup
Container
Magnesium metal reacts with hydrochloric acid according to
the following balance
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2
In an experiment to determine the enthalpy change for this reaction, 0.158 g of Mg metal is combined with enough HCl to make
100.0 mL of solution in a coffee-cup calorimeter. The HCl is sufficiently concentrated so that the Mg completely reacts. The temperature of the solution rises from 25.6°C to 32.8 °C as a result of
the reaction. Find ∆Hrxn for the reaction as written. Use 1.00 g/mL
as the density of the solution and Cs, soln = 4.18 J/g ∙ °C as the specific
heat capacity of the solution.
You are given the mass of magnesium, the volume of solution,
the initial and final temperatures, the density of the solution, and
the heat capacity of the solution. You are asked to find the change
in enthalpy for the reaction.
Cs, soln = 4.18 J/g t °C
1.00 g
=1.00 102 g
1 mL soln
∆T = Tf – Ti = 32.8°C – 25.6°C = 7.2°C
q soln = msoln C s, soln
T
m soln =100.0mL soln
102 g
4.18
q rxn = qsoln = 3.0
qrxn
∆Hrxn =
mol Mg
103 J
=1.00
=
J
7.2°C = 3.0
g t °C
Standard Enthalpies of Formation
103 J
3.0 103 J
1 mol Mg
0.158 g MG
24.31 g Mg
= 4.6
105 J/mol Mg
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The addition of hydrochloric acid to a silver nitrate
solution precipitates silver chloride according to the following reaction:
AgNO3(aq) + HCL(aq) → AgCl(s) + HNO3(aq)
When 50.0 mL of 0.100 M AgNO3 is combined with
50.0 mL of 0.100 M HCl in a coffee-cup calorimeter,
the temperature changes from 23.40 °C to 24.21 °C.
Calculate ∆Hrxn for the reaction as written. Use 1.00 g/
mL as the density of the solution and C = 4.18 J/g ∙ °C as
the specific heat capacity.
3.24
In the application of thermochemistry to a broad range of important calculations we need a convention by which the enthalpy
change for a given reaction, called the enthalpy of reaction, ΔHR, can
be readily calculated. The convention is to define the standard enthalpy of formation, ΔH°f , to specific molecular species, and then
tabulate those values of ΔH°f . Because enthalpy is a state function,
we are concerned only with changes in enthalpy ΔH, so the absolute
scale is not important in such a tabulation. In order to set the scale
for standard enthalpies of formation, the convention is to assign enthalpy values of zero to elements in their standard states. Specifically,
the enthalpy of formation, ΔH°f , is defined as zero for O2, H2, N2 and
C(graphite) in their standard states at one atmosphere pressure, and
25°C. This is shown graphically in Figure 3.15, wherein ΔH°f = 0 sets
the scale for enthalpies of formation for a broad range of molecular
species, both positive (energy required to form a molecular structure from is elements in their standard state) and negative (energy
released in the formation of the species from their standard states).
A great deal of work over time has gone into the determination of
the enthalpies of reaction for hundreds of compounds—information that is now available in tables, specifically Appendix B of this
text. A selection of important examples is shown in Table 3.3.