Download 2A.4 Heat transactions and more Study notes Law in PDF only on Docsity!
42 2 The First Law
the reversible process because increasing the external pressure even infinitesimally at any stage results in compression. It can be inferred from this discussion that, because some pushing power is wasted when p > pex, the maximum work available from a system operating between specified initial and final states is obtained when the change takes place reversibly.
2A.4 Heat transactions
In general, the change in internal energy of a system is
dU = dq + dwexp + dwadd (2A.10)
where dwadd is work in addition (‘add’ for additional) to the ex- pansion work, dwexp. For instance, dwadd might be the electrical work of driving a current of electrons through a circuit. A sys- tem kept at constant volume can do no expansion work, so in that case dwexp = 0. If the system is also incapable of doing any other kind of work (if it is not, for instance, an electrochemical cell connected to an electric motor), then dwadd = 0 too. Under these circumstances:
dU = dq Heat transferred atconstant volume (2A.11a)
This relation can also be expressed as dU = dqV, where the sub- script implies the constraint of constant volume. For a measura- ble change between states i and f along a path at constant volume,
U (^) f U (^) i qv
id^ U^ dq^ V
f i
f
∫ = ∫
−
which is summarized as
ΔU = qV (2A.11b)
Note that the integral over dq is not written as Δq because q, unlike U, is not a state function. It follows from eqn 2A.11b that measuring the energy supplied as heat to a system at con- stant volume is equivalent to measuring the change in internal energy of the system.
(a) Calorimetry
Calorimetry is the study of the transfer of energy as heat dur- ing a physical or chemical process. A calorimeter is a device for measuring energy transferred as heat. The most common de- vice for measuring qV (and therefore ΔU) is an adiabatic bomb calorimeter (Fig. 2A.8). The process to be studied—which may be a chemical reaction—is initiated inside a constant-volume container, the ‘bomb’. The bomb is immersed in a stirred water bath, and the whole device is the calorimeter. The calorimeter is also immersed in an outer water bath. The water in the calo- rimeter and of the outer bath are both monitored and adjusted
to the same temperature. This arrangement ensures that there is no net loss of heat from the calorimeter to the surroundings (the bath) and hence that the calorimeter is adiabatic. The change in temperature, ΔT, of the calorimeter is pro- portional to the energy that the reaction releases or absorbs as heat. Therefore, qV and hence ΔU can be determined by meas- uring ΔT. The conversion of ΔT to qV is best achieved by cali- brating the calorimeter using a process of known output and determining the calorimeter constant , the constant C in the relation q = CΔT (2A.12)
The calorimeter constant may be measured electrically by passing a constant current, I, from a source of known potential difference, Δ ϕ, through a heater for a known period of time, t, for then (The chemist’s toolkit 8) q = ItΔ ϕ (2A.13)
Brief illustration 2A.
If a current of 10.0 A from a 12 V supply is passed for 300 s, then from eqn 2A.13 the energy supplied as heat is q = (10.0 A) × (300 s) × (12 V) = 3.6 × 104 A V s = 36 kJ
The result in joules is obtained by using 1 A V s = 1 (C s−^1 ) V s = 1 C V = 1 J. If the observed rise in temperature is 5.5 K, then the calorimeter constant is C = (36 kJ)/(5.5 K) = 6.5 kJ K−^1.
Alternatively, C may be determined by burning a known mass of substance (benzoic acid is often used) that has a known heat output. With C known, it is simple to interpret an observed temperature rise as a release of energy as heat.
Oxygen input^ Thermometer
Firing leads
Sample Oxygen under pressure
Water
Bomb
Figure 2A.8 A constant-volume bomb calorimeter. The ‘bomb’ is the central vessel, which is strong enough to withstand high pressures. The calorimeter is the entire assembly shown here. To ensure adiabaticity, the calorimeter is immersed in a water bath with a temperature continuously readjusted to that of the calorimeter at each stage of the combustion.
2A Internal energy 43
(b) Heat capacity
The internal energy of a system increases when its tempera- ture is raised. This increase depends on the conditions under which the heating takes place. Suppose the system has a con- stant volume. If the internal energy is plotted against tem- perature, then a curve like that in Fig. 2A.9 may be obtained. The slope of the tangent to the curve at any temperature is called the heat capacity of the system at that temperature. The heat capacity at constant volume is denoted CV and is defined formally as
C
U
V T V
^
^
Heat capacity at constant volume [definition] (2A.14)
(Partial derivatives and the notation used here are reviewed in The chemist’s toolkit 9.) The internal energy varies with the tem- perature and the volume of the sample, but here only its varia- tion with the temperature is important, because the volume is held constant (Fig. 2A.10), as signified by the subscript V.
Internal energy,
U
Temperature, T
A
B
Figure 2A.9 The internal energy of a system increases as the temperature is raised; this graph shows its variation as the system is heated at constant volume. The slope of the tangent to the curve at any temperature is the heat capacity at constant volume at that temperature. Note that, for the system illustrated, the heat capacity is greater at B than at A.
The chemist’s toolkit 8 (^) Electrical charge, current, power, and energy
Electrical charge , Q, is measured in coulombs, C. The funda- mental charge, e, the magnitude of charge carried by a single electron or proton, is approximately 1.6 × 10 −^19 C. The motion of charge gives rise to an electric current , I, measured in cou- lombs per second, or amperes, A, where 1 A = 1 C s−^1. If the electric charge is that of electrons (as it is for the current in a metal), then a current of 1 A represents the flow of 6 × 1018 elec- trons (10 μmol e−) per second. When a current I flows through a potential difference Δ ϕ (measured in volts, V, with 1 V = 1 J A−^1 ), the power, P, is P = IΔ ϕ
It follows that if a constant current flows for a period t the energy supplied is E = Pt = ItΔ ϕ Because 1 A V s = 1 (C s−^1 ) V s = 1 C V = 1 J, the energy is obtained in joules with the current in amperes, the potential difference in volts, and the time in seconds. That energy may be supplied as either work (to drive a motor) or as heat (through a ‘heater’). In the latter case q = ItΔ ϕ
Brief illustration 2A.
In Brief illustration 2A.1 it is shown that the translational con- tribution to the molar internal energy of a perfect monatomic gas is 32 RT. Because this is the only contribution to the internal energy, Um(T) = 32 RT. It follows from eqn 2A.14 that
C
T
V ,m=^ {^32 RT^ }^32 R
The numerical value is 12.47 J K−^1 mol−^1.
Heat capacities are extensive properties: 100 g of water, for instance, has 100 times the heat capacity of 1 g of water (and therefore requires 100 times the energy as heat to bring about the same rise in temperature). The molar heat capacity at constant volume , CV,m = CV/n, is the heat capacity per mole of substance, and is an intensive property (all molar quantities are intensive). For certain applications it is useful to know the
Figure 2A.10 The internal energy of a system varies with volume and temperature, perhaps as shown here by the surface. The variation of the internal energy with temperature at one particular constant volume is illustrated by the curve drawn parallel to the temperature axis. The slope of this curve at any point is the partial derivative (∂U/∂T)V.
Internal energy,
U
Temperature, T Volume, V
Temperature variation of U (^) Slope of U versus T at constant V
2A Internal energy 45
A measurable change of temperature, ΔT, brings about a meas- urable change in internal energy, ΔU, with
ΔU = CVΔT
Internal energy change on heating [constant volume]
(2A.15b)
Because a change in internal energy can be identified with the heat supplied at constant volume (eqn 2A.11b), the last equa- tion can also be written as
qV = CVΔT (2A.16)
This relation provides a simple way of measuring the heat capacity of a sample: a measured quantity of energy is transferred as heat to the sample (by electrical heating, for ex- ample) under constant volume conditions and the resulting increase in temperature is monitored. The ratio of the energy transferred as heat to the temperature rise it causes (qV/ΔT) is
the constant-volume heat capacity of the sample. A large heat capacity implies that, for a given quantity of energy trans- ferred as heat, there will be only a small increase in tempera- ture (the sample has a large capacity for heat).
Brief illustration 2A.
Suppose a 55 W electric heater immersed in a gas in a constant- volume adiabatic container was on for 120 s and it was found that the temperature of the gas rose by 5.0 °C (an increase equivalent to 5.0 K). The heat supplied is (55 W) × (120 s) = 6.6 kJ (with 1 J = 1 W s), so the heat capacity of the sample is
C
6.6kJ V 5.0K 1.3kJK
= = −^1
Property Equation Comment Equation number First Law of thermodynamics (^) ΔU = q + w Convention 2A. Work of expansion (^) dw = −pexdV 2A.5a Work of expansion against a constant external pressure (^) w = −pexΔV pex = 0 for free expansion 2A. Reversible work of expansion of a gas (^) w = −nRT ln(Vf/Vi) Isothermal, perfect gas 2A. Internal energy change ΔU = qV Constant volume, no other forms of work 2A.11b Electrical heating q = ItΔϕ 2A. Heat capacity at constant volume CV = (∂U/∂T)V Definition 2A.
Checklist of concepts
☐ 1. Work is the process of achieving motion against an opposing force.
☐ 2. Energy is the capacity to do work.
☐ 3. An exothermic process is a process that releases energy as heat.
☐ 4. An endothermic process is a process in which energy is acquired as heat.
☐ 5. Heat is the process of transferring energy as a result of a temperature difference.
☐ 6. In molecular terms, work is the transfer of energy that makes use of organized motion of atoms in the sur- roundings and heat is the transfer of energy that makes use of their disorderly motion.
☐ 7. Internal energy , the total energy of a system, is a state function.
☐ 8. The internal energy increases as the temperature is raised. ☐ 9. The equipartition theorem can be used to estimate the contribution to the internal energy of each classically behaving mode of motion. ☐ 10. The First Law states that the internal energy of an iso- lated system is constant. ☐ 11. Free expansion (expansion against zero pressure) does no work. ☐ 12. A reversible change is a change that can be reversed by an infinitesimal change in a variable. ☐ 13. To achieve reversible expansion , the external pressure is matched at every stage to the pressure of the system. ☐ 14. The energy transferred as heat at constant volume is equal to the change in internal energy of the system. ☐ 15. Calorimetry is the measurement of heat transactions.
Checklist of equations
2B.1 The definition of enthalpy
The enthalpy , H, is defined as
H = U + pV Enthalpy[definition] (2B.1)
where p is the pressure of the system and V is its volume. Because U, p, and V are all state functions, the enthalpy is a state function too. As is true of any state function, the change in enthalpy, ΔH, between any pair of initial and final states is independent of the path between them.
(a) Enthalpy change and heat transfer
An important consequence of the definition of enthalpy in eqn 2B.1 is that it can be shown that the change in enthalpy is equal to the energy supplied as heat under conditions of con- stant pressure.
How is that done? 2B.1 (^) Deriving the relation between enthalpy change and heat transfer at constant pressure In a typical thermodynamic derivation, as here, a common way to proceed is to introduce successive definitions of the quantities of interest and then apply the appropriate con- straints. Step 1 Write an expression for H + dH in terms of the defini- tion of H For a general infinitesimal change in the state of the system, U changes to U + dU, p changes to p + dp, and V changes to V + dV, so from the definition in eqn 2B.1, H changes by dH to
H + dH = (U + dU) + (p + dp)(V + dV) = U + dU + pV + pdV + Vdp + dpdV The last term is the product of two infinitesimally small quan- tities and can be neglected. Now recognize that U + pV = H on the right (in blue), so
H + dH = H + dU + pdV + Vdp and hence dH = dU + pdV + Vdp
Step 2 Introduce the definition of dU Because dU = dq + dw this expression becomes
dH = dq + dw + pdV + Vdp
TOPIC 2B Enthalpy
➤ Why do you need to know this material?
The concept of enthalpy is central to many thermody- namic discussions about processes, such as physical trans- formations and chemical reactions taking place under conditions of constant pressure.
➤ What is the key idea?
A change in enthalpy is equal to the energy transferred as heat at constant pressure.
➤ What do you need to know already?
This Topic makes use of the discussion of internal energy (Topic 2A) and draws on some aspects of perfect gases (Topic 1A).
The change in internal energy is not equal to the energy trans- ferred as heat when the system is free to change its volume, such as when it is able to expand or contract under conditions of constant pressure. Under these circumstances some of the energy supplied as heat to the system is returned to the sur- roundings as expansion work (Fig. 2B.1), so dU is less than dq. In this case the energy supplied as heat at constant pressure is equal to the change in another thermodynamic property of the system, the ‘enthalpy’.
Energy as heat
Energy as work
ΔU < q
Figure 2B.1 When a system is subjected to constant pressure and is free to change its volume, some of the energy supplied as heat may escape back into the surroundings as work. In such a case, the change in internal energy is smaller than the energy supplied as heat.
48 2 The First Law
from their molar masses, M, and their mass densities, ρ, by using ρ = M/Vm. The solution The change in enthalpy when the transition occurs is ΔHm = Hm(aragonite) − Hm(calcite) = {Um(a) + pVm(a)} − {Um(c) + pVm(c)} = ΔUm + p{Vm(a) − Vm(c)} where a denotes aragonite and c calcite. It follows by substitut- ing Vm = M/ ρ that
ρ ρ
^
∆H ∆U pM
(a)
m m (c)
Substitution of the data, using M = 100.09 g mol−^1 , gives
− = × ×
× −
−
− −
∆ H ∆U (1.0 10 Pa) (100.09gmol ) 1 2.93gcm
2.71gcm
m m
5 1
3 3
= − 2.8 ×10 Pacm mol 5 3 −^1 = −0.28Pa m mol^3 −^1
Hence (because 1 Pa m^3 = 1 J), ΔHm − ΔUm = −0.28 J mol−^1 , which is only 0.1 per cent of the value of ΔUm. Comment. It is usually justifiable to ignore the difference between the molar enthalpy and internal energy of condensed phases except at very high pressures when pΔVm is no longer negligible.
Self-test 2B.1 Calculate the difference between ΔH and ΔU when 1.0 mol Sn(s, grey) of density 5.75 g cm−^3 changes to Sn(s, white) of density 7.31 g cm−^3 at 10.0 bar. J 4.4− = UΔ − HΔ Answer:
In contrast to processes involving condensed phases, the values of the changes in internal energy and enthalpy might differ significantly for processes involving gases. The enthalpy of a perfect gas is related to its internal energy by using pV = nRT in the definition of H:
H = U + pV = U + nRT (2B.3)
This relation implies that the change of enthalpy in a reaction that produces or consumes gas under isothermal conditions is
ΔH = ΔU + ΔngRT Relation between Δ [isothermal process, perfect gas]H^ and ΔU (2B.4)
where Δng is the change in the amount of gas molecules in the reaction. For molar quantities, replace Δng by Δ νg.
Brief illustration 2B.
In the reaction 2 H 2 (g) + O 2 (g) → 2 H 2 O(l), 3 mol of gas-phase molecules are replaced by 2 mol of liquid-phase molecules,
so Δng = − 3 mol and Δ νg = −3. Therefore, at 298 K, when RT = 2.5 kJ mol−^1 , the enthalpy and internal energy changes taking place in the system are related by ΔHm − ΔUm = (−3) × RT ≈ −7.5 kJ mol−^1
Note that the difference is expressed in kilojoules, not joules as in Example 2B.1. The enthalpy change is smaller than the change in internal energy because, although energy escapes from the system as heat when the reaction occurs, the system contracts as the liquid is formed, so energy is restored to it as work from the surroundings.
2B.2 The variation of enthalpy with temperature
The enthalpy of a substance increases as its temperature is raised. The reason is the same as for the internal energy: mole- cules are excited to states of higher energy so their total energy increases. The relation between the increase in enthalpy and the increase in temperature depends on the conditions (e.g. whether the pressure or the volume is constant).
(a) Heat capacity at constant pressure
The most frequently encountered condition in chemistry is constant pressure. The slope of the tangent to a plot of en- thalpy against temperature at constant pressure is called the heat capacity at constant pressure (or isobaric heat capacity), Cp, at a given temperature (Fig. 2B.3). More formally:
C
H
p T p
^
^
Heat capacity at constant pressure [definition] (2B.5)
Temperature, T
Enthalpy,
H
A
B
Internal energy, U
Figure 2B.3 The constant-pressure heat capacity at a particular temperature is the slope of the tangent to a curve of the enthalpy of a system plotted against temperature (at constant pressure). For gases, at a given temperature the slope of enthalpy versus temperature is steeper than that of internal energy versus temperature, and Cp,m is larger than CV,m.
2B Enthalpy 49
The heat capacity at constant pressure is the analogue of the heat capacity at constant volume (Topic 2A) and is an exten- sive property. The molar heat capacity at constant pressure , Cp,m, is the heat capacity per mole of substance; it is an inten- sive property. The heat capacity at constant pressure relates the change in enthalpy to a change in temperature. For infinitesimal changes of temperature, eqn 2B.5 implies that
dH = CpdT (at constant pressure) (2B.6a)
If the heat capacity is constant over the range of temperatures of interest, then for a measurable increase in temperature
= (^) ∫ = (^) ∫ = −
∆ ∆ H (^) TC (^) pdT C dT C (T T)
T p (^) T
T 1 p^2
2 1
2
T
which can be summarized as
ΔH = CpΔT (at constant pressure) (2B.6b)
Because a change in enthalpy can be equated to the energy supplied as heat at constant pressure, the practical form of this equation is
qp = CpΔT (2B.7)
This expression shows how to measure the constant-pressure heat capacity of a sample: a measured quantity of energy is supplied as heat under conditions of constant pressure (as in a sample exposed to the atmosphere and free to expand), and the temperature rise is monitored. The variation of heat capacity with temperature can some- times be ignored if the temperature range is small; this is an excellent approximation for a monatomic perfect gas (for in- stance, one of the noble gases at low pressure). However, when it is necessary to take the variation into account for other sub- stances, a convenient approximate empirical expression is
C a bT c p,m (^) T^2
= + + (2B.8)
The empirical parameters a, b, and c are independent of tem- perature (Table 2B.1) and are found by fitting this expression to experimental data.
Example 2B.2 (^) Evaluating an increase in enthalpy with temperature
What is the change in molar enthalpy of N 2 when it is heated from 25 °C to 100 °C? Use the heat capacity information in Table 2B.1. Collect your thoughts The heat capacity of N 2 changes with temperature significantly in this range, so you cannot use eqn 2B.6b (which assumes that the heat capacity of the substance is constant). Therefore, use eqn 2B.6a, substitute eqn 2B.8 for the temperature dependence of the heat capacity, and integrate the resulting expression from 25 °C (298 K) to 100 °C (373 K). The solution For convenience, denote the two temperatures T 1 (298 K) and T 2 (373 K). The required relation is
H a bT c T H T d^ dT
H T T
T ( )^ m
( ) m 1 2
m 2 1
2 ∫ =^ ∫ +^ +
^
By using Integral A.1 in the Resource section for each term, it follows that
H T H T a T T b T T c T T
m 2 m 1 2 1 (^12) 2
2 1
2 2 1
^
Substitution of the numerical data results in Hm(373 K) = Hm(298 K) + 2.20 kJ mol−^1 Comment. If a constant heat capacity of 29.14 J K−^1 mol−^1 (the value given by eqn 2B.8 for T = 298 K) had been assumed, then the difference between the two enthalpies would have been calculated as 2.19 kJ mol−^1 , only slightly different from the more accurate value. Self-test 2B.2 At very low temperatures the heat capacity of a solid is proportional to T 3 , and Cp,m = aT 3. What is the change in enthalpy of such a substance when it is heated from 0 to a temperature T (with T close to 0)? 4 aT^14 =^ mHΔ^ Answer:
(b) The relation between heat capacities
Most systems expand when heated at constant pressure. Such systems do work on the surroundings and therefore some of the energy supplied to them as heat escapes back to the sur- roundings as work. As a result, the temperature of the system rises less than when the heating occurs at constant volume. A smaller increase in temperature implies a larger heat capac- ity, so in most cases the heat capacity at constant pressure of a system is larger than its heat capacity at constant volume. As shown in Topic 2D, there is a simple relation between the two heat capacities of a perfect gas: Cp − CV = nR Relation between heat capacities[perfect gas] (2B.9)
Table 2B.1 Temperature variation of molar heat capacities, Cp,m/(J K−^1 mol−^1 ) = a + bT + c/T 2 *
a b/(10 −^3 K −^1 ) c/(10^5 K^2 ) C(s, graphite) 16.86 4.77 −8. CO 2 (g) 44.22 8.79 −8. H 2 O(l) 75.29 0 0 N 2 (g) 28.58 3.77 −0.
- More values are given in the Resource section.
