Transcript week 3
Thermodynamics: applications of the First Law 자연과학부 박영동 교수 the role of enthalpy in chemistry
Standard enthalpy changes
standard enthalpy change,
states.
standard state
the change in enthalpy for a process in which the initial and final substances are in their standard (of a single substance)
,
the pure substance at 1 bar. 𝐻 ⦵ 𝑚 temperature is not a part of the definition of a standard state, and standard states may refer to any temperature (but it should be specified).
Table 3.1 Standard enthalpies of transition at the transition temperature*
Thermodynamics: applications of the First Law
3.1 Physical change 3.1.1 The enthalpy of phase transition 3.1.2 Atomic and molecular change 3.2 Chemical change 3.2.3 Enthalpies of combustion 3.2.4 The combination of reaction enthalpies 3.2.5 Standard enthalpies of formation 3.2.6 Enthalpies of formation and molecular modeling 3.2.7 The variation of reaction enthalpy with temperature
Enthalpy of physical change: Phase Transition
Enthalpy of physical change: Phase Transition
A
phase
is a specific state of matter that is uniform throughout in composition and physical state. solid liquid vapor H 2 O(
s
) → H 2 O(
l
) Δ 𝐻 ⦵ 𝑚 H 2 O(
l
) → H 2 O(
g
) Δ 𝐻 ⦵ 𝑚 = +6.01 kJ = +44 kJ
Enthalpy is a State function
ΔH
forward = -
ΔH
reverse
Δ
sub
H
=
Δ
fus
H
+
Δ
vap
H
Calculate the standard enthalpy of sublimation of ice at 0°C from its standard enthalpy of fusion at 0°C (6.01 kJ mol − 1 ) and the standard enthalpy of vaporization of water at 0°C (45.07 kJ mol − 1 ).
The standard enthalpy of sublimation of magnesium at 25°C is 148 kJ mol − 1 . How much energy as heat (at constant temperature and pressure) must be supplied to 1.00 g of solid magnesium metal to produce a gas composed of Mg 2+ ions and electrons?
Sublimation First ionization, IE 1 Second ionization, IE 2 Sublimation: Mg(
s
) →Mg(
g
) First ionization: Mg(
g
) →Mg + (
g
) + e (
g
) Second ionization: Mg + (
g
) →Mg 2+ (
g
) + e (
g
)
ΔH
⦵ = +148 kJ
ΔH
⦵ = +738 kJ
ΔH
⦵ = +1451 kJ Overall(sum): Mg(
s
) → 3CO 2 (
g
)+ 3H 2 O(
l
)
ΔH
⦵ = +2337kJ
Estimate the standard enthalpy change for the reaction in which liquid methanol is formed from its elements at 25°C.
CH 3 OH(
g
) → CH 3 OH(
l
)
ΔH
⦵ = -38.00 kJ
ΔH
⦵ = (+1837.73 kJ) + (-2059 kJ) + (- 38.00 kJ) = -259 kJ
Enthalpy of Combustion,
Δ c H Δ c 𝐻 = Δ c 𝑈 + Δν gas 𝑅𝑇 NH 2 CH 2 COOH(
s
) + 9 4 O 2 (
g
) →2CO 2 (
g
)+ 5 2 H 2 O(
l
) + 1 2 N 2 (
g
)
ΔH
⦵
Δ c
𝐻 =
Δ c
𝑈 + = - 969.6 kJ mol = - 969.6 kJ mol 1 4 𝑅𝑇 -1 -1 + 1 4 × + 0.62
(8.3145
kJ mol × -1 10 -3 = - 969.0 kJ mol -1 kJ K -1 mol -1 ) × (298.15K)
Hess’s law
The enthalpy changes used in Example 3.4 to illustrate Hess’s law.
Given the thermochemical equations C 3 H 6 (
g
) + H 2 (
g
) →C 3 H 8 (
g
) C 3 H 8 (
g
) + 5O where C 3 H 6 2 (
g
) →3CO 2 (
g
)+ 4H 2 O( is propene and C 3 H 8
l
)
ΔH
⦵ = - 124 kJ
ΔH
⦵ =-2220 kJ is propane, calculate the standard enthalpy of combustion of propene.
The overall reaction is C 3 H 6 (
g
) + 9 2 O 2 the following: (
g
) →3CO 2 (
g
)+ 3H 2 O(
l
)
ΔH
⦵ We can recreate this thermochemical equation from C 3 H 6 (
g
) + H 2 (
g
) →C 3 H 8 (
g
) C 3 H 8 (
g
) + 5O 2 (
g
) →3CO 2 (
g
)+ 4H 2 O(
l
) H 2 O(
l
) → H 2 (
g
) + 1 2 O 2 (
g
) Overall C 3 H 6 (
g
) + 9 2 O 2 (
g
) →3CO 2 (
g
)+ 3H 2 O(
l
)
ΔH ΔH ΔH ΔH
⦵ ⦵ ⦵ ⦵ = - 124 kJ = -2220 kJ = + 286 kJ =-2058 kJ It follows that the standard enthalpy of combustion of propene is − 2058 kJ mol − 1 .
∆ r H ⦵ = Δ f 𝐻 ⦵ 𝑚 (products) Δ f 𝐻 ⦵ 𝑚 (reactants) An enthalpy of reaction may be expressed as the difference between the enthalpies of formation of the products and the reactants.
methylcyclohexane
Δ f 𝐻 ⦵ 𝑚 (axial) Δ f 𝐻 ⦵ 𝑚 (equatorial) = 7.5 kJ/mol
thermo-chemical ‘altitude’ of a compound
The enthalpy of formation acts as a kind of thermo-chemical ‘altitude’ of a compound with respect to the ‘sea level’ defined by the elements from which it is made. Endothermic compounds have positive enthalpies of formation; exothermic compounds have negative energies of formation.
the reaction enthalpy change with temperature
The enthalpy of a substance increases with temperature. Therefore, if the total enthalpy of the reactants increases by a different amount from that of the products, the reaction enthalpy will change with temperature. The change in reaction enthalpy depends on the relative slopes of the two lines and hence on the heat capacities of the substances.
A sample consisting of 1.0 mol CaCO to the atmosphere?
3 (s) was heated to 800°C, when it decomposed. The heating was carried out in a container fitted with a piston that was initially resting on the solid. Calculate the work done during complete decomposition at 1.0 atm. What work would be done if instead of having a piston the container was open
A sample consisting of 1 mol of perfect gas atoms (for which = w , 3 2 R) is taken through the cycle shown in the Figure. (a) Determine the temperature at the points 1,2, and 3. (b) Calculate ΔU , and ΔH for each step and for the overall cycle. If a then write in +, -, 0, or ? as appropriate.
C V,m q numerical answer cannot be obtained from the information given, ,
A sample of 1.00 mol perfect gas molecules at 1.00 atm and 25 ℃ with
C
p,m = 7/2 R is put through the following cycle: (a) constant-pressure heating to twice its initial volume, (b) reversible, adiabatic expansion back to its initial temperature, (c) reversible isothermal compression back to 1.00 atm. Calculate q, w, ΔU, and ΔH for each step and overall.
p
1 2
V p
1 =1atm,
T
1 =298K 3