week 3

applications of the First Law
자연과학부 박영동 교수
the role of enthalpy in chemistry
Standard enthalpy changes
standard enthalpy change, the change in
enthalpy for a process in which the initial
and final substances are in their standard
standard state (of a single substance), the
pure substance at 1 bar. ⦵
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
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.
H2O(s) → H2O(l)
H2O(l) → H2O(g)
= +6.01 kJ
= +44 kJ
Enthalpy is a State function
ΔHforward = -ΔHreverse
ΔsubH = ΔfusH+ ΔvapH
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 Mg2+ ions and electrons?
First ionization, IE1
Second ionization, IE2
Sublimation: Mg(s) →Mg(g)
ΔH⦵= +148 kJ
First ionization: Mg(g) →Mg+(g) + e-(g)
ΔH⦵= +738 kJ
Second ionization: Mg+(g) →Mg2+(g) + e-(g) ΔH⦵= +1451 kJ
Overall(sum): Mg(s) → 3CO2(g)+ 3H2O(l)
ΔH⦵= +2337kJ
Estimate the standard enthalpy change for the reaction
in which liquid methanol is formed from its elements at 25°C.
CH3OH(g) → CH3OH(l)
ΔH⦵= -38.00 kJ
ΔH⦵= (+1837.73 kJ) + (-2059 kJ) + (- 38.00 kJ) = -259 kJ
Enthalpy of Combustion, ΔcH
Δc  = Δc  + Δνgas 
NH2CH2COOH(s) + 94O2(g) →2CO2(g)+ 52 H2O(l) +
Δc  = Δc  + 14 
= - 969.6 kJ mol-1 + 4 ×(8.3145×10-3 kJ K-1 mol-1)×(298.15K)
= - 969.6 kJ mol-1 + 0.62 kJ mol-1
= - 969.0 kJ mol-1
Hess’s law
The enthalpy changes used in Example 3.4 to illustrate Hess’s law.
Given the thermochemical equations
C3H6(g) + H2(g) →C3H8(g)
ΔH⦵= - 124 kJ
C3H8(g) + 5O2(g) →3CO2(g)+ 4H2O(l)
ΔH⦵=-2220 kJ
where C3H6 is propene and C3H8 is propane, calculate
the standard enthalpy of combustion of propene.
The overall reaction is
C3H6(g) + 92O2(g) →3CO2(g)+ 3H2O(l)
We can recreate this thermochemical equation from
the following:
C3H6(g) + H2(g) →C3H8(g)
ΔH⦵= - 124 kJ
C3H8(g) + 5O2(g) →3CO2(g)+ 4H2O(l)
ΔH⦵= -2220 kJ
H2O(l) → H2(g) + 12 O2(g)
ΔH⦵= + 286 kJ
C3H6(g) + 92O2(g) →3CO2(g)+ 3H2O(l)
ΔH⦵=-2058 kJ
It follows that the standard enthalpy of combustion of
propene is −2058 kJ mol−1.
∆rH⦵ = Δf 
(products) - Δf 
An enthalpy of reaction may be
expressed as the difference between the
enthalpies of formation of the products
and the reactants.
(axial) - Δf 
(equatorial) = 7.5 kJ/mol
thermo-chemical ‘altitude’ of a
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
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 CaCO3(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
to the atmosphere?
A sample consisting of 1 mol of perfect gas atoms (for which CV,m
= 32 R) is taken through the cycle shown in the Figure. (a)
Determine the temperature at the points 1,2, and 3. (b) Calculate q,
w, ΔU, and ΔH for each step and for the overall cycle. If a
numerical answer cannot be obtained from the information given,
then write in +, -, 0, or ? as appropriate.
A sample of 1.00 mol perfect gas molecules at 1.00 atm and 25 ℃ with
Cp,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.
p1=1atm, T1=298K

similar documents