Thermodynamics: applica-tions of the First Law

자연과학부 박영동 교수the role of enthalpy in chem-

istry

Standard enthalpy changes

standard enthalpy change, the change in enthalpy for a process in which the initial and final substances are in their standard states. standard state (of a single sub-stance), 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 transi-tion 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

solid liquid vapor

A phase is a specific state of matter that is uniform throughout in composition and physi-cal state.

H2O(s) → H2O(l) Δ= +6.01 kJH2O(l) → H2O(g) Δ= +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 tem-perature and pressure) must be supplied to 1.00 g of solid magnesium metal to produce a gas composed of Mg2+ ions and electrons?

Sublimation

First ionization, IE1

Second ionization, IE2

Sublimation: Mg(s) →Mg(g) ΔH⦵= +148 kJFirst ionization: Mg(g) →Mg+(g) + e-(g) ΔH⦵= +738 kJSecond 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

NH2CH2COOH(s) + O2(g) →2CO2(g)+ H2O(l) + N2(g) ΔH⦵

Δc Δc = - 969.6 kJ mol-1 + ×(8.3145×10-3 kJ K-1 mol-1)×(298.15K)= - 969.6 kJ mol-1 + kJ mol-1

= - 969.0 kJ mol-1

The overall reaction is C3H6(g) + O2(g) →3CO2(g)+ 3H2O(l) ΔH⦵

We can recreate this thermochemical equation from the follow-ing:C3H6(g) + H2(g) →C3H8(g) ΔH⦵= - 124 kJC3H8(g) + 5O2(g) →3CO2(g)+ 4H2O(l) ΔH⦵= -2220 kJH2O(l) → H2(g) + O2(g) ΔH⦵= + 286 kJOverallC3H6(g) + O2(g) →3CO2(g)+ 3H2O(l) ΔH⦵=-2058 kJ

It follows that the standard enthalpy of combustion of propene is −2058 kJ mol−1.

Hess’s lawThe enthalpy changes used in Example 3.4 to illustrate Hess’s law.

Given the thermochemical equationsC3H6(g) + H2(g) →C3H8(g) ΔH⦵= - 124 kJC3H8(g) + 5O2(g) →3CO2(g)+ 4H2O(l) ΔH⦵=-2220 kJwhere C3H6 is propene and C3H8 is propane, calculate the

standard enthalpy of combustion of propene.

rH ⦵ = Δf (products) - Δf (reactants)

An enthalpy of reaction may be ex-pressed as the difference between the enthalpies of formation of the products and the reactants.

methylcyclohexane

Δf (axial) - Δf (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 com-pounds have positive enthalpies of formation; exothermic compounds have negative energies of formation.

the reaction enthalpy change with temperature

The enthalpy of a substance in-creases with temperature. Therefore, if the total enthalpy of the reactants increases by a different amount from that of the products, the reaction en-thalpy will change with temperature. The change in reaction enthalpy de-pends on the relative slopes of the two lines and hence on the heat ca-pacities 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 con-tainer 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 pis-ton the container was open to the atmosphere?

A sample consisting of 1 mol of perfect gas atoms (for which CV,m = R) is taken through the cycle shown in the Figure. (a) Determine the temperature at the points 1,2, and 3. (b) Cal-culate q, w, ΔU, and ΔH for each step and for the overall cy-cle. If a numerical answer cannot be obtained from the in-formation given, then write in +, -, 0, or ? as appropriate.

V

p 1 2

3

p1=1atm, T1=298K

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) re-versible, adiabatic expansion back to its initial temperature, (c) reversible isothermal compression back to 1.00 atm. Cal-culate q, w, ΔU, and ΔH for each step and overall.