CHEMISTRY 161

Chapter 6

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THERMODYNAMICS

HEAT CHANGE

quantitative study of heat and energy changes of a system

the state (condition) of a system is defined by

T, p, n, V, E

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

the state (condition) of a system is defined by

T, p, n, V, E

STATE FUNCTIONS

properties which depend only on the initial and final state, but not on the way how this

condition was achieved

ΔV = Vfinal – Vinitial

Δp = pfinal – pinitial

ΔT = Tfinal – Tinitial

ΔE = Efinal – Einitial

Energy is a STATE FUNCTION

IT DOES NOT MATTER WHICH PATH YOU TAKE

ΔE = m g Δh

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

EN

TH

AL

PY

, H

Reactants

Products

CO2(g) + 2H2O(g)

- 802 kJ

- 88 kJ

- 890 kJ

Hess Law

heat is spontaneous transfer of thermal energy two bodies at different temperatures T1 > T2

spontaneous

T1

T2

Zeroth Law of Thermodynamics

a system at thermodynamical equilibrium

has a constant temperature

First Law of Thermodynamics

energy can be converted from one form to another,

but cannot be created or destroyed

CONSERVATION OF ENERGY

SYSTEM

SURROUNDINGS

THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT

ΔE = ΔEsystem + ΔEsurrounding = 0

-+

ΔΔEEsystemsystem = = ΔΔQ + Q + ΔΔWWΔΔEEsystemsystem = = ΔΔQ + Q + ΔΔWW

First Law of Thermodynamics

ΔΔQ heat changeQ heat changeΔΔQ heat changeQ heat change ΔΔW work doneW work doneΔΔW work doneW work done

Q > 0 ENDOTHERMICQ > 0 ENDOTHERMIC

Q < 0 EXOTHERMICQ < 0 EXOTHERMIC?

ΔΔW = - p W = - p ΔΔVVΔΔW = - p W = - p ΔΔVV

mechanical workmechanical workmechanical workmechanical work

M

the energy of gas goes up

M

the energy of gas goes down

ΔV < 0ΔV > 0

First Law and Enthalpy

ΔΔEEsystemsystem = = ΔΔQ - pQ - pΔΔVVΔΔEEsystemsystem = = ΔΔQ - pQ - pΔΔVV

1.constant pressure → enthalpy change Δ H

2. ideal gas law → p V = n R T

p ΔV = Δn R T

ΔΔEEsystemsystem = = ΔΔQ – R T Q – R T ΔΔnnΔΔEEsystemsystem = = ΔΔQ – R T Q – R T ΔΔnn

ΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHHΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHH

ΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHHΔΔQ = Q = ΔΔEEsystemsystem + R T + R T ΔΔn = n = ΔΔHH

Δ n = nfinal – ninitial

Calculate the energy change of a system for the reaction process at 1 atm and 25C

2 CO(g) + O2(g) → 2 CO2(g) ΔHo = -566.0 kJ

ΔΔEEsystemsystem = = ΔΔHH00 - R T - R T ΔΔn n ΔΔEEsystemsystem = = ΔΔHH00 - R T - R T ΔΔn n

ΔΔEEsystemsystem = -563.5 kJ = -563.5 kJ ΔΔEEsystemsystem = -563.5 kJ = -563.5 kJ

A gas is compressed in a cylinder from a volume of 20 L to 2.0 L by a constant pressure of 10 atm.

Calculate the amount of work done on the system.

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C.

Zn(s) + 2H+(aq) → Zn2+(aq) + H2 (g)

The heat of solution of KCl is +17.2 kJ/mol and the

lattice energy of KCl(s) is 701.2 kJ/mol. Calculate

the total heat of hydration of 1 mol of gas phase

K+ ions and Cl– ions.

Homework

Chapter 6, problems