Chapter 6

Thermochemistry

Final Exam

.

(

May7, 2014

Wednesday)

Instructional Complex 421 10:15 AM. –12:15 PM

Study of energy and its transformation ― thermodynamics

Thermodynamics of chemical reactions ― thermochemistry

Energy ― capacity to do work (W) and transfer heat (Q) unit: J

W and Q ― two ways to transfer energy

Work = Force x Distance

W = F • d

SI unit: W ― J, F ― N, d ― m

J = N • m

Two forms of energy

Kinetic energy

Potential energy

Kinetic energy: energy from motion

2

2

1mvEk

potential energy: energy from the interaction between objects.

depends on the objects and the relative distance.

BrNO + BrNO 2NO +Br2

exothermic reaction

CH4 + 2O2 2H2O + CO2

Exothermic Process

system surroundings

universe

Energy of the universe is constant. Or say energy of the universe is conserved.

First law of thermodynamics

∆E = Q + W

E: energy of the system, including Ek and Ep. Or called internal energy of the system.

∆E: change of internal energy of the system.

∆E = Ef − Ei

System absorbs heat from surroundings, Q > 0.

System releases heat to surroundings, Q < 0.

System does work on surroundings, W < 0.

Surroundings do work on system, W > 0.

Calculate ΔE for a system undergoing an endothermic process

in which 15.6 kJ of heat flows and where 1.4 kJ of work is

done on the system.

State Function

A property of a system whose value is determined only by

specifying the system’s state, not on how the system arrived

at that state.

The state of a system is specified by parameters such as

temperature, pressure, concentration and physical states (solid,

liquid, or gas).

Ei

Ef

∆E = Ef − Ei

E: State FunctionsSystem: a battery

State functions: E, P, V, T, do not depend on path.

W, Q: not state functions, depend on process.

W = −P∆V

PV work:

To inflate a balloon you must do pressure–volume

work on the surroundings. If you inflate a balloon

from a volume of 0.100 L to 1.85 L against an

external pressure of 1.00 atm, how much work is

done (in joules)?

Example 6.4, page 245

Enthalpy: H = E + PV

Only PV work, constant P ∆H = Qp

∆H = Qp

∆H > 0 ↔ endothermic

∆H < 0 ↔ exothermic

Recall ….

EnthalpyEnthalpy

Identify each process as endothermic or exothermic

and indicate the sign of H.

(a) sweat evaporating from skin

(b) water freezing in a freezer

(c) wood burning in a fire

Example 6.6, page 250

Q = c m ∆T

c: specific heat (capacity)

m: mass

∆T: change of temperature

g

°C

J·°C−1·g−1

same sign as Q

Q = c m ∆T

C = c m

C: heat capacity, J ·°C−1

Q = C ∆T

Suppose you find a copper penny (minted before 1982, when pennies were almost entirely copper) in the snow. How much heat is absorbed by the penny as it warms from the temperature of the snow, which is –8.0 C, to the temperature of your body, 37.0 C? Assume the penny is pure copper and has a mass of 3.10 g.

Example 6.2, page 242

A 32.5-g cube of aluminum initially at 45.8 C is submerged into 105.3 g of water at 15.4 C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)

Example 6.3, page 243

A constant-pressure coffee-cup Calorimeter

ΔHrxn = − Qsoln

= − c m ΔT

Read example 6.8, page 253

∆E = Q + W

PV work: W = − P∆V

PV work only, constant P: ∆H = Qp

PV work only, constant V:

∆E = Qv

Read example 6.5,page 248

A constant-volumeCalorimeter

∆Erxn = −Qcal

= −Ccal ΔT

2H2(g) + O2(g) 2H2O(g) ∆H = −483.6 kJ

∆H = Hproducts − Hreactants

1. It is important to specify the state of each species in a thermochemical reaction.

∆H: enthalpy (change) of reaction, heat of reaction

2. ∆H of the reverse reaction is the negative of the original reaction.

3. ∆H depends on how the reaction is written.

Characteristics of enthalpy change of a reaction

An LP gas tank in a home barbeque contains 13.2 kg of propane, C3H8. Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank.

Example 6.7, page 252

HA

HB

∆HAB = HB − HA

HC

∆HAB = ∆HAC + ∆HCB

∆HAC = HC − HA

∆HCB = HB − HC

H1 = H2 + H3

Hess’s Law: ∆H for the overall reaction is equal to the sum ofthe enthalpy changes of each individual step.

The enthalpy of reaction for the combustion of C to CO2 is−393.5 kJ/mol C, and the enthalpy for the combustion of COto CO2 is −283.0 kJ/mol CO:

C(s) + O2(g) CO2(g) ∆H1 = −393.5 kJ

CO(g) + ½ O2(g) CO2(g) ∆H2 = −283.0 kJ

Using these data, calculate the enthalpy for the combustionof C to CO:

C(s) + ½ O2(g) CO(g) ∆H = ?

Calculate the ∆H for the reaction

2C(s) + H2(g) C2H2(g)

Given the following chemical equations and their respectiveenthalpy changes:

C2H2(g) + 5/2 O2(g) 2CO2(g) + H2O(l) ∆H1 = −1299.6 kJ

C(s) + O2(g) CO2(g) ∆H2 = −393.5 kJ

H2(g) + ½ O2(g) H2O(l) ∆H3 = −285.8 kJ

Try example 6.9 and for practice 6.9, page 256-257

Sea level

Atlanta, GA

Reno, NV

1000 ft

4500 ft

3500 ft

Standard0 ft

H = H (P, T, phase), phase = s, l, g

Standard state

P = 1 atmT = temperature of interest, often 25 °Cstate = most stable form

The standard enthalpy of formation of a pure substance,∆H°f, is the change in enthalpy for the reaction that formsone mole of the substance from its elements, with allsubstances in their standard states.

“Sea level” : free elements at standard state

Calculate the standard enthalpy change for the following reaction

4NH3(g) + 5O2(g) 4NO (g) + 6H2O(g)

Example 6.11, page 261

ΔHrxn° = Σnp ΔHf°(products) − Σnr ΔHf° (reactants)

Calculate the standard enthalpy change for the following reaction

2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)

Al(s) + ½ Fe2O3(s) ½ Al2O3(s) + Fe(s)

For practice 6.11, page 261