chapter 8chem.yonsei.ac.kr/chem/upload/che1001-04-00/111214809611824.pdf · department, harcourt,...

Copyright © 2001 by Harcourt, Inc. All rights reserved. 8.1

Chapter 8

Thermochemistry

Copyright © 2001 by Harcourt, Inc.All rights reserved. Requests for permission to make copies of any part of the work should be mailed to the following address: Permissions Department, Harcourt, Inc. 6277 Sea Harbor Drive, Orlando, Florida 32887-6777


Thermochemistry

Basic concepts

Calorimetry

Thermochemical Equations


Basic Concepts

Define:systemsurroundingsstate property

Basic Equation for heat flow:

q = c × m × ∆t (c = specific heat)


Basic Concepts

Suppose 652 J of heat is added to 15.0 g of water

(c = 4.18 J/g ⋅ °C), originally at 20°C.

What is the final temperature?

∆t = = 10.4°C; final t = 30.4°C652 J4.18 J/g ⋅ °C × 15.0 g


Basic Concepts

Note that if heat is absorbed by the system (q is positive), temperature increases; if q is negative, temperature drops.

For a reaction at constant T and P:endothermic: q = ∆H > 0 system absorbs heatexothermic: q = ∆H < 0 system evolves heat


Enthalpy reaction for exothermic reactions


Enthalpy reaction for endothermic reactions


Calorimetry

Coffee cup calorimeter:

∆H of the reaction = -q water

Heat given off by reaction is absorbed by the water in the coffee cup.


Calorimetry (cont.)

Suppose heat is absorbed by 412 g of water, increasing the temp from 20.12 to 29.86°C. What is the ∆H?

q water = 4.18 × 4.12 g × 9.74°C = 1.68 × 103 J

∆H = –1.68 kJ

Jg⋅°C


Calorimetry (cont.)

Bomb calorimeter: Some heat is absorbed by the metal as well as the surrounding water.


Calorimetry (cont.)

Equation for bomb calorimeter:

q reaction = -q calorimeter = -(Ccalorimeter) × ∆t

where Ccalorimeter is the total heat capacity of the bomb and water.


Calorimetry (cont.)

Suppose combustion of 1.60 g CH4 in bomb calorimeter raises the temperature by 5.14°C (Ccalorimeter = 17.2 kJ/°C)

q reaction = -17.2 kJ/°C × 5.14°C = -88.4 kJ


Thermochemical Equations

H2(g) + Cl2(g) → 2HCl(g); ∆H= -185 kJ185 kJ of heat evolved when 2 moles of HCl are formed.

2HgO(s) → 2Hg(l) + O2(g); ∆H= +182 kJ182 kJ of heat must be absorbed to decompose 2 moles HgO.


Rules of Thermochemistry

∆H is directly proportional to amount of reactants or products.

When one mole of ice melts, 6.00 kJ of heat is absorbed, ∆H = +6.00 kJ.

If one gram of ice melts, ∆H = 6.00 kJ/18.02 = +0.333 kJ.

In general, ∆H can be related to amount by the conversion factor approach.


Rules of Thermochemistry (cont.)

H2(g) + Cl2(g) → 2HCl(g) ∆H = -185 kJ

When 1.00 g of Cl2 reacts:

∆H = 1.00 g Cl2 × × = -2.61 kJ1 mol Cl2

70.90 g Cl2

-185 kJ1 mol Cl2



• ∆H for a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reaction.

H2O(s) → H2O(l); ∆H = +6.00 kJ; 6.00 kJ absorbed

H2O(l) → H2O(s); ∆H = -6.00 kJ; 6.00 kJ evolved


∆H independent of path



• Hess’ law:

If equation 1 + equation 2 = equation 3, then

∆H3 = ∆H1 + ∆H2

Often used to calculate ∆H for one step, knowing ∆H for all other steps and for the overall reaction.



C(s) + 1/2O2(g) → CO(g) ∆H1 = ?

CO(g) + 1/2O2 (g) → CO2(g) ∆H2 = -283.0 kJ

C(s) + O2(g) → CO2(g) ∆H3 = -393.5 kJ

∆H1 = -110.5 kJ


Heats of Formation

∆Hf° of a compound = ∆H when one mole of compound is formed from the elements in their stable states.

2Ag(s) + Cl2(g) → 2AgCl(s) ∆H° = -254.0 kJ

∆Hf° AgCl(s) = -127.0 kJ


Heats of Formation (cont.)

HgO(s) → Hg (l) + 1/2O2 (g) ∆H° = +90.8 kJ

∆Hf° HgO(s) = -90.8 kJ



For any thermochemical equation:

∆H° = Σ∆Hf° products - Σ∆Hf° reactants

The heat of formation for an element in a stable state is zero.



CH4(g) + 2O2(g) → CO2(g) + 2H2O

∆H° = ∆Hf° CO2(g) + 2∆Hf° H2O(g ) - ∆Hf° CH4(g)

= -393.5 kJ + 2(-285.8 kJ) - (-74.8 kJ) = -890.3 kJ



Can apply to ions:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

set ∆Hf° H+(aq) = 0

∆H° = ∆Hf° Zn2+(aq) = 152.4 kJ


Bond Energies

B.E. = ∆H when one mole of bonds is broken in a gaseous state.

Cl2(g) → 2Cl(g) ∆H = B.E. Cl–Cl = 243 kJ

N2(g) → N(g) ∆H = B.E. N N = 941 kJ

In general, multiple bonds are stronger than single bonds:C–C = 347 kJ C=C = 612 kJ C C = 820 kJ


Comparison of bond energies in H2 and Cl2


First Law

∆E = q + w∆E = change in energy of system

q = heat flow into systemw = work done on system

∆H = ∆E + ∆(PV)To calculate ∆(PV), ignore liquids and solids, use ideal gas law

to find PV for gases.

CH4(g) + O2(g) → CO2(g) + H2O(l)

∆H = ∆E - RT ; RT = 2.5 kJ at 25°C


First law of thermodynamics

chapter 8chem.yonsei.ac.kr/chem/upload/che1001-04-00/111214809611824.pdf · department, harcourt,...

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