Energy Changes in Chemical Energy Changes in Chemical ReactionsReactions

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Most reactions give off or absorb energy Energy is the capacity to do work or supply

heat.◦ Heat: transfer of thermal (kinetic) energy

between two systems at different temperatures (from hot to cold)

Metal bar in water Metal bar drilled

Work (w): energy transfer when forces are applied to a system

Heat (q): energy transferred from a hot object to a cold one◦ Radiant energy heat from the sun◦ Thermal energy associated with motion of

particles◦ Potential energy energy associated with

object’s position or substance’s chemical bonds◦ Kinetic energy energy associated with object’s

motion

Describe the difference between the two. SI unit of energy: J

1 watt = 1 J/s, so a 100 Watt bulb uses 100 J each second

We often use the unit of kJ to refer to chemical heat exchanges in a reaction. 1 kJ = 1000 J

Energy is also reported in calories:◦ Amount of energy needed to raise 1 gram of water by

1oC◦ 1 cal = 4.184 J; 1 Cal = 4184 J ◦ Cal (or kcal) is used on food labels

Molecular heat transfer2

2

s

m kg 1 J 1

Heat: form of energy transferred from object at higher temperature to one at lower temperature (from hot object to cold object)

Thermochemistry: study of heat changes in chemical reactions, in part to predict whether or not a reaction will occur

Thermodynamics: study of heat and its transformations

First Law of Thermodynamics: Energy can be converted from one form to another but cannot be created or destroyed

System loses heat (negative); gains heat (positive)

Endothermic reaction: q is positive (q > 0)◦ Reaction (system) absorbs heat◦ Surroundings feel cooler

Exothermic reaction: q is negative (q < 0)◦ Reaction releases heat◦ Surroundings feel warmer

Determine if the following processes are endothermic or exothermic…◦ Combustion of methane◦ Reacting Ba(OH)2 with NH4Cl◦ Neutralization of HCl◦ Melting◦ CaCO3 (s) CaO (s) + CO2 (g)

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◦ Combustion of methane exothermic◦ Reacting Ba(OH)2 with NH4Cl endothermic◦ Neutralization of HCl exothermic◦ Melting endothermic◦ CaCO3 (s) CaO (s) + CO2 (g) endothermic

◦ Combustion, neutralization, and combination reactions tend to be exothermic

◦ Decomposition reactions tend to be endothermic◦ Melting, boiling, and sublimation are endothermic

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Specific heat (sp. ht.): amount of heat required to raise 1 gram of substance by 1oC

Use mass, specific heat, and T to calculate the amount of heat gained or lost:

q = msT ms = C q = CT◦ Heat capacity (C): amount of heat required to

raise the temperature of a given quantity of a substance by 1oC; C = q / T = J / oC

◦ Molar heat capacity (Cm): amount of heat that can be absorbed by 1 mole of material when temperature increases 1oC; q = (Cm) x (moles of substance) x (T) = J / mol • oC

Calculate the amount of heat transferred when 250 g of H2O (with a specific heat of 4.184 J/g·oC) is heated from 22oC to 98oC.

q = msT Is heat being put into the system or given off by

the system?

If a piece of hot metal is placed in cold water, what gains heat and what loses heat? Which one will have a positive q value and which will have a negative q value?

34.8 g of an unknown metal at 25.2oC is mixed with 60.1 g of H2O at 96.2 oC (sp. ht. = 4.184 J/g·oC). The final temperature of the system comes to 88.4oC. Identify the unknown metal.

Specific heats of metals:◦ Al 0.897 J/g·oC◦ Fe 0.449 J/g·oC◦ Cu 0.386 J/g·oC◦ Sn 0.228J/g·oC

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Heat changes in a reaction can be determined by measuring the heat flow at constant pressure

Apparatus to do this is called a calorimeter.

Heat evolved by a reaction is absorbed by water; heat capacity of calorimeter is the heat capacity of water.

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A 28.2 gram sample of nickel is heated to 99.8oC and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5oC. After the metal cools, the final temperature of the metal and water is 25.0oC.

qabsorbed + qreleased = 0 Which substance absorbed heat? Which substance released heat? Calculate the heat absorbed by the

substance you indicated above.

A hot piece of copper (at 98.7oC, specific heat = 0.385 J/g•oC) weighs 34.6486 g. When placed in room temperature water, it is calculated that 915.1 J of heat are released by the metal.

What gains heat? What loses heat? What is the final temperature of the metal? Watch signs!!!!

Enthalpy (H) describes heat flow into and out of a system under constant pressure

Enthalpy (a measure of energy) is heat transferred per mole of substance.

At constant pressure, ◦ qpH = Hproducts – Hreactants

◦ H > 0 endothermic (net absorption of energy from environment; products have more internal energy)

◦ H < 0 exothermic (net loss of energy to environment; reactants have more internal energy)

Why does T become constant during melting and evaporating?

Melting, vaporization, and sublimation are endothermic

We can calculate total heat needed to convert a 15 gram piece of ice at -20oC to steam at 120oC.

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2.09 J/g2.09 J/gooCC

4.184 J/g4.184 J/gooCC

2.080 J/g2.080 J/gooCC

334 J/g334 J/g

2250 J/g2250 J/g

Heat of fusion (Hfus): Amount of heat required to melt (solid liquid)

Heat of vaporization (Hvap): Amount of heat required to evaporate (liquid gas)

Heat of sublimation (Hsub): Amount of heat required to sublime (solid gas)

Why are there no values for Hfreezing, Hcondendsation, or Hdeposition?

Shows both mass and enthalpy relationships

2Al (s) + Fe2O3 (s) 2Fe (s) + Al2O3 (s) Ho = -852 kJ

Amount of heat given off depends on amount of material:◦ 852 kJ of heat are released for every 2 mol Al, 1

mol Fe2O3, 2 mol Fe, and 1 mol Al2O3

2Al (s) + Fe2O3 (s) 2Fe (s) + Al2O3 (s) Ho = -852 kJ

How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al?

What if we reversed the reaction? Heat would have to be put in to make the

reaction proceed:◦ 2Fe (s) + Al2O3 (s) 2Al (s) + Fe2O3 (s) Ho = +852 kJ

If a compound cannot be directly synthesized from its elements, we can add the enthalpies of multiple reactions to calculate the enthalpy of reaction in question.

Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps

Look at direction of reaction and amount of reactants/products

Value changes sign with direction

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Figure 8.5

Values of enthalpy change◦ For a reaction in the reverse direction, enthalpy is

numerically equal but opposite in sign Reverse direction, heat flow changes; endothermic

becomes exothermic (and vice versa); sign of H changes

◦ Proportional to the amount of reactant consumed Twice as many moles = twice as much heat; half as

many moles = half as much heat

HT = H1 + H2 + H3 + ….

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Thermochemical equation:◦H2(g) + I2(s) 2HI(g) H =

+53.00 kJ Two possible changes: Reverse the equation:

◦2HI(g) H2(g) + I2(s) H = -53.00 kJ

Double the amount of material:◦2H2(g) + 2I2(s) 4HI(g) H =

+106.00 kJ2626

Calculate Ho for 2NO (g) + O2 (g) N2O4 (g) Ho = ?

N2O4 (g) 2NO2 (g) Ho = 57.2 kJ

NO (g) + ½ O2 (g) NO2 (g) Ho = -57.0 kJ

We can use known values of Ho to calculate unknown values for other reactions

P4 (s) + 3 O2 (g) P4O6 (s) H = -1640.1 kJ

P4 (s) + 5 O2 (g) P4O10 (s) H = -2940.1 kJ

What is Ho for the following reaction?P4O6 (s) + 2 O2 (g) P4O10 (s) H = ?

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Given: 2NH3(g) N2H4(l) + H2(g) H = 54 kJ N2(g) + H2(g) NH3(g) H = -69 kJ CH4O(l) CH2O(g) + H2(g) H = -195 kJ

Find the enthalpy for the following reaction: N2H4(l) + CH4O(l) CH2O (g) + N2(g) + 3H2(g)

H = ? kJ

2

1

2

3

Given the following equations: 2CO2 (g) O2 (g) + 2CO (g) H = 566.0

kJ ½ N2 (g) + ½ O2 (g) NO (g) H = 90.3 kJ

Calculate the enthalpy change for: 2CO (g) + 2NO (g) 2CO2 (g) + N2 (g) H =

?

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Standard heat of formation Hof): heat

needed to make 1 mole of a substance from its stable elements in their standard states

Hof = 0 for a stable (naturally occurring)

element Which of these have Ho

f = 0?◦ CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s)

Do the following equations represent standard enthalpies of formation? Why or why not?◦ 2Ag (l) + Cl2 (g) 2AgCl (s)

◦ Ca (s) + F2 (g) CaF2 (s)

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Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book)

Horxn = nHo

f (products) – nHof (reactants)

◦ sum; n = number of moles (coefficients) Direct calculation of enthalpy of reaction if

the reactants are all in elemental form◦ Sr (s) + Cl2 (g) SrCl2 (g)

◦ Horxn = [Ho

f (SrCl2)] – [Hof (Sr) + Ho

f (Cl2)] = -828.4 kJ/mol

Some Common Substances (25oC)

Horxn = Ho

f,products - Hof,reactants

Calculate values of Ho for the following rxns: 1) CaCO3 (s) CaO (s) + CO2 (g) 2) 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

Hof values:

CaCO3: -1207.1 kJ/mol; CaO: -635.5 kJ/mol; CO2: -393.5 kJ/mol; C6H6: 49.0 kJ/mol; H2O(l): -285.8 kJ/mol

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Use Standard Heat of Formation values to calculate the enthalpy of reaction for:

C6H12O6(s) C2H5OH(l) + CO2(g) Hint: Is the equation balanced?

Hof (C6H12O6(s)) = -1260.0 kJ/mol

Hof (C2H5OH(l)) = -277.7 kJ/mol

Hof (CO2(g)) = -393.5 kJ/mol

Bond Dissociation Energy (or Bond Energy, BE): energy required to break a bond in 1 mole of a gaseous molecule

Reactions generally proceed to form compounds with more stable (stronger) bonds (greater bond energy)

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H2 Bond Energy

Bond energies vary somewhat from one mole- cule to another so we use average bond dissociation energy (D)

H-OH 502 kJ/mol Avg O-H = 453

H-O 427 kJ/mol kJ/mol H-OOH 431 kJ/mol

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Horxn = BE (reactants) + - BE (products)

endothermic exothermic energy input energy

released

BE(react) > BE(prod) endothermic BE(react) < BE(prod) exothermic

Use only when heats of formation are not available, since bond energies are average values for gaseous molecules

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Use bond energies to calculate the enthalpy change for the following reaction:

N2(g) + 3H2(g) 2NH3(g) Hrxn = [BEN N + 3BEH-H] + [-6BEN-H] Hrxn = [945 + 3(436)] – [6(390)] = -87 kJ measured value = -92.2 kJ Why are the calculated and measured

values different?

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Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl3 (g) N2 (g) + Cl2 (g)

How many distinct bond types are there in each molecule?

How many of each bond type do we need to calculate Hrxn? ◦ BE(N-Cl) = 200 kJ/mol

◦ BE(N≡N) = 945 kJ/mol

◦ BE(Cl-Cl) = 243 kJ/mol4242

6(N-Cl) + -1(N N) + -3(Cl-Cl) 6(200) + -(945) + -3(243) = -474 kJ

≡

Use q = msT (s = J/g·oC) If given mass of reactant, convert to moles

and multiply by enthalpy to find total heat transferred

If given multiple equations with enthalpies, use Hess’s Law

If given Hof values: products – reactants

If given bond energy (BE) values: +reactants + -products

Identify how to set up the following problems:

Calculate the Ho of reaction for:◦ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)

◦ Hof C3H8(g): -103.95 kJ/mol; Ho

f CO2(g): -393.5 kJ/mol; Ho

f H2O(l): -285.8 kJ/mol 8750 J of heat are applied to a 170 g sample of

metal, causing a 56oC increase in its temperature.  What is the specific heat of the metal? Which metal is it?

C2H4(g ) + 6F2(g) 2CF4(g) + 4HF(g) Ho = ?◦ H2 (g) + F2 (g) 2HF (g) Ho = -537 kJ

◦ C (s) + 2F2 (g) CF4 (g) Ho = -680 kJ

◦ 2C (s) + 2H2 (g) C2H4 (g) Ho = 52.3 kJ

Use average bond energies to determine the enthalpy of the following reaction◦ CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)

(BEC-Cl = 328 kJ/mol)