I. Reaction Energy

Section 17-1 and 17-2

Ch. 17 – Reaction Energy and Reaction Kinetics

A. Reviewing…

Heat and Temperature Temperature Joule Calorimeter Heat

A. Reviewing… Heat Capacity and Specific Heat

Specific Heat – used to compare heat absorption of different materials

Heating and Cooling Curves q

= m · ∆Hf or v

q = Cp · m · ∆T

B. Energy of Reactions

Reaction Pathway Shows the change in energy during a

chemical reaction

B. Energy of Reactions Exothermic reaction that

releases energy

products have lower PE than reactants

2H2(l) + O2(l) 2H2O(g) + energy

energyreleased

B. Energy of Reactions Endothermic

reaction that absorbs energy

reactants have lower PE than products

2Al2O3 + energy 4Al + 3O2

energyabsorbed

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The heat that is released or absorbed in a chemical reaction

Equivalent to HC + O2(g) CO2(g) + 393.5 kJ

C + O2(g) CO2(g) H = -393.5 kJ

In thermochemical equation, it is important to indicate the physical state

H2(g) + ½ O2 (g) H2O(g) H = -241.8 kJ

H2(g) + ½ O2 (g) H2O(l) H = -285.8 kJ

B. Energy of Reaction

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Exothermic this is why gummy bears are widely disliked!!

Gummy Bear SacrificeGummy Bear Sacrifice and another

EndothermicMr. Wizard

B. Energy of Reactions

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The heat content a substance has at a given temperature and pressure

Can’t be measured directly because there is no set starting point

The reactants start with a heat content The products end up with a heat

content So we can measure how much

enthalpy changes ΔΔHH

C. Enthalpy

C. EnthalpyEnthalpy Change in Reaction H = total of all forms of energy (bonds, PE, KE, phases,

etc.) ΔH = heat of reaction = change in enthalpy during

chemical reaction ΔH > 0

thermodynamically stable - does not spontaneously decompose at room temperature

ΔH < 0thermodynamically unstable - spontaneous decomposition at room temperature

-ΔH favors spontaneous change!

(kinetically stable - overall negative energy but imperceptible change due to slowness of reaction)

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Ene

rgy

Reactants Products

Change is downΔH is <0Exothermic

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Ene

rgy

Reactants Products

Change is upΔH is > 0Endothermic

D. Heat of FormationHeat of Formation – energy transferred when

one mole of a compound is formed from its elements

ΔHf° is symbol depends on temperature, pressure and

phase ΔHf° - standard heat of formation @25°C

and 101.3 kPa Appendix A-14 page 902

Stable = large negativeUnstable = small negative or positive

D. Heat of Formation2HgO(s) --> 2Hg(l) + O2(g) ΔH = 182 kJWhat’s the standard heat of formation of HgO?

(stable or unstable?)

D. Heat of Formation

4Fe(s) + 3O2(g)  --> 2Fe2O3(s) ΔH = -1643 kJ

What is the heat of formation of rust? (stable or unstable?)

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The standard heat of The standard heat of formation of an element formation of an element in it’s standard state is in it’s standard state is arbitrarily set at “0 kJ”arbitrarily set at “0 kJ”

This includes the This includes the diatomics!diatomics!

D. Heat of Formation

E. Calculating Heat of Reactionmethod 1:

ΔH = Hfinal - Hinitial   (state function)

heat of reaction = ∑ ΔHformation of products - ∑ΔHformation of reactants

multiply heat of formation by coefficient if more than one mole being used in balanced equation

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CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

H= - 802.4 kJ

E. Calculating Heat of Reaction

E. Calculating Heat of ReactionCl2(g) + 2HBr(g)  2HCl(g) + Br2(g)

2NO(g) + O2(g) 2NO2(g)  

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method 2: Hess’s LawIf you add two or more If you add two or more thermochemical equations to give a thermochemical equations to give a final equation, then you can also final equation, then you can also add the heats of reaction to give add the heats of reaction to give the final heat of reactionthe final heat of reaction.

E. Calculating Heat of Reaction

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If you turn an equation around, you change the sign:H2(g) + 1/2 O2(g) H2O(g) H=-285.5 kJ

then H2O(g) H2(g) + 1/2 O2(g) H =+285.5 kJ

If you multiply the equation by a number, you multiply the heat by that number:2 H2O(g) H2(g) + O2(g) H =+571.0 kJ

Or, you can just leave the equation “as is”

E. Calculating Heat of Reaction

E. Calculating Heat of Reaction   CuO(s) + H2(g)    Cu(s) + H2O(g)              ΔH

= ?                   Cu(s) + ½ O2(g) CuO ΔH1 = -155 kJH2 (g) + ½ O2  H2O(g)     ΔH2 = -

242 kJ

rearranging:CuO(s)   --> Cu(s) + ½ O2(g)                  155 kJH2 (g) + ½ O2  --> H2O(g)                   -242 kJ                                                            ΔH = -87 kJ

E. Calculating Heat of ReactionCalculate the enthalpy for the formation of C5H12.

5C(s) + 6H2(g) C5H12 ΔH = ?

C(s) + O2(g) CO2(g) ΔH = -393.5 kJH2(g) + ½ O2(g) H2O(l) ΔH = -285.8

kJC5H12(s) + 8O2(g) 5CO2(g) + 6H2O(l) ΔH = -

3535.6 kJ

EXAMPLE (a little harder)

B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g) ΔH = +2035 kJ H2O(l) → H2O(g) ΔH = +44 kJ H2(g) + ½ O2(g) → H2O(l) ΔH = -286 kJ 2B(s) + 3H2(g) → B2H6(g) ΔH = +36 kJ

2B(s) + 3/2 O2(g) → B2O3(s) ΔH =?

SOLUTIONAfter the multiplication and reversing of the equations (and their enthalpy changes), the result is:

B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) ΔH = -2035 kJ 3H2O(g) → 3H2O(l) ΔH = (-44 x 3) = -132

kJ 3H2O(l) → 3H2(g) + 3/2 O2(g) ΔH = (+286 x 3) = +858

kJ 2B(s) + 3H2(g) → B2H6(g) ΔH = +36 kJ

Adding these equations and canceling out the common terms on both sides, we get

2B(s) + 3/2 O2(g) → B2O3(s) ΔH = -1273 kJ

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Heat of Combustion

The heat from the reaction that completely burns 1 mole of a substance:

C + O2(g) CO2(g) + 393.5 kJsame as saying:

C + O2(g) CO2(g) H = -393.5 kJ