CH16
REACTION
ENERGY
11 Geng Tiffany Wang
SECTION 1 THERMOCHEMISTRY
• Thermochemistry is the study of the transfers of energy as heat that accompany chemical
reactions and physical changes
• Ex: The burning of a match
• A calorimeter measures energy absorbed as heat
• Temperature is a measure of the average kinetic energy (KE) of the
particles in a sample of matter (More KE in in particles=hotter)
• Joule (J) is the SI unit of heat
HEAT &SPECIFIC HEAT
• Heat is the energy transferred between samples of matter because of a difference in their
temperatures
*Energy always moves from higher temperature to lower temperature
• Specific Heat is the amount of energy required to raise the temperature
of one gram of a substance by one Celsius degree (1°C)
or one Kelvin (1K)
EXAMPLE PROBLEM 1
Question: A 500-gram cube of lead is heated from 25 °C to 75 °C. How much energy was
required to heat the lead? The specific heat of lead is 0.129 J/g°C.
SOLUTION
m = 500 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C
Plug these values into the specific heat equation from above.
Q = mcΔT
Q = (500 grams)·(0.129 J/g°C)·(50 °C)
Q = 3225 J
ENTHALPY
• The energy absorbed as heat during a chemical reaction at constant pressure is represented by
ΔH (enthalpy)
*Only changes in enthalpy can be measured
• The enthalpy of reaction is the quantity of energy transferred as heat during a chemical
reaction
• Endothermic Reaction: Absorb heat
2H2O (g) + 483.6 kJ
2H2 (g) + O2 (g)
ΔH= +483.6 kJ
• Exothermic Reaction: Release heat
2H2 (g) + O2 (g)
2H2O (g) + 483.6 kJ
ΔH= -483.6 kJ
*Exothermic compounds tend to be very stable
• The basis for calculating enthalpies of reaction is known as
Hess’s law: The overall enthalpy change in a reaction us
equal to the sum of enthalpy changes
for the individual steps of the process
EXAMPLE PROBLEM 2
Question: Find the enthalpy change for the reaction
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
when:
C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol
C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol
SOLUTION
Our reaction needs one CO2 in the product and the first reaction also has one CO2 product.
C(s) + O2(g) → CO2(g) ΔHf = -393.5 kJ/mol
This reaction gives us the CO2 needed on the product side and one of the O2 needed on the reactant
side. The other two O2 can be found in the second reaction.
S(s) + O2(g) → SO2(g) ΔHf = -296.8 kJ/mol
Since only one O2 is in the reaction, multiply the reaction by two to get the second O2. This doubles the
ΔHf value.
2 S(s) + 2 O2(g) → 2 SO2(g) ΔHf = -593.6 kJ/mol
Combining these equations gives
2 S(s) + C(s) + 3 O2(g) → CO2(g) + SO2(g)
The enthalpy change is the sum of the two reactions: ΔHf = -393.5 kJ/mol + -593.6 kJ/mol = 987.1 kJ/mol
This equation has the product side needed in the problem but contains an extra two S and one C
atom on the reactant side. Fortunately, the third equation has the same atoms. If the reaction is
reversed, these atoms are on the product side. When the reaction is reversed, the sign of the
change in enthalpy is reversed.
CS2(l) → C(s) + 2 S(s); ΔHf = -87.9 kJ/mol
Add these two reactions together and the extra S and C atoms cancel out. The remaining reaction
is the reaction needed in the question. Since the reactions were added together, their ΔHf values
are added together.
ΔHf = -987.1 kJ/mol + -87.9 kJ/mol
ΔHf = -1075 kJ/mol
Answer: The change in enthalpy for the reaction
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) is ΔHf = -1075 kJ/mol.
DRIVING FORCE OF REACTIONS
• Entropy (S) is a measure of the disorder or randomness in a system
*The natural process of the universe is to go toward more entropy
FREE ENERGY
• Gibbs free energy: determines if a reaction is endothermic or exothermic
EXAMPLE SOLUTION 3
Calculate ∆G at 290 K for the following reaction:
2NO(g)+O2(g)→2NO2(g)(4)(4)2NO(g)+O2(g)→2NO2(g)
Given
∆H = -120 kJ
∆S = -150 JK -1
SOLUTION
now all you have to do is plug in all the given numbers into the above equation that was
introduced earlier. expect you have to convert ΔSΔS so the same unit was ΔHΔH.
ΔS=-150J/K(1kJ1000J)=-0.15kJ/KΔS=150J/K(1kJ1000J)=-0.15kJ/K
∆G=-120kJ-(290K)(-0.150kJ/K)
=-120kJ+43kJ
=77kJ∆G=