Endothermic and Exothermic Reactions
-Chemistsdeinethetotalinternalenergyofasystematconstant
pressureasEnthalpy.
-EnthalpyisgiventhesymbolΔH.
-Chemistshavenomethodofmeasuringthetotalenergyofasystem
easily,buttheycanmeasurethechangesthatoccurinthisenergy.
Theydothisbyexaminingwhathappenstothesurroundingsofthe
system.
-ThesechangesaretermedΔHorenthalpychange.Weusethe
symboldelta(Δ)torepresentchange.Therefore,wewillgenerally
alwaysrefertothechangeinenthalpy,orΔH.
ENDOTHERMIC:
Energy is absorbed
The enthalpy change is positive (+H).
Examples: Melting of ice to liquid water, flowering lily, cold
pack on your arm, baking bread, mixing water and ammonium
nitrate, photosynthesis.
EXOTHERMIC:
Energy is released
The enthalpy change is negative (-H).
Examples: Match burning, dynamite, fireworks, rusting of iron,
candle flame, liquid water to ice.
A typically exothermic reaction would
be written like this:
Cu(s) + Cl2(g) → CuCl2(g) + 220.1 kJ
Exothermic reactions release energy because the heat content of the
reactants is greater than that of the products.
The "excess" energy is released to the surroundings during the
reaction.
-When the amount of enthalpy in a system decreases, a negative sign
is used to indicate the drop.
-Thus, we could re-write the equation shown on the last slide in H
notation where the heat energy is written separate from the equation
as:
Cu(s) + Cl2(g)
CuCl2(g)
∆ H = -220.1 kJ
During endothermic reactions, energy is
absorbed from the surroundings. A typical
endothermic reaction is written with the
energy term on the reactant side of the
equation:
2H2O(g) + C(s) + 132 kJ → CO2(g) + 2H2(g)
In other words:
YOU NEED TO ADD HEAT FOR THE REACTION TO OCCUR
As with the exothermic reaction, we can remove the energy term
from the equation and record the enthalpy change with a positive
value, indicating that enthalpy increased during the reaction:
2H2O(g) + C(s)
CO2(g) + 2H2(g)
∆ H = +132 kJ
1) For each of the following, re-write the equation in "∆H "
notation, for one mole of the underlined substance.
Fe2O3(s) + 3CO(g)
3CO2(g) + 2Fe(s) + 25 KJ
2) For each of the following reactions written in "∆H"
notation, re-write the equation using whole number
balances and energy as a term in the equation.
3FeCl3(s)
3FeCl2(s) + 3/2Cl2(g)
∆H = + 173 KJ
Using Heats of Formation
The ∆H of the reaction is the difference between the total heat
(enthalpy) of the products and that of the reactants.
∆H = Heat of Products - Heat of Reactants
∆H = ∑∆Hf (products) – ∑∆Hf
(reactants)
The symbol sigma ( ) stands for “summation” or “the sum of”.
A few things to keep in mind:
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Pay attention to states of each substance – this will
ensure you are using the correct Hf.
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Pay attention to the balancing coefficients in the
equation, as you must multiply the Hf values by these
coefficients.
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Be very careful with + and – values.
Heats of formations of various substances are listed in a
tables of heats of formation. Heats of formation are given the
symbol ∆Hf.
Note : The heats of formation of elements in their standard
state are zero and they do not often appear in these tables.
Steps to Calculating Using Heats of Formation:
To calculate H reaction with this formula we need to;
1) Balance the equation.
2) Locate the heats of formation from the table.
3) Plug the heats of formations into the formula and
solve
Example : Calculate the ∆Hf of the following reaction using heats of
formations
___C3H8(g) + ___O2(g)
___CO2(g) + ___H2O(g)