� A calorimeter can be used to measure the heat evolved or absorbed during the reaction
� However, for reactions that occur over millions of years it is virtually impossible to measure enthalpy change
� Chemists can use a theoretical way to determine change in H
Hess�s Law
� Hess�s law states that if you can add two more thermochemical equations to produce a final equation for the reaction, then the sum of the enthalpy change for the individual reactions is the enthalpy change for the final reaction
Applying Hess�s Law
� Ways of applying Hess�s Law�..
EXAMPLE: calculate the energy reaction that produces SO3?
GIVEN: 2S(s) + 3O2( g) ----------> 2SO3(g) change in H = ?
A) S(s) + O2(g) ----------> SO2(g) change in H = -297 kJ
B) 2SO3(g) ----------> 2SO2(g) + O2(g) change in H = 198 kJ
Because the given equation is written with 2 mols of sulfur reacting, equation A must also be written with 2 mols of sulfur reacting. So, you must multiply all the coefficients in equation A by 2. The enthalpy change must also be doubled because 2 mols of sulfur will create twice the energy. Equation A now becomes equation C.
C) 2S(s) + 2O2(g)----------> 2SO2(g) change in H = -594 kJ
Equation B must be reversed because the desired product is SO3 , NOT the reactant. Remember that when you reverse an equation, the sign of change in H changes. The reverse of equation B is equation D.
D) 2SO2(g) + O2(g) ---------> 2SO3(g) change in H = -198 kJ
Now add equations C and D. Any terms that are common on both sides of the equati?on should be canceled.
C) 2S(s) + 2O2(g)----------> 2SO2(g) change in H = -594 kJ
+ D) 2SO2(g) + O2(g) ---------> 2SO3(g) change in H = -198 kJ
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2S(s) + 3O2(g) ----------> 2SO3(g) change in H = -792 kJ
For additional examples click HERE
Standard Enthalpy (Heat) of Formation
� The standard state of a substance means the normal physical state of the substance at one atmospheric pressure and at 298 K (25�C).
� The change in H value for these reactions is called the standard enthalpy (heat) of formation of the compound
� The standard enthalpy of formation (change in H�f) is based on an arbitrary standard
� Every free element in its standard state is assigned a standard enthalpy of formation of exactly 0.0 kJ
� This is similar to 0.0 �C being the arbitrary assignment for the freezing point of water
� The formation equations used to find the standard enthalpy of formation can be added together by applying Hess�s Law and find the enthalpy change for a reaction
EXAMPLE: Find the change in H�rxn for the equation H2S(g) + 4F2(g) -----------> 2HF(g) + SF6(g)
You are given the following equations
A) 1/2 H2(g) + 1/2 F2(g) ----------> HF(g) change in H�f = -273 kJ
B) S(s) + 3F2(g) -------------> SF6(g) change in H�f = -1220 kJ
C) H2(g) + S(s) ------------>H2S(g) change in H�f = -21 kJ
Because there are two moles of HF are required, equation A must be doubled.
A) H2(g) + F2(g) ----------> 2HF(g) change in H�f = -546 kJ
In the desired equation, H2S is a reactant. Therefore, equation C must be reversed.
C) H2S(g) --------------> H2(g) + S(s) change in H�f = 21 kJ
Now the three equations can be added.
A) H2(g) + F2(g) ----------> 2HF(g) change in H�f = -546 kJ
B) S(s) + 3F2(g) -------------> SF6(g) change in H�f = -1220 kJ
C) H2S(g) --------------> H2(g) + S(s) change in H�f = 21 kJ
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H2S(g) + 4F2(g) ------------> 2HF(g) + SF6(g) change in H�rxn = -1745 kJ
This process can be summed up in the equation
?H�rxn = the sum of the change in H�f (products) - the sum of the change in H�f (reactants)