Analogy for Hess's Law
• There is an old Chinese proverb which says: There are many ways to the top of a mountain, but the view from the top is always the same.
State Functions
Red’s House
Grandma’s HouseThe pathway
doesn’t matter; the altitude change is
the same.
Hess’s Law
the heat evolved or absorbed in a chemical process is the same whether the process takes
place in one or several steps.
Hess’s Lawif two or more chemical equations can be added
together to produce an overall equation, the sum of the enthalpy equals the enthalpy change of the
overall equation.
B. How to Use Hess’s Law1. The goal is to obtain the desired reaction by adding up two or
more reactions.2. Work backwards. For each reaction, line up reactants and
products to the desired reaction. If they are not already lined up, flip the equation and change the sign of ΔH by multiplying by -1.
3. Skip reactants or products that appear in more than one reaction.
4. Make sure that the coefficients match. If they do not, multiply the coefficients of the entire reaction by the necessary number (sometimes a fraction) and also multiply ΔH by that number.
5. Always write down state symbols (s, l, g, aq). Some problems will ask that you simply cancel out certain states.
6. Cancel substances that are on both sides of the arrow.7. On occasion, multiple steps of multiplying and reversing the
reactions are needed to solve the problem.
Determine the heat of reaction for the reaction:
C2H4(g) + H2(g) C2H6(g)
Use the following reactions:
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ
C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) H = -1550 kJ
H2(g) + 1/2O2(g) H2O(l) H = -286 kJ
Determine the heat of reaction for the reaction:
C2H4(g) + H2(g) C2H6(g)
Use the following reactions:
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ
C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) H = -1550 kJ
H2(g) + 1/2O2(g) H2O(l) H = -286 kJ
1st check to see if they are lined up and then check coefficient
-1
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ
2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ
H2(g) + 1/2O2(g) H2O(l) H = -286 kJC2H4(g) + H2(g) C2H6(g) H = -137 kJ
Calculate H for this reaction:
6C(s)+6H2(g)+3O2(g)C6H12O6(s)
using the following three equations:
C (s)+O2(g)CO2 (g) H= -393.51 kJ
H2(g)+ ½O2(g)H2O(l) H= -285.83 kJ
C6H12O6(s)+6O2 (g)6CO2 (g)+6H2O(l) H= -2803.02 kJ
66
-1
6C (s)+6O2(g)6CO2 (g) H=-2361.06 kJ
6H2(g)+ 3O2(g)6H2O(l) H=-1714.98 kJ
6CO2 (g)+6H2O(l) C6H12O6(s)+6O2 (g) H= +2803.02 kJ
6C(s)+6H2(g)+3O2(g)C6H12O6(s) ∆H = -1273.02 kJ
Extra Problem
• Na(s) + ½Cl2(g) NaCl(s)
2Na(s) + 2HCl(g) 2NaCl(s) + H2(g) ΔH = –637.4 kJ
H2(g) + Cl2(g) 2HCl(g) ΔH = –184.6 kJ
With your partner, determine what each reaction should be multiplied by. Do not solve.
• Na(s) + ½Cl2(g) NaCl(s)
2Na(s) + 2HCl(g) 2NaCl(s) + H2(g) ΔH = –637.4 kJ
H2(g) + Cl2(g) 2HCl(g) ΔH = –184.6 kJ
½ ½
Top Related