I II III
Chapter 16 Hess’s Law
HESS’S LAW
If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.
Hess’s law can be used to determine the
enthalpy change for a reaction
that cannot be measured directly!
HNET = H1 + H2
N2(g) + O2(g) 2NO(g) ΔH1= +181kJ
2NO(g) + O2(g) 2NO2(g) ΔH2= -113kJ
ADD THEM UP ALEGBRAICALLY
N2(g) + O2(g) 2NO(g) ΔH1= +181kJ
2NO(g) + O2(g) 2NO2(g) ΔH2= -113kJ
N2(g) + 2O2(g)
First, add up the chemical
equations.
+ 2NO(g) 2NO(g) + 2NO2(g)
Notice that 2NO(g) is on both the reactants and
products side and can be cancelled
out.N2(g) + O2(g) 2NO(g) ΔH1= +181kJ
2NO(g) + O2(g) 2NO2(g) ΔH2= -113kJ
N2(g) + 2O2(g) + 2NO(g) 2NO(g) + 2NO2(g)
Write the net equation:
N2(g) + 2O2(g) + 2NO(g) 2NO(g) + 2NO2(g)
N2(g) + 2O2(g) 2NO2(g)
HNET = H1 + H2
ΔH1= +181kJ
ΔH2= -113kJ
Apply Hess’s Law to calculate the enthalpy for the
reaction.
HNET = H1 + H2
ΔHNET = (+181kJ) + (-113kJ)
ΔHNET = +68kJOverall, the formation of NO2
from N2 and O2 is an endothermic process, although one of the steps is exothermic.
ΔH
Reaction Progress
N2(g) + 2O2(g)
2NO(g) + O2(g)
2NO2(g)
ΔHNET = +68kJ
ΔH1 = +181kJ
ΔH2 = -113kJ
RULES for Hess’s Law Problems
1.If the coefficients are multiplied by a factor, then the enthalpy value MUST also be multiplied by the same factor.
2.If an equation is reversed, the sign of ΔH MUST also be reversed.
C(s) + ½O2(g) CO(g) ΔH1= -110.5kJCO(g) + ½O2(g) CO2(g) ΔH2= -283.0kJ
C(s) + O2(g) + CO(g) CO(g) + CO2(g)
Practice Problem: #1
C(s) + O2(g) CO2(g)
HNET = H1 + H2
HNET = (-110.5kJ) + (-283.0kJ)HNET = -393.5kJ
Net Equation
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)
Practice Problem: #3
CH3OCH3(l) + 3O2(g) 2CO2(g) + 3H2O(g)
ΔH1= -1234.7kJ
ΔH2= -1328.3kJ
You have to REVERSE equation 2to get the NET equation.
DON’T forget to change the sign Of ΔH2
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)
Practice Problem: #3
2CO2(g) + 3H2O(g) CH3OCH3(l) + 3O2(g)
ΔH1= -1234.7kJ
ΔH2= +1328.3kJ
C2H5OH(l) + 3O2(g) + 2CO2(g) + 3H2O(g) 2CO2(g) +
3H2O(g) + CH3OCH3(l) + 3O2(g)
Net EquationC2H5OH(l) CH3OCH3(l)
Net EquationC2H5OH(l) CH3OCH3(l)
HNET = H1 + H2
HNET = (-1234.7kJ) + (+1328.3kJ)
HNET = +93.6kJ
H2(g) + F2(g) 2HF(g) ΔH1= -542.2kJ2H2(g) + O2(g) 2H2O(g) ΔH2= -571.6kJ
Practice Problem: #5
You have to REVERSE equation 2to get the NET equation.
DON’T forget to change the sign Of ΔH2
H2(g) + F2(g) 2HF(g) ΔH1= -542.2kJ
2H2O(g) 2H2(g) + O2(g) ΔH2= +571.6kJ
Practice Problem: #5
You will need to multiply the first equation by 2.
DON’T forget to multiply the ΔH by 2 also.
2H2(g) + 2F2(g) 4HF(g) ΔH1= -1084.4kJ2H2O(g) 2H2(g) + O2(g) ΔH2= +571.6kJ
Practice Problem: #5
Net Equation
2H2(g) + 2F2(g) + 2H2O(g) 4HF(g) + 2H2(g) + O2(g)
2F2(g) + 2H2O(g) 4HF(g) + O2(g)
HNET = H1 + H2
HNET = (-1084.4kJ) + (+571.6kJ)HNET = -512.8kJ
Hess’s Law
Start Finish
Enthalpy is Path independent.
Both lines accomplished the same result, they went from start to finish. Net result = same.
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