Chapter 16 Hess’s Law
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Transcript of Chapter 16 Hess’s Law
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I II III
Chapter 16 Hess’s Law
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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.
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Hess’s law can be used to determine the
enthalpy change for a reaction
that cannot be measured directly!
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HNET = H1 + H2
N2(g) + O2(g) 2NO(g) ΔH1= +181kJ
2NO(g) + O2(g) 2NO2(g) ΔH2= -113kJ
ADD THEM UP ALEGBRAICALLY
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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)
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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)
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Write the net equation:
N2(g) + 2O2(g) + 2NO(g) 2NO(g) + 2NO2(g)
N2(g) + 2O2(g) 2NO2(g)
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HNET = H1 + H2
ΔH1= +181kJ
ΔH2= -113kJ
Apply Hess’s Law to calculate the enthalpy for the
reaction.
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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.
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ΔH
Reaction Progress
N2(g) + 2O2(g)
2NO(g) + O2(g)
2NO2(g)
ΔHNET = +68kJ
ΔH1 = +181kJ
ΔH2 = -113kJ
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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.
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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
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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
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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)
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Net EquationC2H5OH(l) CH3OCH3(l)
HNET = H1 + H2
HNET = (-1234.7kJ) + (+1328.3kJ)
HNET = +93.6kJ
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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
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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.
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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
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Hess’s Law
Start Finish
Enthalpy is Path independent.
Both lines accomplished the same result, they went from start to finish. Net result = same.