Chapter 16 Hess’s Law

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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.