Hess’s Law

EQ: Why is Hess’s Law a useful tool in solving for ∆Hrxn?

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.

Develop an analogy for soccer and scoring a goal.

Develop an analogy for soccer and scoring a goal.

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

Practice problems

• Do Page 508 PPQ # 28-29