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 House The pathway doesn’t matter; the altitude change is the same.

Develop an analogy for soccer and scoring a goal.

the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or several steps.

if 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 Law 1.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 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: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Use the following reactions: C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ

Determine the heat of reaction for the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Use the following reactions: C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ 1 st check to see if they are lined up and then check coefficient

C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ 2CO 2 (g) + 3H 2 O(l)  C 2 H 6 (g) + 7/2O 2 (g)  H = kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = -137 kJ

Calculate  H for this reaction: 6C(s)+6H 2 (g)+3O 2 (g)  C 6 H 12 O 6 (s) using the following three equations: C (s)+O 2 (g)  CO 2 (g)  H  = kJ H 2 (g)+ ½O 2 (g)  H 2 O(l)  H  = kJ C 6 H 12 O 6 (s)+6O 2 (g)  6CO 2 (g)+6H 2 O(l)  H  = kJ 6 6

6 C (s)+ 6 O 2 (g)  6 CO 2 (g)  H  = kJ 6 H 2 (g)+ 3 O 2 (g)  6 H 2 O(l)  H  = kJ 6CO 2 (g)+6H 2 O(l)  C 6 H 12 O 6 (s)+6O 2 (g)  H  = kJ 6C(s)+6H 2 (g)+3O 2 (g)  C 6 H 12 O 6 (s)∆H = kJ

Extra Problem Na(s) + ½Cl 2 (g)  NaCl(s) 2Na(s) + 2HCl(g)  2NaCl(s) + H 2 (g) ΔH = –637.4 kJ H 2 (g) + Cl 2 (g)  2HCl(g) ΔH = –184.6 kJ With your partner, determine what each reaction should be multiplied by. Do not solve.

Na(s) + ½Cl 2 (g)  NaCl(s) 2Na(s) + 2HCl(g)  2NaCl(s) + H 2 (g) ΔH = –637.4 kJ H 2 (g) + Cl 2 (g)  2HCl(g) ΔH = –184.6 kJ ½ ½

Practice problems Do Page 508 PPQ # 28-29