The basis for calculating enthalpies of reaction is known as Hess’s law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process. This means that the energy difference between reactants and products is independent of the route taken to get from one to the other.

If you know the reaction enthalpies of individual steps in an overall reaction, you can calculate the overall enthalpy without having to measure it experimentally.

-137KJ +125KJ +87KJ -193KJ +102KJ -163KJ +52KJ -147KJ -269KJ +7KJ Regardless of the route the climber and the miner took they ended up having the same amount of potential energy!!

5 Hess’s Law Start Finish A State Function: Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.

∆H target = ∑∆H known ∆H target =∆H rxn1 +∆H rxn2 +∆H rxn Hess’s Law Equation:

For example: C + O 2  CO 2 occurs as 2 steps C + ½O 2  CO  H  = – kJ CO + ½O 2  CO 2  H  = – kJ C + CO + O 2  CO + CO 2  H  = – kJ C + O 2  CO 2  H  = – kJ

 What is the enthalpy change for the formation of two moles of nitrogen monoxide from its elements? This reaction may be called the target equation to distinguish it clearly from other equations N 2 (g) + O 2 (g) 2NO (g) 1.½ N 2 (g) + O 2 (g) NO 2 (g) ΔH 1 θ = +34 kJ 2.NO(g) + ½ O 2 (g) NO 2 (g) ΔH 2 θ = - 56 kJ

If we work with these two equations, which may be called known equations, and then add them together, we obtain the chemical equation for the formation of nitrogen monoxide N 2 (g) + O 2 (g) 2NO (g) If we look at the target eqn., it has 1 mole of N 2 as reactant. How do we make it 1 mole?

2 x (½ N 2 (g) + O 2 (g) NO 2 (g) ΔH 1 θ = 2(+34 )kJ 2 x (NO 2 (g) NO(g) + ½ O 2 (g) ΔH 2 θ = 2(+ 56 )kJ N 2 (g) + 2O 2 (g) 2 NO 2 (g) ΔH 1 θ = 2(+34 )kJ 2 NO 2 (g) 2NO(g) + O 2 (g) ΔH 2 θ = 2(+ 56 )kJ N 2 (g) + 2O 2 (g) + 2NO 2 (g) 2NO 2 (g) + 2NO(g) + O 2 (g) N 2 (g) + O 2 (g) 2NO(g)

∆H θ = 2(+ 34) kJ + 2(+ 56) kJ = + 68 kJ kJ ∆H θ = kJ

To demonstrate how to apply Hess’s law, we will work through the calculation of the enthalpy of formation for the formation of methane gas, CH 4, from its elements, hydrogen gas and solid carbon: C(s) + 2H 2 (g) → CH 4 (g)

The component reactions in this case are the combustion reactions of carbon, hydrogen, and methane: H 2 (g) + ½O 2 (g) → H 2 O(l) C(s) + O 2 (g) → CO 2 (g) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) The overall reaction involves the formation rather than the combustion of methane, so the combustion equation for methane is reversed, and its enthalpy changed from negative to positive: CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ∆ H 0 = kJ

Because 2 moles of water are used as a reactant in the above reaction, 2 moles of water will be needed as a product. Therefore, the coefficients for the formation of water reaction, as well as its enthalpy, need to be multiplied by 2: 2H 2 (g) + O 2 (g) → 2H 2 O(l) We are now ready to add the three equations together using Hess’s law to give the enthalpy of formation for methane and the balanced equation. 2H 2 (g) + O 2 (g) → 2H 2 O(l) C(s) + O 2 (g) → CO 2 (g) C(s) + 2H 2 (g) → CH 4 (g) CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ΔH θ = kJ

Using Hess’s law, any enthalpy of reaction may be calculated using enthalpies of formation for all the substances in the reaction of interest, without knowing anything else about how the reaction occurs. Mathematically, the overall equation for enthalpy change will be in the form of the following equation: ∆ H 0 = sum of [( of products) × (mol of products)] – sum of [( of reactants) × (mol of reactants)]

Hess’s law allows us to add equations. We add all reactants, products, &  H  values. We can also show how these steps add together via an “enthalpy diagram” …

Exothermic Enthalpy Diagram

Endothermic Enthalpy Diagram

Steps in drawing enthalpy diagrams 1.Balance the equation(s). 2.Sketch a rough draft based on  H  values. 3.Draw the overall chemical reaction as an enthalpy diagram (with the reactants on one line, and the products on the other line).

4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line. 5. Check arrows: Start: two leading away Finish: two pointing to finish Intermediate: one to, one away 6.Look at equations to help complete balancing (all levels must have the same # of all atoms). 7.Add axes and  H  values.

C + O 2  CO 2  H  = – kJ Reactants Intermediate Products C + O 2 CO 2 CO Enthalpy Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.  H  = – kJ  H  = – kJ  H  = – kJ + ½ O 2 C + ½ O 2  CO  H  = – kJ CO + ½ O 2  CO 2  H  = – kJ

How much energy is lost in the formation of C 2 H 5 OH (l) ? C 2 H 4(g) + H 2 O (l)  C 2 H 5 OH (l) Given: 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 5 OH (l) + 3O 2(g)  H  = kJ

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 5 OH(l) + 3O 2 (g)  H  = kJ C 2 H 4 (g) + H 2 O(l)  C 2 H 5 OH(l) Reactants Products Intermediate C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) 2CO 2 (g) + 3H 2 O(l) Enthalpy  H  = – kJ  H  = kJ  H  = – 44.0 kJ + 3O 2 (g)  H  = – 44.0 kJ Solution:

Two Rules to Follow: 1. If a chemical equation is reversed, then the sign of ∆H changes 2. If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the value of ∆H is altered in the same way

We may need to manipulate equations further: 2Fe + 1.5O 2  Fe 2 O 3  H  =?, Given: Fe 2 O 3 + 3CO  2Fe + 3CO 2  H  = – kJ CO + ½ O 2  CO 2  H  = – kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 1.5O 2  Fe 2 O 3 3CO O 2  3CO 2  H  = – kJ 2Fe + 3CO 2  Fe 2 O 3 + 3CO  H  = kJ CO + ½ O 2  CO 2  H  = – kJ  H  = – kJ Flip equation =Flip sign Multiply coefficients and values

Calculate the enthalpy of reaction for the combustion of nitrogen monoxide gas, NO, to form nitrogen dioxide gas, NO 2, as given in the following equation. NO(g) + ½O 2 (g) → NO 2 (g) Use the enthalpy-of-formation data. Solve by combining the known thermochemical equations. Given: ½ N 2 (g) + ½ O 2 (g) NO(g) ∆H f θ kJ ½ N 2 (g) + O 2 (g) NO 2 (g) ∆H f θ kJ Unknown: ∆H θ for NO(g) + ½ O 2 (g) NO 2 (g) Solution: Using Hess’s law, combine the given thermochemical equations in such a way as to obtain the unknown equation, and its ∆ H 0 value.

The desired equation is: The other equation should have NO 2 as a product, so we can use the second given equation as is: Reversing the first given reaction and its sign yields the following thermochemical equation:

We can now add the equations and their ∆H 0 values to obtain the unknown ∆H 0 value.

When carbon is burned in a limited supply of oxygen, carbon monoxide is produced: The above overall reaction consists of two reactions: 1) carbon is oxidized to carbon dioxide 2) carbon dioxide is reduced to give carbon monoxide.

Because these two reactions occur simultaneously, it is not possible to directly measure the enthalpy of formation of CO(g) from C(s) and O 2 (g). We do know the enthalpy of formation of carbon dioxide and the enthalpy of combustion of carbon monoxide: We reverse the second equation because we need CO as a product. Adding gives the desired enthalpy of formation of carbon monoxide.

Determine the heat of reaction for the reaction: Target  4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g)  H= ? Using the following sets of reactions: (1) N 2 (g) + O 2 (g)  2NO(g)  H = kJ (2) N 2 (g) + 3H 2 (g)  2NH 3 (g)  H = kJ (3) 2H 2 (g) + O 2 (g)  2H 2 O(g)  H = kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the  H values must be treated accordingly.

4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Using the following sets of reactions: (1) N 2 (g) + O 2 (g)  2NO(g)  H = kJ (2) N 2 (g) + 3H 2 (g)  2NH 3 (g)  H = kJ (3) 2H 2 (g) + O 2 (g)  2H 2 O(g)  H = kJ Goal: NH 3 : O2 O2 : NO: H 2 O: (2)( Reverse and x 2) 4NH 3  2N 2 + 6H 2  H = kJ Found in more than one place, SKIP IT (its hard). (1) ( Same x2) 2N 2 + 2O 2  4NO  H = kJ (3)( Same x3 ) 6H 2 + 3O 2  6H 2 O  H = kJ

4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Goal: NH 3 : O 2 : NO: H 2 O: Reverse and x2 4NH 3  2N 2 + 6H 2  H = kJ Found in more than one place, SKIP IT. x2 2N 2 + 2O 2  4NO  H = kJ x3 6H 2 + 3O 2  6H 2 O  H = kJ Cancel terms and take sum. 4NH 3 + 5O 2  4NO + 6H 2 O  H = kJ Is the reaction endothermic or exothermic?  H = kJ kJ + (-1451kJ)

Determine the heat of reaction for the reaction: TARGET  C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = ? Use the following reactions: (1) C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ (2) C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = kJ (3) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ Consult your neighbour if necessary.

Determine the heat of reaction for the reaction: Goal: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = ? Use the following reactions: (1) C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ (2) C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = kJ (3) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 4 (g) :use 1 as is C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ H 2 (g) :# 3 as is H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 6 (g) : rev #2 2CO 2 (g) + 3H 2 O(l)  C 2 H 6 (g) + 7/2O 2 (g)  H = kJ C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = -137 kJ

Hess’s Law Example Problem Calculate  H for the combustion of methane, CH 4 using the equations above Ans = kJ Reaction  H f o C + 2H 2  CH kJ C + O 2  CO kJ H 2 + ½ O 2  H 2 O kJ