Conservation of Energy in chemical reactions, Hess’s Law (Text reference 17.4)
Consider the following reaction:
C (s, graphite) + O2(g) ---> CO2(g) H° = -394 kJ
What kind of change is occurring, and what elements are involved?
(By the way, notice the presence of the sign, °, on the enthalpy. This indicates that the reaction is
happening under standard conditions.)
Now consider this reaction:
CO2(g) ---> C (s, graphite) + O2(g) H° = ???kJ
What kind of change is occurring this time, and what elements are involved?
What is the critical difference?
What would H be in the second reaction? _____
Why does this make sense?
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There are many chemical reactions that are difficult to study directly because the energy produced is very
high, or because the reactants are difficult to obtain or handle. Hess’s Law (named after the scientist who
proposed it) helps us to calculate H for a reaction by using data from other reactions that are related to
it. Hess’s Law states:
If a chemical equation can be written as the ‘sum’ of several other chemical equations, the
enthalpy change of the first chemical equation equals the sum of the enthalpy changes of
the other chemical equations.
Here is an example of how Hess’s Law can be used to figure out the H for a nearly impossible reaction
to do in the lab… forming diamonds from ordinary graphite ! Here is the equation:
C (s, graphite) ---> C (s, diamond) H° = ??? kJ
Hess' Law offers a way out. If we could show this reaction as the sum of a sequence of other reactions,
then we can add their enthalpies to get what we want. Here are two known reactions we can “add”:
C (s, graphite) + O2(g) ---> CO2(g)
H° = -394 kJ
C (s, diamond) + O2(g) ---> CO2(g) H° = -396 kJ
Look at each one of them and see that their reactants and products have the parts we are interested in
combining into a new equation. Notice that both reactions are exothermic. And notice that both equations
have some of the same formulas.
Now look at what happens when you ‘reverse’ the second equation. This puts C(s, diamond) on the product
side.
C (s, graphite) + O2(g) ---> CO2(g)
H° = -394 kJ
CO2(g) ---> C (s, diamond) + O2(g) H° = +396 kJ
Notice the other change that occurs. The H° value changes from positive to negative. This is an
absolute requirement when using Hess' Law. Reversing an equation means reversing the sign on
the enthalpy value. If a reaction is exothermic in the forward direction, it must be endothermic in the
reverse direction.
Now combine these two equations, including the enthalpies. Here is the combined equation:
CO2(g) + C (s, graphite) + O2(g) ---> CO2(g) + C (s, diamond) + O2(g) H° = (-394 kJ) + (+396 kJ)
Can you find formulas that are the same on both sides? If so, cancel them. (This is similar to cancelling
out spectator ions when you write net ionic equations):
CO2(g) + C (s, graphite) + O2(g) ---> CO2(g) + C (s, diamond) + O2(g) H° = (-394 kJ) + (+396 kJ)
Re-write what’s left:
C (s, graphite) ---> C (s, diamond) H° = +2 kJ
We now have the value H for an experiment by building it from two other relatively easy ‘experiments’.
And we avoid performing a tricky, expensive, possibly dangerous experiment.
Lab Application:
Determine the value for the molar heat of reaction for the combustion of magnesium:
Mg(s) + ½ O2(g) ---> MgO(s)
(equation # 1)
You might remember this reaction – it is highly exothermic, so we really can’t use ordinary calorimetry
methods to determine ΔH. So we will determine the heats of reaction for several other reactions that will
then be used to “add up” ΔH values. The 3 other reactions are:
Mg(s) + 2HCl(aq) ---> MgCl2(aq) + H2(g)
(equation #2)
MgO(s) + 2HCl(aq) ---> MgCl2(aq) + H2O(l)
(equation #3)
H2(g) + ½ O2(g) ---> H2O(l)
(equation #4)
Procedure
Reaction for Equation #2: Magnesium metal with Hydrochloric Acid
1) Obtain approx. 0.5 g of magnesium ribbon. Polish the ribbon with steel wool, and then obtain the
exact mass. (Coil the ribbon loosely.)
2) Place 100.0-mL of 1.0M HCl into a coffee-cup calorimeter. Measure carefully. Since the density of
the acid is nearly the same water, we will accept this as being 100.0g. Record the initial
temperature of the acid.
3) While stirring gently with the thermometer, add the Mg ribbon to the acid.
4) Make sure that the contents of the calorimeter are evenly mixed, and that the Mg reacts
completely. Record the highest temperature reached in your data table.
5) Discard the solution down the drain with the water running. Rinse the calorimeter and the
thermometer.
Mass of Mg used:
Mass of HCl in the calorimeter:
Initial temp. of HCl
Final temp. of HCl
______ g
______ g (dilute HCl has the same density as H2O)
______ °C
______ °C
Reaction for Equation #3: Magnesium Oxide with Hydrochloric Acid
6) Obtain approximately 1.00-g of magnesium oxide. Use a weighing boat or other appropriate
container. Record the exact mass you use.
7) Now, repeat steps 2 - 5 above with the MgO instead of the Mg.
Mass of MgO used:
Mass of HCl in the calorimeter:
Initial temp. of HCl
Final temp. of HCl
______ g
______ g (dilute HCl has the same density as H2O)
______ °C
______ °C
Equation #4: Hydrogen gas with Oxygen gas
8) For safety reasons, this reaction will not be done in class. The Heat of Formation of water (called
Hfo) has been professionally determined, and the value is listed in standard reference sources.
You can find this value in your textbook, pg. 530. Be sure to use the value for H2O(l), as seen in
the 4th equation above.
Analysis: Complete calculations neatly in the spaces provided.
1) Calculate H values. Remember that dilute HCl has the same thermal properties as water.
For Reaction #2:
Energy transferred in the calorimeter:
Moles Mg reacted:
H, in kJ/mol (be sure to attach the algebraic sign):
Fill-in the equation:
Mg(s) + 2HCl(aq) ---> MgCl2(aq) + H2(g)
ΔH = ______ kJ/mol
For Reaction #3:
Energy transferred in the calorimeter:
Moles MgO reacted:
H, in kJ/mol (be sure to attach the algebraic sign):
Fill-in the equation:
MgO(s) + 2HCl(aq) ---> MgCl2(aq) + H2O
(l)
ΔH = ______ kJ/mol
2) Look up the value of H for Reaction #4. It is the Hf for H2O(l) found in your textbook.
Fill-in the equation:
H2(g) + ½ O2(g) ---> H2O(l) )
ΔH = ______ kJ/mol
3) Look back at equation #1 again. Each of its reactants and products is found somewhere in the
other 3 equations. We are looking to have Mg and ½ O2 as the only reactants, and MgO as the
only product. All of the other formulas will be cancelled when the equations are written correctly.
Re-write the equations for reactions 2, 3, and 4 (along with the H values – watch algebraic signs!)
so that they can be ‘added up’ to produce Equation 1. This requires you to think about which
equation(s) to reverse.
2)
3)
4)
Mg(s) + ½ O2(g) ---> MgO(s)
ΔH = ______ kJ/mol
4) Calculate percent of error for your result. The accurate value is
597.7 kJ/mol.