Passage:Heats of Formation

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Germain Henri Hess, a swiss born, russian chemist, became famous for his formulation of what is now called Hess's Law, a principle fundamental to the development of thermochemistry. Simply stated, in a series of chemical reactions the total energy gained or lost depends only on the initial and final states, regardless of the number of steps. This is also known as the law of constant heat summation.

The law states that because enthalpy is a state function, the enthalpy change of a reaction is the same regardless of what pathway is taken to achieve the products. In other words, only the start and end states matter to the reaction, not the individual steps between. This allows the change in enthalpy for a reaction to be calculated even when it cannot be measured directly. Often enthalpy is recorded for numerous substances, as seen in figure 1, which can then be used as a reference point to determine the enthalpies of other unknown substances.

The concepts of Hess's law can be expanded to include changes in entropy and in free energy, which are also state functions. Such extensions are especially helpful because entropy and free energy are not measured directly, and therefore must be calculated through alternative paths.

Substance Formula ΔHfɵ /KJ.mol-1
Methane CH4 (g) -75
Methanol CH3OH (l) -238.6
Acetylene C2H2 (g) +226.7
Ethylene C2H4 (g) +52.3
Ethane C2H6 (g) -84.7
Carbon Dioxide CO2 (g) -394
Water H2O (l) -286

Alternatively, enthalpies can be given for a reaction pathway. In this case the enthalpy change represents the endothermic or exothermic properties of the reaction.

Reaction ΔHfɵ /KJ.mol-1
B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g) ΔH = +2035 kJ
H2O(l) → H2O(g) ΔH = +49 kJ
H2(g) + (1/2)O2(g) → H2O(l) ΔH = -293 kJ
2B(s) + 3H2(g) → B2H6(g) ΔH = +46 kJ


1. A reaction in which ΔH = +300 kJ/mol is considered to be?

Endothermic.
Exothermic.
→ ΔH is negative for exothermic reactions has it releases heat and lowers the final enthalpy of the system.
Spontaneous
→ ΔH alone says nothing about the spontaneity of a reaction.
Nonspontaneous

2. What is the enthalpy change for the reaction, 2B(s) + (3/2)O2(g) → B2O3(s)?

-1423 KJ/mol
+1423 KJ/mol
-2727 KJ/mol
→ {To answer this question we must use the four reactions given in the passage. Reversing the first (thus changing the sign of ΔH) and multiplying the second and third reactions by 3 (thus multiplying the ΔH by 3) will allow us to derive the overall reaction we are looking for. Thus if we sum these new reactions together we get the reaction in question and likewise we can do the same for the ΔH's.
+2727 KJ/mol

3. What type of reaction is 2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g)?

Decomposition
Isomerisation
Acid-Base
Displacement

4. What is the theoretical maximum mass of carbon dioxide that can be produced when 2 moles of acetylene undergoes combustion?

176 g
→ If we first need to determine what the reaction is. In this case, Acetylene, C2H2, will burn with O2 to form carbon dioxide and water. Because we are dealing with two carbons from the acetylene, it must create two moles of carbon dioxide for every one mole burned. 2 moles, thus would burn to create 4 moles of carbon dioxide, which, when we convert to grams using the MW of CO2 (44 g/mol) we get 176g.
88 g
44 g
22 g

5. How will a reaction behave if its change in enthalpy is positive while its change in entropy is negative?

It will always proceed.
It will never proceed.
→ Looking at ΔG = ΔH - TΔS we can see that it will always be positive regardless of what values are put in as long as ΔH is positive and ΔS is negative.
It will proceed at low temperature.
It will proceed at high temperature.

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