Some H o f values are tabulated in Table 8.2 (above)

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1 8.10 Standard Heats of Formation We cannot have a table for the H values for every reaction there is, because there are too many of them. However, as we saw with Hess s Law, we can express any reaction as a set of step reactions, and calculate the enthalpy changes based on the enthalpy changes of sum of the steps. It makes sense to define a standard set of formation reactions. The standard heat of formation ( H f ) is the enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states. Some H o f values are tabulated in Table 8.2 (above) From the table we can state that: The standard molar enthalpy of formation of water (liq) at 1atm and 25 o C refers to the equation: H 2 (g) (1 atm, 25 o C) + ½ O 2 (g) (1 atm, 25 o C) H 2 O(l) (1 atm, 25 o C) 1mole of water has a value H o f ( H 2O) = -285 kj/mol For MgO(s): (from another H o f table ) Mg(s) + ½ O 2 (g) ----> MgO(s) (MgO) = -601 kj/mol H o f How about H o f for O 2 and other elements? H f for an element = 0 by definition. H f (O 2 ) = O 1

2 We use (a) standard enthalpies of formation ( H o f ), (b) along with Hess Law to calculate the enthalpy for any reaction ( H o reaction ) It can be shown using Hess Law that H o for a given reaction is equal to the sum of ( H o for the products) minus the sum of ( H o f for the reagents). H 0 reaction = Ho f (products) - Ho f (reagents) For the general reaction: a A + b B c C + d D we will find H Reaction = [c H f (C) + d H f (D) ] [a H f (A) + b H f (B)] Problem E: Calculate H o for the combustion of propane according to the equation: C 3 H 8 (g) + 5 O 2 (g) ----> 3 CO 2 (g) + 4 H 2 O (l) Given: H o f (C 3 H 8 ) = kj H o f (CO 2 ) = kj H o f (H 2O) = kj Ans: H o reaction = Ho f (products) - Ho f (reagents) = [3x H o f (CO 2 ) + 4 x Ho f (H 2 O) ] [ Ho f (C 3 H 8 ) + Ho f (O 2 ) ] = [3 ( kj) + 4 ( kj)] - [( kj + 0 )] = kj 2

3 Problem 8.16 Use the information in Table 8.2 to calculate H (in kj) for the reaction of ammonia and O 2 to yield nitric oxide (NO) and H 2 O (g). We haven t been given a balanced equation, so we need to figure this out. Show for yourself that: 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) Ans: solution to problem done in class 8.11 Bond Dissociation Energies One problem with trying to find values of H o is that only about a few thousand values of H o f are known and there are over 18 million chemical compounds. Since the energy of a chemical reaction involves the breaking of bonds in the reactants and the formation of new bonds to create products, we can use values of bond dissociation energies, D, and Hess s Law to determine approximate values of H o f for any compound. Average bond dissociation energies are tabulated in table 7.1 on page

4 The bond dissociation energy, D, is the value of H o for the reaction X Y ----> X + Y H o ~ D Values are always +ve since energy must always be supplied to break a bond. Values are slightly variable since the energy of a given bond depends also on what other bonds are involved with the given atomic pair. Values of H o calculated using values of D are only approximate. Using Hess s Law, it can be shown that H o = (D (bonds broken)) - (D (bonds formed) Example:Using bond energies calculate the H o for: C(s) + H 2 O(g) > CO(g) + H 2 (g) Ans: -- you need to break 2 H-O bonds and create 1 C-O and 1 H-H bond -- using D values from Table 7.1 and -- H o = (D(bonds broken)) - (D(bonds formed) = 2(460kJ) - (1(350kJ) + 1(436kJ)) = 134kJ 4

5 8.12 Fossil Fuels, Fuel Efficiency and Heats of Combustion A very common reaction is combustion of fuel to produce heat energy 2 C 8 H 18 (g) + 25 O 2 (g) -----> 16 CO 2 (g) + 18 H 2 O(g) H o c = kj H 2 (g) + ½ O 2 (g) -----> H 2 O(g) H o c = kJ/mol Values of the enthalpy of combustion, H o c are given in Table 8.3 in kj/ mol and also in kj/ g or kj/ ml. Fuels are chosen for their energy content and also for their practicality in use, e.g. liquid fuels in cars (mass and volume) and hydrogen in a rocket where weight is important...hydrogen has the highest energy/unit mass, With the exception of hydrogen, all common fuels are organic compounds where the energy in the compound was initially created by the sun via photosynthetic production of carbohydrates in green plants. The net reaction is the conversion of carbon dioxide and water into glucose and oxygen. 6CO 2 (g) + 6H 2 O -----> C 6 H 12 O 6 (s) + 6O 2 (g) H o = 2816kJ The glucose is used to synthesize cellulose and starch which is used to form structural parts of the plants and also as food for animals. The production of glucose is highly endothermic and requires a lot of solar energy. Estimates give the amount of solar radiation absorbed by plants as about kj which would produce about 5x10 14 kg of glucose per year! 5

6 Fossil fuels, coal, petroleum and natural gas, are the decayed remains of ancient organisms and plants and are complex mixtures of compounds. Coal is similar to graphite (C) and is derived primarily from plant material, petroleum is a viscous liquid of marine origin, composed primarily of hydrocarbons. Coal is normally burned just as it is mined but petroleum is refined using distillation and cracking techniques. So-called straight run gasoline is the distillation fraction boiling between about 30 and 200 o C and contains C5 to C11 hydrocarbons Kerosene (diesel fuel) is the C11 to C14 fraction boiling between about 175 to 300 o C gas oil is the C14 to C25 fraction ( o C) Lubricating oils and waxes contain the higher molecular weight hydrocarbons that will still distill at reasonable temperatures. The left over materials (asphalt) is melted and used to tar roads or is broken down (cracked) into smaller, more volatile materials using the appropriate catalysts and procedures. As deposits of petroleum become depleted, other fuels must be found...hydrogen, methanol, ethanol, etc. Methanol can be produced from natural gas (still need a supply of that) Ethanol can be produced by fermentation of plant material which is abundant. Non-combustion alternatives are also being investigated (fuel cell technology to produce electricity for electric motors) 6

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