In determining the direction of a chemical reaction we must rely on an understanding based on a number of factors, and contributions that are not always easy to assess. Although the assessment is prone to error, but it is usually reliable where it seems reasonable to try, and that is certainly to answer ignorance.
In predicting chemical reactions there are several known factors such as, if the free energy of a reaction is known, then there is no problem predicting a chemical reaction. Then if only enthalpy changes are known, then predictions usually apply to room temperature but are more or less reliable as well for higher temperatures. If the reaction occurring in the solution and the oxidation potential of the involved compound is known then the prediction is relatively simple, and this oxidation potential buys a rough guide for possible reactions in the absence of a solvent. If the equilibrium constant is known, relating to ΔG0 = - RH in K gives us a change of free energy. But information like this is still lacking, so we must rely on our understanding of the preceding principles.
For chemical reactions at room temperature the expected entropy changes are not so great that the relative strength of the reactants and the products produced will guide the course of the reaction. Here are the rules that may be useful and will be used.
The reaction tends to occur where the bonds of the orbitals and some of the electrons are available and allow for attractive tensile interactions between atoms.
There is the possibility of electrons to be divided, and it always happens with energy changes. Therefore we can predict with certainty that all atoms that have empty or full orbitals will join the ability of the atoms under the applicable conditions whether produced by the same or different elements. The only elements of atomic forests do not join in ordinary circumstances are those that do not contain atoms of low external energy ie: "inert" elements or helium groups. Even this, when the circumstances are created, then through the influence of highly electronegative elements such as fluorine can be united by chemical bonds.
When that possibility exists, the tendency of atoms to form strong bonds may be formed, if at room temperature or at higher temperatures, then on the outcome there is little influence from the circumstances. In predicting reactions at ordinary temperatures, we can consider the preceding principles or bond strength and seek to decide whether the total bond strength will be greater in the reactants or products. On this basis we can often make prediction rules for definite possibilities or on reactions.
The types of reactions that are predicted by this kind are most only some of them as follows:
(1) Synthesis - a direct combination of elements or compounds,
(2) Substitution - transfer of one element or compound, from excessive complex combinations with other elements or compounds, and
(3) Metathesis - double decomposition, or exchange partner.
Synthesis
Almost all the different elements in electronegativity, the bond between two different elements of eletronegaivitas almost always polar. As already indicated, that polar tends to be stronger than nonpolar bonds. The bonds between two different elements of their electonegativity tend to show the strength of the average bond or bond on the free element. The heat of formation in larger compounds of binary compounds is largely negative (exothermic), and generally it is better to have different electronegativity and resultant bond polarities.
Part of the lure of chemistry is that things don’t always work out the way you expect. You plan a reaction, anticipate the products and, quite often, the results astound you! The exercise, then, is trying to figure out what was formed, why, and whether your observation leads to other useful generalizations. The first step in this process of discovery is anticipating or predicting the products which are likely to be formed in a given chemical reaction. The guidelines we describe here will accurately predict the products of most classes of simple chemical reactions. As your experience in chemistry grows, however, you will begin to appreciate the unexpected!
In simple synthesis reactions involving reaction of elements, such as aluminum metal reacting with chlorine gas, the product will be a simple compound containing both elements. In this case, it is easiest to consider the common charges that the elements adopt as ions and build your product accordingly. Aluminum is a Group III element and will typically form a +3 ion. Chlorine, being Group VII, will accept one electron and form a monoanion. Putting these predictions together, the product is likely to be AlCl3. In fact, if aluminum metal and chlorine gas are allowed to react, solid AlCl3 is the predominant product.
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
The synthesis reaction involving the non-metals hydrogen gas and bromine can be approached similarly. The product will contain both elements. Hydrogen, Group I, has one valence electron and will form one covalent bond. Bromine, Group VII, has seven valence electrons and will form one covalent bond. The likely product is therefore HBr, with one covalent bond between the hydrogen and the bromine.
H2 (g) + Br2 (g) < → 2 HBr (g)
For a single-replacement reaction, recall that (in general) metals will replace metals and non-metals will replace non-metals. For the reaction between lead(IV) chloride and fluorine gas, the fluorine will replace the chlorine, leading to a compound between lead and fluorine and the production of elemental chlorine. The lead can be viewed as a “spectator” in the reaction and the product is likely to be lead(IV) fluoride. The complete equation is shown below.
PbCl4 (s) + 2 F2 (g) → PbF4 (s) + 2 Cl2 (g)
In single-replacement reactions in which metals (or carbon or hydrogen) are expected to replace metals, first you should check the activity series to see if any reaction is anticipated. Remember that metals can only replace metals that are less active than themselves (to the right in the Table). If the reaction is predicted to occur, use the same general guidelines that we used above. For example, solid iron reacting with aqueous sulfuric acid (H2SO4). In this reaction the question is whether iron will displace hydrogen and form hydrogen gas. Consulting the activity series, we see that hydrogen is to the right of iron, meaning that the reaction is expected to occur. Next, we reason that iron will replace hydrogen, leading to the formation of iron sulfate, where the sulfate is the “spectator” ion. The formation of hydrogen gas requires a change in oxidation number in the hydrogen of +1 to zero. Two hydrogen atoms must therefore be reduced (a decrease in oxidation number) and the two electrons required for the reduction must come from the iron. The charge on the iron is therefore most likely to be +2 (it starts off at zero and donates two electrons to the hydrogens). The final product is therefore most likely iron(II) sulfate. The complete equation is shown below.
Fe (s) + H2SO4 (aq) → FeSO4 (aq) + H2 (g)
Decomposition reactions are the most difficult to predict, but there are some general trends that are useful. For example, most metal carbonates will decompose on heating to yield the metal oxide and carbon dioxide.
NiCO3 (s) → NiO (s) + CO2 (g)
Metal hydrogen carbonates also decompose on heating to give the metal carbonate, carbon dioxide and water.
2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
Finally, many oxygen-containing compounds will decompose on heating to yield oxygen gas and “other compounds”. Identifying these compounds and building an understanding of why and how they are formed is one of the challenges of chemistry. Some examples:
H2O2 (aq) → O2 (g) + H2O (l)
2 HgO (s) → O2 (g) + 2 Hg (l)
2 KClO3 (s) → 3 O2 (g) + 2 KCl (s)
The potential products in double-replacement reactions are simple to predict; the anions and cations simply exchange. Remember, however, that one of the products must precipitate, otherwise no chemical reaction has occurred. For the reaction between lead(II) nitrate and potassium iodide, the products are predicted to be lead(II) iodide and potassium nitrate. No redox occurs, and the product, lead iodide, precipitates from the solution as a bright yellow solid. The question of how do you predict this type of solubility trend is addressed in the next section.
Pb(NO3)2 (aq) + 2 KI(aq)< → PbI2 (s) + 2 KNO3 (aq)
