A chemical equation is a symbolic representation of a chemical reaction. It is a formula used to display different stages for chemical reactions, where chemical substances are changed into other substances. The elements and/or compounds to the left of the arrow in a chemical equation represent the reactants, the arrow represents the transition stage, and the species to the right of the arrow represent the products. The four basic chemical equations are:
For example, the combustion of methane (in oxygen) is depicted as:
and the reversible reaction of the Haber process is shown as
A chemical equation should represent the stoichiometry observed in the chemical reaction. When the net amount of atoms on both sides of the equation is identical the equation is said to be a balanced equation.
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Simple chemical equations can be balanced by inspection, that is, by trial and error. Generally, it is best to balance the most complicated molecule first. Hydrogen and oxygen are usually balanced last.
Ex #1. Na + O2 → Na2O
In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Na's on the right. This problem is solved by putting a 2 in front of the Na on the left hand side:
In this equation there are 2 Na atoms on the left and 2 Na atoms on the right. In the next step the oxygen atoms are balanced as well. On the left hand side there are 2 O atoms and the right hand side only has one. This is still an unbalanced equation. To fix this a 2 is added in front of the Na2O on the right hand side. Now the equation reads:
Notice that the 2 on the right hand side is "distributed" to both the Na2 and the O. Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. The right hand side has 4 Na's total and 2 O's. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this 2 more Na's are added on the left side. The equation will now look like this:
This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides of the equation.
Ex #2. P4 + O2 → P4O10
This equation is not balanced because there is an unequal amount of O's on both sides of the equation. The left hand side has 4 P's and the right hand side has 4 P's. So the P atoms are balanced. The left hand side has 2 O's and the right hand side has 10 O's. To fix this unbalanced equation a 5 in front of the O2 on the left hand side is added to make 10 O's on both sides resulting in
The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.
Ex #3. C2H5OH + O2 → CO2 + H2O
This equation is more complex than the previous examples and requires more steps. The most complicated molecule here is C2H5OH, so balancing begins by placing the coefficient 2 before the CO2 to balance the carbon atoms.
Since C2H5OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H2O:
Finally the oxygen atoms must be balanced. Since there are 7 oxygen atoms on the right and only 3 on the left, a 3 is placed before O2, to produce the balanced equation:
In reactions involving many compounds, balancing may get harder, we can then try to balance equation using algebraic method, based on solving set of linear equations:
1. Assign variables to each coefficient:
2. We must have the same quantities of each atom in each side of the equation. So, for each element, count its atoms and equal both sides:
3. Solving the system (usually direct substitution is the best way)
which means that we have all coefficients depending on a parameter a, just choose a=1 (a number that will make all of them small whole numbers) and you'll have:
4. And the balanced equation at last:
To speed up the process, one can combine both methods to get a more practical algorithm:
1. Identify elements which occur in one compound in each member (this is very usual)
2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, let's say a.
3. Well, K2SO4 has to be 2a (because of K), and also, FeSO4 has to be 1a (because of Fe), CO has to be 6a (because of C) and (NH4)2SO4 has to be 3a (because of N). Well, this takes out the first four equations of the system! We already now that, whatever the coefficients are, those proportions must hold:
4. We can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) we could continue by noticing that adding the Sulfurs we get 6a for H2SO4 and finally by adding the hydrogens (or the oxygens) we get the lasting 6a for H2SO4.
5. Again, having a convenient value for a (in this case 1 will do, but if a gets fractionary values in the other coefficients you will like to cancel the denominators) we get the result:
Reaktionsgleichung | Ecuación química | Équation chimique | Equazione chimica | Mokokano | Kémiai egyenlet | 化学反応式 | Chemical equation | Ekwasyong kimikal | Хімічне рівняння
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