Chemical equation

A chemical equation is a symbolic representation of a chemical reaction, wherein one set of substances, called the reactants, is converted into another set of substances, called the products. The reactants and products are shown using their chemical formulas, and an arrow is used to indicate the direction of the reaction. The reactants are usually placed to the left of the arrow, and the products are placed to the right. If the reaction is irreversible, a single arrow is used; if the reaction is reversible, a double arrow (pointing in opposite directions) is used. The first chemical equation was diagrammed by Jean Beguin in 1615.

Examples

The combustion of methane in oxygen may be shown by the following equation:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

This equation represents an irreversible reaction in which one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

The reaction of sodium metal with oxygen produces sodium oxide, as follows:

4Na + O₂ → 2Na₂O

In this equation, four sodium atoms react with one oxygen molecule to produce two molecules of sodium oxide.

The synthesis of ammonia from nitrogen and hydrogen by the Haber process is a reversible reaction:

N₂(g) + 3H₂(g) ↔ 2NH₃(g)

The parenthetical "g" after a substance indicates that that substance is in the gaseous form.

Reading chemical equations

When reading a chemical equation, several points need to be considered:

• Each side of the equation represents a mixture of chemicals. The mixture is written as a set of chemical formulas of the atoms and molecules involved in the reaction, separated by + symbols.

• The two sides of the equation are separated by an arrow. If the reaction is irreversible, a right-arrow (→) is used, indicating that the left side represents the reactants (mixture of chemicals before the reaction) and the right side represents the products (mixture obtained after the reaction). For a reversible reaction, a two-way arrow is used.

• The formula of each reactant and product is normally preceded by a scalar number called the
  

  Vapors of hydrogen chloride in a beaker and ammonia in a test tube meet to form a cloud of a new substance, ammonium chloride.

stoichiometric number or stoichiometric coefficient. (The absence of a scalar number implies that the number is 1.) The stoichiometric numbers indicate the relative quantities of the molecules (or moles) participating in the reaction. For instance, the string 2H₂O + 3CH₄ represents a mixture containing 2 molecules of H₂O for every 3 molecules of CH₄.

• A chemical equation does not imply that all reactants are consumed in a chemical process. For instance, a limiting reactant determines how far a reaction can go.

Balancing chemical equations

In a chemical reaction, the quantity of each element does not change. Thus, each side of the equation must represent the same quantity of any particular element. In other words, the number of atoms of a given element in the products must equal the number of atoms of that element in the reactants. This is known as the "conservation of mass" in a chemical reaction. The process of equalizing these numbers in a chemical equation is known as "balancing the equation."

Also, in case of net ionic reactions the same charge must be present on both sides of the hiddly unbalanced equation, one may balance it by changing the scalar number for each molecular formula.

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:

2Na + O₂ → Na₂O

In this 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:

2Na + O₂ → 2Na₂O

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:

4Na + O₂ → 2Na₂O

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. 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.

P₄ + O₂ → 2P₂O₅

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

P₄ + 5O₂ → 2P₂O₅

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.

C₂H₅OH + O₂ → 2CO₂ + H₂O

Since C₂H₅OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H₂O:

C₂H₅OH + O₂ → 2CO₂ + 3H₂O

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:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

Linear system balancing

In reactions involving many compounds, equations may then be balanced using an algebraic method, based on solving set of linear equations.

1. Assign variables to each coefficient. (Coefficients represent both the basic unit and mole ratios in balanced equations.):

• a K₄Fe(CN)₆ + b H₂SO₄ + c H₂O → d K₂SO₄ + e FeSO₄ + f (NH₄)₂SO₄ + g CO

2. There must be the same quantities of each atom on each side of the equation. So, for each element, count its atoms and let both sides be equal.

• K: 4a = 2d
• Fe: 1a = 1e
• C: 6a = g
• N: 3a = f
• H: 2b+2c = 8f
• S: b = d+e+f
• O: 4b+c = 4d+4e+4f+g

3. Solve the system (Direct substitution is usually the best way.)

• d=2a
• e=a
• g=6a
• f=3a
• b=6a
• c=6a

which means that all coefficients depend on a parameter a, just choose a=1 (a number that will make all of them small whole numbers), which gives:

• a=1 b=6 c=6 d=2 e=1 f=3 g=6

4. And the balanced equation at last:

• K₄Fe(CN)₆ + 6 H₂SO₄ + 6 H₂O → 2 K₂SO₄ + FeSO₄ + 3 (NH₄)₂SO₄ + 6 CO

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, such as a.

• a K₄Fe(CN)₆ + H₂SO₄ + H₂O → K₂SO₄ + FeSO₄ + (NH₄)₂SO₄ + CO

3. 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). This removes the first four equations of the system. It is already known that whatever the coefficients are, those proportions must hold:

• a K₄Fe(CN)₆ + H₂SO₄ + H₂O → 2a K₂SO₄ + a FeSO₄ + 3a (NH₄)₂SO₄ + 6a CO

4. One can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) one could continue by noticing that adding the Sulfurs yields 6a for H₂SO₄ and finally by adding the hydrogens (or the oxygens) one can find the lasting 6a for H₂SO₄.

5. Again, having a convenient value for a (in this case 1 will do, but if a results in fractional values in the other coefficients, one would like to cancel the denominators) The result is

• K₄Fe(CN)₆ + 6 H₂SO₄ + 6 H₂O → 2 K₂SO₄ + FeSO₄ + 3 (NH₄)₂SO₄ + 6 CO

See also

• Atom
• Chemical formula
• Chemical reaction
• Molecule

References

ISBN links support NWE through referral fees

• Chang, Raymond. 2006. Chemistry. 9th ed. New York: McGraw-Hill Science/Engineering/Math. ISBN 0073221031.

• Cotton, F. Albert, and Geoffrey Wilkinson. 1980. Advanced Inorganic Chemistry. 4th ed. New York: Wiley. ISBN 0471027758.

• McMurry, J., and R.C. Fay. 2004. Chemistry. 4th ed. Upper Saddle River, NJ: Prentice Hall. ISBN 0131402080.

External links

• Classic Chembalancer FunBased Learning. (A free, online game to learn how to balance equations by inspection.) Retrieved September 2, 2008.
• Stoichiometric equations. ScienceSoft.at. (Online calculator to determine the coefficients of a chemical equation.) Retrieved September 2, 2008.
• Chemical equation balancer. adomas.org. (Solves chemical equations.) Retrieved September 2, 2008.
• Balancing chemical equation using Equation Balancing and Stoichiometry calculator. ChemBuddy. (Works off-line, calculates stoichiometry and limiting reagents, balances charge.) Retrieved September 2, 2008.
• Balance Chemical Equation - Online Balancer. (Balances equation of any chemical reaction, full or half-cell, in one click.) Retrieved September 2, 2008.

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