|
When working with almost any chemistry problem, the first place to start is
with a balanced chemical equation. This is an equation that shows how the
atoms rearrange in the reaction. A balanced equation is one where the left and
the right side of the equation have the same atoms: if there are five hydrogens
and two carbons on the left, there must be five hydrogens and two carbons on the
right. This is done by adjusting the coefficients in front of the molecules in
the equation, not by adjusting the atom subscripts in the molecules.
To balance an equation, begin by writing down the skeleton equation:
this is the unbalanced equation with just the molecules on each side: an example
is the combustion of methane
- CH4(g) + O2(g) -> H2O(g) +
CO2(g)
This reaction has the proper products and reactants,
but the number of atoms differs: there are four hydrogens on the left and only
two on the right, for example.
Next, pick an atom to balance. In general, if you have an atom that is
present by itself in one of the products or reactants, such as O2 in
the above reaction, save it for last. Other than that, you can choose any atom.
Let's start with carbon: there is one on the left and one on the right. Carbon
is already balanced.
Next, let's move to hydrogen. There are four on the left and two on the
right. If we place a coefficient of 2 in front of the water on the right, there
are now four on each side
- CH4(g) + O2(g) -> 2H2O(g) +
CO2(g)
Next, we balance the oxygen. There are two on the left and four on the right.
(One each on two waters and two on the carbon dioxide.) Place a 2 in front of
the oxygen on the left:
- CH4(g) + 2O2(g) -> 2H2O(g) +
CO2(g)
The reaction is now balanced.
It is possible to end up with a situation where it is not obvious how to
balance the equation: a good example is the combustion of methanol,
- CH3OH(l) + O2(g) -> H2O(g) +
CO2(g)
Oxygen is present alone, so leave it for last. We
start with carbon, one on each side, so carbon is balanced. There are four
hydrogens on the left and two on the right, so place a two in front of the water
on the right
- CH3OH(l) + O2(g) -> 2H2O(g) +
CO2(g)
When we try to balance the oxygen, we find three on
the left and four on the right. We want to stay away from non-integral numbers
of atoms in an equation, so how do we balance this?
When you have an odd number of atoms on one side of an equation and an even
on the other, multiply both sides of the equation through by two. This will give
an even number on both sides and make the equation easy to balance.
- 2CH3OH(l) + 2O2(g) -> 4H2O(g) +
2CO2(g)
The carbon and hydrogens are still balanced, and now
there are six oxygens on the left and eight on the right. Changing the
coefficient in front of the oxygens to 3 gives eight on the left, so
- 2CH3OH(l) + 3O2(g) -> 4H2O(g) +
2CO2(g)
The equation is now balanced.
Example: Balance the following equation:
- C2H8N2(s) + N2O4(g)
-> N2(g) + CO2(g) + H2O(g)
Solution: Since nitrogen is present solo on the right, leave it for
last. Start with carbon. There are two on the left and one on the right: place a
2 in front of the carbon dioxide on the right to get two on both
- C2H8N2(s) + N2O4(g)
-> N2(g) + 2CO2(g) + H2O(g)
Next,
balance the hydrogens. There are eight on the left and two on the right. Place a
four in front of the water and you have eight on both sides
- C2H8N2(s) + N2O4(g)
-> N2(g) + 2CO2(g) + 4H2O(g)
Next,
the oxygens. There are four on the left and eight on the right. Multiply the
N2O4 by two and both sides wil have eight.
- C2H8N2(s) +
2N2O4(g) -> N2(g) + 2CO2(g) +
4H2O(g)
Last, adjust the
nitrogens. There are six on the
left and two on the right: place a 3 in front of the nitrogens.
- C2H8N2(s) +
2N2O4(g) -> 3N2(g) + 2CO2(g) +
4H2O(g)
The equation is now balanced. |