Chemical Equations and Chemical Reactions
Chemical equations give information in two major areas.
First, they tell us the chemical formulae for the reactants - substances reacting (those being used up) and the chemical formulae for the products - substances are produced (those being made).
Chemical equations do not come balanced. The equation must be balanced before the equation can be used in a meaningful way.
Here is an example of an equation:
H2(g) + O2(g) → H2O(l)
The reactants are on the left side of a chemical equation and the products are on the right side.
However, you might ask, "On the left and right side of what?" Answer - the arrow.
This is an unbalanced equation (also called a skeleton equation – shows the correct chemical formulae for the reactants and products).
To calculate the number of atoms in a formula; subscripts are multiplied and then multiplied by the coefficient (number in front of the formula). How many H atoms in 2(NH4)3PO4? Start with the inner most subscript and multiply by the subscript outside the brackets and then multiply by the coefficient.
4 x 3 x 2 = 24 H atoms.
The unbalanced equation has unequal numbers of atoms on each side of the arrow. A balanced equation must have equal numbers of each type of atom on each side of the equation. Why?
All chemical reactions follow the Law of Conservation of Mass which is the rationale for balancing a chemical equation. "Matter is neither created nor destroyed during an ordinary chemical reaction." Therefore, we must finish our chemical reaction with as many atoms of each element as when we started (atoms are not created or destroyed during chemical reactions).
In the above example, there are two atoms of hydrogen on each side, but there are two atoms of oxygen on the left side and only one on the right side. When an even number on one side and uneven number on the other side multiply the uneven by 2 to make the number even and then balance the other side.
H2(g) + O2(g) → 2H2O(l) (O balanced – 2 atoms on each side – but not H)
2H2(g) + O2(g) → 2H2O(l) (now H is balanced – 4 atoms on each side)
Remember this: a balanced equation must have equal numbers of each type of atom on both sides of the arrow.
An equation is balanced by changing coefficients (numbers in front of the formula). It is important to note that only the coefficients can be changed, never a subscript.
Three things you CANNOT do when balancing an equation.
1) You cannot change a subscript.
You cannot change the oxygen's subscript in water from one to two, as in:
H2(g) + O2(g) → H2O2(l)
True, this balances the equation, but you have changed the substances in it. H2O2 is a completely different substance from H2O.
2) You cannot place a coefficient in the middle of a formula.
The coefficient goes at the beginning of a formula, not in the middle, as in:
H2(g) + O2(g) → H22O(l)
Water only comes as H2O and you can only use whole formula units of it.
3) Your final coefficients are all whole numbers with no common factors other than one. For example, this equation is balanced:
4 H2(g) + 2 O2(g) → 4 H2O(l)
However, all the coefficients have the common factor of two. Divide through to eliminate common factors.
The equation just above is correctly balanced, but it is not the best answer. The best answer has all common factors greater than one removed.
Second, the coefficients of a balanced equation tell us in what ratio the substances react or are produced.
This point has practical consequences whenever chemicals react. For example, the large middle tank of the space shuttle actually has two smaller tanks in it - one holding liquid oxygen and the other holding liquid hydrogen. The tank with the hydrogen holds twice as much as the oxygen-holding tank. Why?
Answer - hydrogen and oxygen react in a 2:1 ratio (balanced equation above). For any of oxygen used by the shuttle; there needs to be twice as much hydrogen. If the two tanks were equal sizes, the hydrogen tank would run dry when the oxygen tank was still half-filled.
Practice:
1. Fill in the blank:
1 How many hydrogen atoms are shown in 2CH3COOH? ____
2 How many oxygen atoms are indicated: 3Fe(NO3)2? ____
3 How many atoms in total are shown in 2 Al2(CO3)3? ____
2. Balance these equations:
1) ____H2(g) + ____Cl2(g) → ____HCl(g)
2) ____O2(g) → ____O3(g)
3) ____S8(s) + ____F2(g) → ____SF6(s)
4) ____C2H6(g) + ____O2(g) → ____CO2(g) + ____H2O(l)
5) ____Ca(H2PO4)2(aq) + ____CaSO4(aq) + ____HF(aq) → ____Ca10F2(PO4)6(s) + ____H2SO4(aq)
6) ____Fe(s) + ____O2(g) → ____Fe2O3(s)
Reaction Types
Synthesis (Composition) means that two pieces join together to produce one, more complex compound (product). Written using generic symbols, it is usually shown as: A + B → AB. These pieces A and B can be elements or simpler compounds. The complex product compound has more atoms than the reactant molecules.
These are some examples (skeleton equations given - equations are not balanced):
Mg(s) + O2(g) → MgO(s)
H2(g) + O2(g) → H2O(g)
K(s) + Cl2(g) → KCl(s)
Fe(s) + O2(g) → Fe2O3(s)
CaO(s) + CO2(g) → CaCO3(s)
Na2O(s) + CO2(g) → Na2CO3(s)
KCl(s) + O2(g) → KClO3(s)
There is only one substance on the right-hand (product) side. This is not always the case in a synthesis reaction. Sometimes there will be two products. Here's an example:
CO2(g) + H2O(l) → C6H12O6(aq) + O2(g) You might recognize this as the photosynthesis equation.
Decomposition one compound splits apart into two (or more pieces). These pieces can be elements or simpler compounds. Written using generic symbols, it is usually shown as: AB → A + B. These pieces A and B can be elements or simpler compounds.
These are some examples (skeleton equations given - equations are not balanced):
HgO(s) → Hg(l) + O2(g)
H2O(l) → H2(g) + O2(g)
MgCl2(s) → Mg(s) + Cl2(g)
FeS(s) → Fe(s) + S(s)
Decomposition can also split one compound into two simpler compounds (or compound and an element) as in these examples (skeleton equations given - equations are not balanced)::
CaCO3(s) → CaO(s) + CO2(g)
Na2CO3(s) → Na2O(s) + CO2(g)
KClO3(s) → KCl(s) + O2(g)
Ba(ClO3)2(s)→ BaCl2(s) + O2(g)
1) Binary compounds (like the first example set above) will break down into their elements.
2) Carbonates (like the first two in the second example set above) break down to the metal oxide and carbon dioxide.
3. Chlorates (like KClO3 and Ba(ClO3)2 in the example) will break down to the binary chloride salt and oxygen.
Single Replacement (Displacement) During single replacement, one element replaces another element in a compound. There are two different possibilities:
1. One cation replaces another. Written using generic symbols, it is:
AX + B ---> BX + A
Element B has replaced A (in the compound AX) to form a new compound BX and the free element A. Remember that A and B are both cations (postively-charged ions) in this example.
These are some examples (skeleton equations given - equations are not balanced):
Cu(s) + AgNO3(aq) → Ag(s) + Cu(NO3)2(aq)
Fe(s) + Cu(NO3)2(aq) → Fe(NO3)2(aq) + Cu(s)
Ca(s) + H2O(l) → Ca(OH)2(aq) + H2(g)
Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)
Notice how, when hydrogen gets displaced, it is written as a diatomic.
2. One anion replaces another. Written using generic symbols, it is:
X + AY → AX + Y
Element X has replaced Y (in the compound AY) to form a new compound AX and the free element Y. Remember that X and Y are both anions (negatively-charged ions) in this example.
Here are two examples (skeleton equations given - equations are not balanced)::
Cl2 (g) + NaBr (aq) → NaCl(aq) + Br2 (l)
Br2 (aq)+ KI (aq) → KBr (aq) + I2 (aq)
In single replacement, one reactant is always an element. It does not matter if the element is written first or second on the reactant side. The other reactant will be a compound.
Double Replacement (Displacement)
During double replacement, the cations and anions of two different compounds switch places.
Written using generic symbols, it is:
AX + BY → AY + BX
A and B are the cations (postively-charged ions) in this example, with X and Y being the anions (negatively-charged ions).
These are some examples (skeleton equations given - equations are not balanced):
KOH(aq) + H2SO4(aq) → K2SO4(aq) + H2O(l)
FeS(s) + HCl(aq) → FeCl2(aq) + H2S(g)
NaCl(s) + H2SO4(aq) → Na2SO4(aq) + HCl(g)
AgNO3(aq) + NaCl(aq)→ AgCl(s) + NaNO3(aq)
There are some special cases of double replacement reactions. An example is
CaCO3(s) + HCl(aq) → CaCl2(aq) + H2CO3(aq) but H2CO3(aq) is not stable and will immediately decompose as shown:H2CO3(aq) → H2O(l) + CO2(g). The reaction is double replacement followed by decomposition, the reaction is best represented by: CaCO3(s) + HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
More examples:
CaCO3(s) + HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)
K2SO3(s) + HNO3 → KNO3(aq) + SO2(g) + H2O(l)
NH4Cl(aq) + NaOH(aq)→ NaCl(aq) + NH3(g) + H2O(l)
Notice how, one of the two product compounds decomposes. Whenever H2CO3, H2SO3, or NH4OH is a product formula, the correct technique is to write the products as done in the examples.
Combustion written as a general equation complete combustion is usually shown as:
CxHy + O2 → CO2 + H2O
These are some examples (skeleton equations given - equations are not balanced):
CH4 (g) + O2 (g) → CO2 (g) + H2O (g)
C2H6 (g) + O2 (g) → CO2 (g) + H2O (g)
C6H12O6 (g)+ O2 (g)→ CO2 (g)+ H2O (g)
C2H5OH (l) + O2 (g) → CO2 (g)+ H2O (g)
Notice that some compounds contain carbon, hydrogen AND oxygen. However, the products are all the same, in every reaction. Isn't that great? We could vary it a bit by adding nitrogen (burns to form NO2) to the compound formula or sulfur (burns to form SO2). Like this:
C21H24N2O4(s) + O2 (g) → CO2 (g) + H2O (g) + NO2 (g)
C2H5SH(g) + O2 (g)→ CO2 (g) + H2O (g) + SO2 (g)
Balance each of the following equations place the appropriate coefficient in front of each.
Identify the reaction type composition, decomposition, single replacement, double replacement, combustion.
1. H2(g) + O2(g) → H2O(g)
2. Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)
3. S8(s) + O2(g) → SO3(g)
4. Na(s) + H2O(l) → NaOH(aq) + H2(g)
5. Fe(s) + O2(g) → Fe2O3(s)
6. K(s) + Br2(g) → KBr(s)
7. C2H2(g) + O2(g) → CO2(g) + H2O(g)
8. H2O2(aq) → H2O(l) + O2(g)
9. C7H16(l) + O2(g) → CO2(g) + H2O(g)
10. SiO2(s) + HF(aq) → SiF4(aq) + H2O(l)
11. KClO3(s) → KCl(s) + O2(g)
12. P4O10(s) + H2O(l) → H3PO4(aq)
13. Sb(s) + O2(g) → Sb4O6(s)
14. PCl5(s) + H2O(l) → HCl(g) + H3PO4(aq)
15. H2S(g) + Cl2(g) → S8(s) + HCl(g)
16. Na2O(s) + H2O(l) → NaOH(aq)
17. Fe(s) + H2O(l) → Fe2O3(s) + H2(g)
18. N2(g) + H2(g) → NH3(g)
19. Fe2(SO4)3(aq) + KOH(aq) → K2SO4(aq) + Fe(OH)3(s)
20. N2(g) + O2(g) → N2O(g)
21. Al(OH)3(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2O(l)
22. Al(s) + FeO(s) → Al2O3(s) + Fe(s)
23. K2O(s) + H2O(l) → KOH(aq)
24. FeCl3(aq) + NH4OH(aq) → Fe(OH)3(s) + NH4Cl(aq)
25. Al(s) + O2(g) → Al2O3(s)
26. Al2(SO4)3(aq) + Ca(OH)2(aq) → Al(OH)3(s) + CaSO4(aq)
27. N2O5(g) + H2O(l) → HNO3(aq)
28. Al(s) + HCl(aq) → AlCl3(aq) + H2(g)
29. Mg(s) + N2(g) → Mg3N2(s)
30. Na2CO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)