1.4 Chemical Reactions: Synthesis, Balancing, and Net Ionic Equations

Types of Chemical Reactions

In chemical situations, it is often possible to predict the type of reaction that will occur when two substances are combined or to determine if a reaction will fail to take place entirely. For the four most common reaction types, specific rules guide these predictions:

• 1. Synthesis:     * Condition for occurrence: A synthesis reaction will occur provided the two elements involved are capable of demonstrating opposite valences.     * Formula Representation: $\text{Zn}^{+2} + \text{Cl}^{-1} \rightarrow \text{ZnCl}_2$

• 2. Decomposition:     * Condition for occurrence: The reaction will occur provided specific conditions are supplied, such as a change in temperature ($\Delta$).     * Example: $2\text{HgO} \xrightarrow{\Delta} 2\text{Hg} + \text{O}_2$

• 3. Single Displacement (Replacement):     * Condition for occurrence: The reaction will occur if the element (or cation) that is tasked with displacing or exchanging is more active than the one currently in the compound. Activity is determined by an activation chart.     * Example #1 (No Reaction): $\text{Ag} + \text{Cu}(\text{NO}_3)_2 \rightarrow \text{NR}$ (No Reaction). This occurs because copper is the more active element and is already in the compound; silver cannot displace it.     * Example #2 (Reaction): $\text{Sn} + \text{CuSO}_4 \rightarrow \text{Cu} + \text{SnSO}_4$. This reaction occurs because Tin (Sn) is higher on the activity chart than Copper (Cu).

• 4. Double Displacement (Replacement):     * Condition for occurrence: This reaction has criteria similar to single replacement but is primarily dependent on the solubility of the involved substances.     * Example: $\text{Ba}(\text{NO}_3)_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaNO}_3$

• 5. Combustion:     * Definition: A reaction specifically between a hydrocarbon and oxygen ($\text{O}_2$).     * Products: Always produces water ($\text{H}_2\text{O}$) and carbon dioxide ($\text{CO}_2$).     * Example: $\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$

Characterizing Chemical Equations

Chemists utilize two primary types of equations to describe chemical transformations:

• Word Equations: These describe the reactants (substances that react) and the products (substances formed), including their physical states.

• Formula Equations: A shorthand method involving chemical symbols. These are categorized into two subtypes:     * Skeleton Equations: Unbalanced equations that list the correct chemical formula and state for each reactant and product.     * Balanced Equations: These list the correct formulas and states while balancing the equation to account for the total number of atoms present. This process adheres to the Law of Conservation of Mass, ensuring the same number and type of atoms exist in both the reactants and products. In advanced chemistry (Chemistry 30), only balanced equations should be used.

Rules for Balancing Equations

• Methodology: Equations are balanced by changing the coefficients (numbers placed in front of a substance).

• Strict Restriction: One must NEVER change the subscripts or the formulas of substances to balance an equation.

• Counting Atoms: Accurate atom counting is critical. The coefficient is multiplied by the subscripts in each formula to determine the total atoms represented. If atoms of the same element appear in multiple reactants or products, their totals are added together.     * Example 1 (Reactants): $2\text{SO}_2 + \text{O}_2$         * Sulfur: $2$         * Oxygen: $(2 \times 2) + 2 = 6$     * Example 2 (Products): $4\text{CO}_2 + 6\text{H}_2\text{O}$         * Carbon: $4$         * Oxygen: $(4 \times 2) + 6 = 14$         * Hydrogen: $6 \times 2 = 12$

Balancing by Inspection: Techniques and Hints

Balancing by inspection is essentially a trial-and-error process, facilitated by the following strategies:

• 1. Priority Atoms: Balance atoms that appear only once in the reactants and only once in the products first. Save atoms that appear in multiple locations for last.     * Example (Propane Combustion): $\text{C}_3\text{H}_8(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g)$     * Step 1: Balance Carbon ($3$ on left, $1$ on right) and Hydrogen ($8$ on left, $2$ on right).     * Step 2: Finally, balance Oxygen ($O$) which appears in multiple products.

• 2. Polyatomic Ions: Balance polyatomic ions (e.g., $\text{SO}_4^{2-}$) as a single group rather than individual atoms. This only applies if the ion remains identical on both sides of the equation.     * Example: $\text{Ca}(\text{NO}_3)_2(aq) + \text{Na}_3\text{PO}_4(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + \text{NaNO}_3(aq)$     * Step 1: Balance $Ca$ and $Na$.     * Step 2: Balance the $\text{PO}_4$ and $\text{NO}_3$ groups.

• 3. The Doubling Rule: If an element possesses an odd number of atoms on the reactant side but must result in an even number on the product side (due to subscripts), double the coefficients of all previously balanced substances and continue.     * Example: $\text{CuFeS}_2(s) + \text{O}_2(g) \rightarrow \text{Cu}(s) + \text{FeO}(s) + \text{SO}_2(g)$     * Step 1: Balance once-appearing atoms ($Cu$, $Fe$, $S$).     * Step 2: Check Oxygen. If there are 5 product Oxygen atoms and it is impossible to balance with $\text{O}_2$ (which must be even), double the existing balances for other elements to reach an even Oxygen count.

Writing Net Ionic Equations

In many aqueous reactions, the traditional formula equation can be misleading regarding what physically occurs in the solution. This is due to the solubility of ionic substances.

• Solubility and Precipitates: Soluble products remain dissolved as ions. Insoluble products form a solid precipitate.

• Ionic Equations: Show individual ions as they exist in solution.

• Spectator Ions: Ions that do not participate in the chemical change or change state during the reaction.

• Net Ionic Equation: An equation that removes all spectator ions to show only the species involved in the chemical change.

The Three-Step Process for Net Ionic Equations

• Example Scenario: A solution of Barium chloride ($\text{BaCl}_2$) reacts with Sodium Carbonate ($\text{Na}_2\text{CO}_3$) to form Barium Carbonate ($\text{BaCO}_3$) precipitate and Sodium Chloride ($\text{NaCl}$) solution.

• Step #1: Balanced Formula Equation     * $\text{BaCl}_2(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow \text{BaCO}_3(s) + 2\text{NaCl}(aq)$

• Step #2: Total Ionic Equation     * Write dissolved species (aq) as ions. Solids (s), liquids (l), and gases (g) remain as molecules/compounds.     * $\text{Ba}^{2+}(aq) + 2\text{Cl}^{-}(aq) + 2\text{Na}^{+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{BaCO}_3(s) + 2\text{Na}^{+}(aq) + 2\text{Cl}^{-}(aq)$

• Step #3: Net Ionic Equation     * Cancel out ions appearing on both sides (spectator ions: $2\text{Cl}^{-}$ and $2\text{Na}^{+}$).     * $\text{Ba}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{BaCO}_3(s)$

Assignment Problems and Practice

1. Atom Counting Practice

• Reactant Set 1: $\text{PbS} + 2\text{PbO}$     * $\text{Pb} = 3$, $\text{S} = 1$, $\text{O} = 2$

• Reactant Set 2: $2\text{NH}_4\text{NO}_3 + \text{H}_2\text{S}$     * $\text{N} = 4$, $\text{H} = 10$, $\text{O} = 6$, $\text{S} = 1$

• Reactant Set 3: $\text{Fe}(\text{NO}_3)_3 + 3\text{LiOH}$     * $\text{Fe} = 1$, $\text{N} = 3$, $\text{O} = 12$, $\text{Li} = 3$, $\text{H} = 3$

• Reactant Set 4: $\text{Ca}_3(\text{PO}_4)_2 + 3\text{H}_2\text{SO}_4$     * $\text{Ca} = 3$, $\text{P} = 2$, $\text{O} = 20$, $\text{H} = 6$, $\text{S} = 3$

2. Equations to Balance by Inspection

• a) $\text{K}_2\text{O} + \text{H}_2\text{O} \rightarrow \text{KOH}$

• b) $\text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + \text{H}_2\text{O}$

• c) $\text{CaCl}_2 + \text{HNO}_3 \rightarrow \text{Ca}(\text{NO}_3)_2 + \text{HCl}$

• d) $\text{FeCl}_3 + \text{KOH} \rightarrow \text{KCl} + \text{Fe}(\text{OH})_3$

• e) $\text{NaClO} + \text{H}_2\text{S} \rightarrow \text{NaCl} + \text{H}_2\text{SO}_4$

• f) $\text{Ag}_2\text{S} + \text{HNO}_3 \rightarrow \text{AgNO}_3 + \text{NO} + \text{S} + \text{H}_2\text{O}$

• g) $\text{I}_2 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + \text{NaI}$

• h) $\text{Fe} + \text{CuNO}_3 \rightarrow \text{Fe}(\text{NO}_3)_2 + \text{Cu}$

• i) $\text{MgCl}_2 + \text{NH}_4\text{NO}_3 \rightarrow \text{Mg}(\text{NO}_3)_2 + \text{NH}_4\text{Cl}$

• j) $\text{Al} + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2$

3. Ionic Equation Practice

• a) $\text{Zn}(s) + \text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(aq)$

• b) $\text{FeCl}_3(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgCl}(s) + \text{Fe}(\text{NO}_3)_3(aq)$

• c) $\text{KOH}(aq) + \text{H}_3\text{PO}_4(aq) \rightarrow \text{K}_3\text{PO}_4(aq) + \text{H}_2\text{O}(l)$

• d) $\text{HCl}(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow \text{H}_2\text{O}(l) + \text{CO}_2(g) + \text{NaCl}(aq)$

• e) $\text{Ba}(\text{OH})_2(aq) + \text{Fe}_2(\text{SO}_4)_3(aq) \rightarrow \text{Fe}(\text{OH})_3(s) + \text{BaSO}_4(aq)$

4. Reaction Prediction Scenarios

For each combination, determine (i) expected reaction type, (ii) if it occurs, (iii) reasons for non-occurrence, and (iv) the balanced equation:

• a) Tin and copper (II) sulphate

• b) Iron (III) nitrate and sodium chromate

• c) Calcium and iodine

• d) Magnesium and hydrochloric acid

• e) Calcium oxide (Decomposition/Electrolysis)

• f) Carbon and oxygen

• g) Sodium carbonate and sulfuric acid

• h) Iron (II) sulfide (Decomposition/Electrolysis)

• i) Platinum and lead(II) nitrate

• j) Propane and oxygen gas