Writing and Balancing Chemical Equations

Learning Objectives

• Derive chemical equations from narrative descriptions of chemical reactions.

• Write and balance chemical equations in molecular, total ionic, and net ionic formats.

Core Components of Chemical Equations

• Representation of Chemical Species:

  • Element symbols represent individual atoms.

  • When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas representing these species.

  • Chemical equations extend this symbolism to represent both the identities and relative quantities of substances undergoing a chemical or physical change.

• Reaction Example (Combustion of Methane):

  • Reactants: One methane molecule ($CH_4$) and two diatomic oxygen molecules ($O_2$).

  • Products: One carbon dioxide molecule ($CO_2$) and two water molecules ($H_2O$).

  • Chemical Equation: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

• Four Fundamental Aspects of Chemical Equations:

  • Reactants: Substances undergoing reaction, whose formulas are placed on the left side of the equation.

  • Products: Substances generated by the reaction, whose formulas are placed on the right side of the equation.

  • Symbols: Plus signs ($+$) separate individual reactant and product formulas; an arrow ($\rightarrow$) separates the reactant (left) and product (right) sides.

  • Coefficients: Numbers placed immediately to the left of each formula representing relative numbers of reactant and product species. A coefficient of $1$ is omitted by convention.

• Quantitative Interpretation of Coefficients:

  • Smallest possible whole-number coefficients are standard practice.

  • Coefficients represent relative numbers of species and can be interpreted directly as numerical ratios.

  • Methane and oxygen react to yield carbon dioxide and water in a $1:2:1:2$ ratio.

  • This ratio is satisfied by absolute molecule counts such as $1-2-1-2$, $2-4-2-4$, or $3-6-3-6$.

  • Equivalent amount unit interpretations include:

  • One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.

  • One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.

  • One mole of methane molecules and $2\text{ moles}$ of oxygen molecules react to yield $1\text{ mole}$ of carbon dioxide molecules and $2\text{ moles}$ of water molecules.

Conservation of Matter and Equation Balancing

• Law of Conservation of Matter:

  • A chemical equation must be balanced, meaning equal numbers of atoms for each element involved are represented on both reactant and product sides.

  • Verified by summing the total number of atoms for each element on either side and comparing sums for equality.

  • Atomic Count Calculation: Multiply the formula coefficient by the element's subscript within that formula. If an element appears in multiple formulas on one side, compute each individual count and sum them.

  • Example calculation for oxygen atoms on the product side of $CO_2 + 2H_2O$: $(1 \times 2) + (2 \times 1) = 4\text{ oxygen atoms}$.

• Atom Count Verification for Methane Reaction ($CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$):

  • Carbon ($C$): Reactants $1 \times 1 = 1$; Products $1 \times 1 = 1$; Balanced: Yes ($1 = 1$).

  • Hydrogen ($H$): Reactants $1 \times 4 = 4$; Products $2 \times 2 = 4$; Balanced: Yes ($4 = 4$).

  • Oxygen ($O$): Reactants $2 \times 2 = 4$; Products $(1 \times 2) + (2 \times 1) = 4$; Balanced: Yes ($4 = 4$).

• Method of Balancing by Inspection:

  • Process of deriving a balanced equation from a qualitative reaction description by systematically adjusting coefficients.

  • Subscript Rule: Formula subscripts define chemical identity and must never be altered. Changing subscripts changes the substance itself (e.g., changing $H_2O$ to $H_2O_2$ changes water to hydrogen peroxide).

  • Step-by-Step Example (Decomposition of Water):

  • Unbalanced qualitative equation: $H_2O \rightarrow H_2 + O_2$

  • Initial Atom Counts:

    • Hydrogen ($H$): Reactants $1 \times 2 = 2$; Products $1 \times 2 = 2$; Balanced: Yes ($2 = 2$).

    • Oxygen ($O$): Reactants $1 \times 1 = 1$; Products $1 \times 2 = 2$; Balanced: No ($1 \neq 2$).

  • Step 1 (Balance Oxygen): Change coefficient of $H_2O$ to $2$ to yield $2H_2O \rightarrow H_2 + O_2$

    • Hydrogen ($H$): Reactants $2 \times 2 = 4$; Products $1 \times 2 = 2$; Balanced: No ($4 \neq 2$).

    • Oxygen ($O$): Reactants $2 \times 1 = 2$; Products $1 \times 2 = 2$; Balanced: Yes ($2 = 2$).

  • Step 2 (Rebalance Hydrogen): Change coefficient of $H_2$ product to $2$ to yield $2H_2O \rightarrow 2H_2 + O_2$

    • Hydrogen ($H$): Reactants $2 \times 2 = 4$; Products $2 \times 2 = 4$; Balanced: Yes ($4 = 4$).

    • Oxygen ($O$): Reactants $2 \times 1 = 2$; Products $1 \times 2 = 2$; Balanced: Yes ($2 = 2$).

  • Final Balanced Equation: $2H_2O \rightarrow 2H_2 + O_2$

Advanced Balancing Techniques and Conventions

• Intermediate Fractional Coefficients:

  • Fractions can be used as convenient intermediate coefficients during the balancing process.

  • Once balanced, all equation coefficients are multiplied by a whole number to clear denominators and generate integer coefficients without upsetting balance.

  • Example (Combustion of Ethane):

  • Unbalanced equation: $C_2H_6 + O_2 \rightarrow CO_2 + H_2O$

  • Balance $C$ and $H$ first: $C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O$

  • Calculate product oxygen atoms: $(2 \times 2) + (3 \times 1) = 7\text{ atoms}$ (odd number).

  • Apply fractional coefficient to $O_2$ reactant: $C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$

  • Multiply entire equation by $2$ for integer coefficients: $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

• Smallest Whole-Number Convention:

  • Convention dictates using the smallest whole-number coefficients.

  • Equations balanced with larger multiples must be simplified by dividing all coefficients by their greatest common factor (GCF).

  • Example: The balanced equation $3N_2 + 9H_2 \rightarrow 6NH_3$ has coefficients divisible by $3$.

  • Dividing each coefficient by $3$ yields the preferred equation: $N_2 + 3H_2 \rightarrow 2NH_3$

• Interactive Practice Tutorial:

  • Additional equation balancing practice is accessible at: http://openstax.org/l/16BalanceEq

Physical States and Reaction Conditions

• Parenthetical Physical State Abbreviations:

  • Indicates physical states of reactants and products, placed immediately after chemical formulas.

  • Standards:

  • $(s)$: Solids

  • $(l)$: Liquids

  • $(g)$: Gases

  • $(aq)$: Aqueous solutions (substances dissolved in water)

  • Example Reaction (Sodium in Water):

  • Reaction description: Solid sodium metal reacts with liquid water to produce molecular hydrogen gas and dissolved ionic sodium hydroxide.

  • Notated equation: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$

• Designation of Special Reaction Conditions:

  • Special conditions required for a reaction are written above or below the reaction arrow.

  • Thermal conditions: A reaction carried out by heating is indicated by the uppercase Greek letter delta ($\Delta$) over the arrow ($\xrightarrow{\Delta}$).

Equations for Ionic Reactions

• Aqueous Media and Types of Chemical Equations:

  • Molecular Equation:

  • Represents ionic reactants and products as intact, neutral molecular formulas without explicitly showing dissolved ions.

  • Example: Reaction between aqueous calcium chloride ($CaCl_2$) and aqueous silver nitrate ($AgNO_3$) yielding aqueous calcium nitrate ($Ca(NO_3)_2$) and solid silver chloride ($AgCl$).

  • Balanced Molecular Equation: $CaCl_2(aq) + 2AgNO_3(aq) \rightarrow Ca(NO_3)_2(aq) + 2AgCl(s)$

  • Complete Ionic Equation:

  • Explicitly represents all dissolved ionic species as dissociated ions dispersed in solution.

  • Dissociation representations:

    • $CaCl_2(aq) \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)$

    • $2AgNO_3(aq) \rightarrow 2Ag^+(aq) + 2NO_3^-(aq)$

    • $Ca(NO_3)_2(aq) \rightarrow Ca^{2+}(aq) + 2NO_3^-(aq)$

    • $AgCl(s)$ does not dissolve significantly and retains its solid state notation.

  • Complete Ionic Equation: $Ca^{2+}(aq) + 2Cl^-(aq) + 2Ag^+(aq) + 2NO_3^-(aq) \rightarrow Ca^{2+}(aq) + 2NO_3^-(aq) + 2AgCl(s)$

  • Spectator Ions:

  • Ions present in identical form on both reactant and product sides of the equation.

  • Maintain charge neutrality but undergo no chemical or physical change.

  • Spectator ions in the calcium chloride/silver nitrate reaction: $Ca^{2+}(aq)$ and $NO_3^-(aq)$.

  • Net Ionic Equation:

  • Equation produced by eliminating spectator ions from a complete ionic equation to show only the active chemical change.

  • Unsimplified Net Ionic Equation: $2Cl^-(aq) + 2Ag^+(aq) \rightarrow 2AgCl(s)$

  • Simplified Net Ionic Equation (smallest integer coefficients): $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$

  • Indicates solid silver chloride forms from dissolved silver(I) and chloride ions regardless of their initial compound source.

Comprehensive Worked Examples and Solutions

• Example 7.1: Formation of Dinitrogen Pentoxide

  • Problem: Write a balanced equation for the reaction of molecular nitrogen ($N_2$) and oxygen ($O_2$) to form dinitrogen pentoxide ($N_2O_5$).

  • Step 1 (Unbalanced Equation): $N_2 + O_2 \rightarrow N_2O_5$

  • Step 2 (Initial Atom Count):

  • Nitrogen ($N$): Reactants $1 \times 2 = 2$; Products $1 \times 2 = 2$; Balanced: Yes ($2 = 2$).

  • Oxygen ($O$): Reactants $1 \times 2 = 2$; Products $1 \times 5 = 5$; Balanced: No ($2 \neq 5$).

  • Step 3 (Balance Oxygen): Least common multiple of subscripts $2$ and $5$ is $10$. Adjust coefficients of $O_2$ to $5$ and $N_2O_5$ to $2$:

  • Equation: $N_2 + 5O_2 \rightarrow 2N_2O_5$

  • Nitrogen ($N$): Reactants $1 \times 2 = 2$; Products $2 \times 2 = 4$; Balanced: No ($2 \neq 4$).

  • Oxygen ($O$): Reactants $5 \times 2 = 10$; Products $2 \times 5 = 10$; Balanced: Yes ($10 = 10$).

  • Step 4 (Restore Nitrogen Balance): Change coefficient of $N_2$ reactant to $2$:

  • Equation: $2N_2 + 5O_2 \rightarrow 2N_2O_5$

  • Nitrogen ($N$): Reactants $2 \times 2 = 4$; Products $2 \times 2 = 4$; Balanced: Yes ($4 = 4$).

  • Oxygen ($O$): Reactants $5 \times 2 = 10$; Products $2 \times 5 = 10$; Balanced: Yes ($10 = 10$).

  • Final Answer: $2N_2 + 5O_2 \rightarrow 2N_2O_5$

• Check Your Learning 7.1: Decomposition of Ammonium Nitrate

  • Problem: Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last because it appears in multiple products).

  • Unbalanced Equation: $NH_4NO_3 \rightarrow N_2 + O_2 + H_2O$

  • Answer: $2NH_4NO_3 \rightarrow 2N_2 + O_2 + 4H_2O$

• Example 7.2: Reaction of Carbon Dioxide with Sodium Hydroxide

  • Problem: When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.

  • Unbalanced Form: $CO_2(aq) + NaOH(aq) \rightarrow Na_2CO_3(aq) + H_2O(l)$

  • Molecular Equation: Balance by changing coefficient of $NaOH$ to $2$:

  • $CO_2(aq) + 2NaOH(aq) \rightarrow Na_2CO_3(aq) + H_2O(l)$

  • Complete Ionic Equation: Represent dissolved ionic species ($NaOH$ and $Na_2CO_3$) as dissociated ions:

  • $CO_2(aq) + 2Na^+(aq) + 2OH^-(aq) \rightarrow 2Na^+(aq) + CO_3^{2-}(aq) + H_2O(l)$

  • Net Ionic Equation: Eliminate spectator ion $Na^+(aq)$ from both sides:

  • $CO_2(aq) + 2OH^-(aq) \rightarrow CO_3^{2-}(aq) + H_2O(l)$

• Check Your Learning 7.2: Industrial Electrolysis of Brine

  • Problem: Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced along with diatomic hydrogen via the electrolysis of brine, according to the unbalanced equation: $NaCl(aq) + H_2O(l) \rightarrow Cl_2(g) + H_2(g) + NaOH(aq)$. Write balanced molecular, complete ionic, and net ionic equations.

  • Answer:

  • Balanced Molecular Equation: $2NaCl(aq) + 2H_2O(l) \rightarrow Cl_2(g) + H_2(g) + 2NaOH(aq)$

  • Complete Ionic Equation: $2Na^+(aq) + 2Cl^-(aq) + 2H_2O(l) \rightarrow Cl_2(g) + H_2(g) + 2Na^+(aq) + 2OH^-(aq)$

  • Net Ionic Equation: $2Cl^-(aq) + 2H_2O(l) \rightarrow Cl_2(g) + H_2(g) + 2OH^-(aq)$