Chemical Equations, Balancing, and Reaction Types

General Principles of Chemical Equations

• Formulas and Coefficients:

  • Formulas: Describe the identity of a compound or element (e.g., $(H<em>2)$, $(CO</em>2)$, $(H<em>2O)$). The formula is self-contained and cannot be changed for a given substance. For instance, hydrogen's formula is $(H</em>2)$, oxygen's is $(O<em>2)$ (both are diatomic), and water's is $(H</em>2O)$—these are fixed.

  • Subscripts: Indicate the number of atoms of a specific element within one molecule of a compound (e.g., the subscript $<em>2$ in $(H</em>2O)$ means there are $2$ hydrogen atoms in one water molecule).

  • Coefficients: Are numbers placed before a chemical formula (e.g., $2H<em>2O$). They tell you how much of each reactant or product is involved in the reaction, specifically the ratio of molecules or moles. If no coefficient is written, it is implied to be $1$. A coefficient of $2$ in front of $(H</em>2O)$ means there are two water molecules, leading to $(2 imes 2 = 4)$ hydrogen atoms and $(2 imes 1 = 2)$ oxygen atoms in total from the water.

  • Example Reaction: For the combustion of hydrogen: $(2H<em>2(g) + O</em>2(g) ightarrow 2H_2O(l))$

    • This means $2$ molecules of hydrogen react with $1$ molecule of oxygen to produce $2$ molecules of water.

    • This represents a direct combustion of hydrogen.

• States of Matter and Reaction Conditions:

  • Importance of States of Matter: Indicating the state of matter is crucial because it affects the interaction and energy involved in reactions (thermochemistry). For example, making liquid water versus gaseous water involves different energies.

  • Standard Indications:

    • $(s)$: solid

    • $(l)$: liquid

    • $(g)$: gas

    • $(aq)$: aqueous (dissolved in water, implying a solution). This is always used for acids and can be used for other compounds.

  • General State Rules:

    • Ionic compounds are typically solids at room temperature unless specified as $(aq)$.

    • Diatomic elements ($(H<em>2, N</em>2, O<em>2, F</em>2, Cl<em>2, Br</em>2, I<em>2)$) have specific states, e.g., $(H</em>2, N<em>2, O</em>2, F<em>2, Cl</em>2)$ are gases, $(Br<em>2)$ is liquid, $(I</em>2)$ is solid. However, instructors may provide specific states for elements.

  • Separators: Multiple reactants or products are separated by a plus ($(+)$) sign, meaning "and" (e.g., $(CaO + CO_2)$ means "calcium oxide and carbon dioxide"). The order of reactants or products on their respective sides does not matter.

  • Conditions: Necessary conditions like heating, pressure, or specific temperatures must be stated above or below the reaction arrow.

    • $( riangle)$ (delta): indicates heating in general.

    • Specific temperatures (e.g., $300^ ext{o}C$) or pressures can also be noted.

• Description to Chemical Equation Example:

  • Description: "Solid calcium carbonate is heated and undergoes decomposition to solid calcium oxide and carbon dioxide gas."

  • Step 1: Identify Substances and Formulas (using nomenclature knowledge):

    • Calcium carbonate: $(CaCO<em>3)$ ($(Ca^{2+})$, $(CO</em>3^{2-})$)

    • Calcium oxide: $(CaO)$ ($(Ca^{2+})$, $(O^{2-})$)

    • Carbon dioxide: $(CO_2)$

  • Step 2: Identify Reactants and Products: "decomposes to" indicates calcium carbonate is the reactant, and calcium oxide and carbon dioxide are products.

  • Step 3: Write Formulas with States and Conditions:

    • $(CaCO<em>3(s) riangle ightarrow CaO(s) + CO</em>2(g))$

  • Step 4: Balance the Equation (check conservation of mass):

    • Count atoms on both sides:

      • Left: $1$ Ca, $1$ C, $3$ O

      • Right: $1$ Ca, $1$ C, $(1 + 2 = 3)$ O

    • This equation is already balanced.

• Balancing Chemical Equations (Conservation of Mass):

  • Goal: To ensure the number of atoms of each element is the same on both the reactant and product sides of the equation, reflecting the conservation of mass.

  • Method: Add coefficients in front of chemical formulas. Never change the subscripts within a formula.

  • Process Steps:

    1. Count Atoms: Determine the number of moles of each element on both sides of the equation. Be careful if an element appears in more than one compound on a side.

    2. Identify Unbalanced Elements: Find elements with different numbers of moles on each side.

    3. Add Coefficients (Iterative Process): Go through one element at a time, adding coefficients. This often requires going back and forth because adding a coefficient to a compound affects all elements within that compound.

    4. General Order for Balancing:

       • Start with metals, then nonmetals.

       • If an element is by itself (e.g., $(O_2)$), save it for last.

       • Polyatomic Ions: If a polyatomic ion appears identically on both sides (e.g., phosphate $(PO_4^{3-})$), treat it as a single unit and balance it first.

       • Combustion Reactions (hydrocarbon + $(O_2)$): Balance carbon (C) first, then hydrogen (H), and finally oxygen (O).

    5. Double Check: After adding all coefficients, recount every atom of every element on both sides to ensure everything is balanced.

    6. Smallest Whole Number Ratio: The final balanced equation must have the smallest possible whole number coefficients. This means at least one coefficient in the equation must be an odd number. If all coefficients are even, divide all coefficients by $2$ (or their greatest common divisor).

• Balancing Example: Magnesium Nitride + Water $\rightarrow$ Magnesium Hydroxide + Ammonia

  • Unbalanced: $(Mg<em>3N</em>2(s) + H<em>2O(l) ightarrow Mg(OH)</em>2(aq) + NH_3(aq))$

  • Initial Counts:

    • Left: Mg ($3$), N ($2$), H ($2$), O ($1$)

    • Right: Mg ($1$), N ($1$), H ($5$ = $2$ from $(OH)<em>2$ + $3$ from $(NH</em>3)$), O ($2$)

  • Balance Mg: Place $3$ in front of $(Mg(OH)_2)$.

    • $(Mg<em>3N</em>2 + H<em>2O ightarrow 3Mg(OH)</em>2 + NH_3))$

    • New Counts (Right): Mg ($3$), N ($1$), H ($9$ = $3 imes 2$ from $(OH)<em>2$ + $3$ from $(NH</em>3)$), O ($6$ = $3 imes 2$)

  • Balance N: Place $2$ in front of $(NH_3)$.

    • $(Mg<em>3N</em>2 + H<em>2O ightarrow 3Mg(OH)</em>2 + 2NH_3))$

    • New Counts (Right): Mg ($3$), N ($2$), H ($12$ = $3 imes 2$ from $(OH)<em>2$ + $2 imes 3$ from $(NH</em>3)$), O ($6$ = $3 imes 2$)

  • Balance O/H: Place $6$ in front of $(H_2O)$.

    • $(Mg<em>3N</em>2(s) + 6H<em>2O(l) ightarrow 3Mg(OH)</em>2(aq) + 2NH_3(aq))$

    • Final Counts (Left): Mg ($3$), N ($2$), H ($12$), O ($6$)

    • Final Counts (Right): Mg ($3$), N ($2$), H ($6 + 6 = 12$), O ($6$)

  • The equation is balanced. Note the coefficients $1, 6, 3, 2$ include odd numbers, so it's the smallest whole number ratio.

• Balancing Example: Ammonium Dichromate Decomposition

  • Description: "Solid ammonium dichromate decomposes to form nitrogen gas, water vapor, and chromium (III) oxide."

  • Unbalanced Equation: $( (NH<em>4)</em>2Cr<em>2O</em>7(s) <br>ightarrow N<em>2(g) + H</em>2O(g) + Cr<em>2O</em>3(s) )$

  • Initial Counts:

    • Left: N ($2$), H ($8$), Cr ($2$), O ($7$)

    • Right: N ($2$), H ($2$), Cr ($2$), O ($4$ = $1$ from $(H<em>2O)$ + $3$ from $(Cr</em>2O_3)$) (N and Cr are already balanced)

  • Balance H: Left has $8$ H, Right has $2$ H. Place $4$ in front of $(H_2O)$.

    • $( (NH<em>4)</em>2Cr<em>2O</em>7(s) <br>ightarrow N<em>2(g) + 4H</em>2O(g) + Cr<em>2O</em>3(s) )$

    • New Counts (Right): N ($2$), H ($8$), Cr ($2$), O ($7$ = $4$ from $(H<em>2O)$ + $3$ from $(Cr</em>2O_3)$) (O is now also balanced).

  • The equation is balanced.

• Special Considerations for Combustion Reactions (Hydrocarbon + $(O_2)$): Order C, H, O.

  • Oxygen Balancing Trick: If $(O<em>2)$ is the only source of oxygen on the reactant side, and balancing leads to an odd number of oxygen atoms on the product side (e.g., from $(H</em>2O)$, which has one O per molecule), you must double all coefficients in the equation. This ensures an even number of oxygens on the right, allowing $(O_2)$ on the left to be balanced with a whole number.

  • Example: Octane ($(C<em>8H</em>{18})$) Combustion

    • Unbalanced: $(C<em>8H</em>{18}(l) + O<em>2(g) ightarrow CO</em>2(g) + H_2O(g))$

    • Balance C: $8$ C on left, $1$ C on right. Put $8$ in front of $(CO_2)$.

      • $(C<em>8H</em>{18} + O<em>2 ightarrow 8CO</em>2 + H_2O)$

    • Balance H: $18$ H on left, $2$ H on right. Need to multiply $(H_2O)$ by $9$ ($(18/2)$).

      • At this point, if we place $9$ in front of $(H<em>2O)$, the total oxygen on the right would be $(8 imes 2) + (9 imes 1) = 16 + 9 = 25$. Since $(O</em>2)$ is the only source of oxygen on the left, and $25$ is an odd number, we cannot get a whole number coefficient for $(O_2)$ ($25/2 = 12.5$).

      • Solution: Double all current coefficients (where the coefficient for $(C<em>8H</em>{18})$ is implicitly $1$).

        • Double $(C<em>8H</em>{18})$ to $2C<em>8H</em>{18}$.

        • Double $(8CO<em>2)$ to $16CO</em>2)$.

        • Double $(9H<em>2O)$ to $18H</em>2O)$.

      • Current equation: $(2C<em>8H</em>{18} + O<em>2 ightarrow 16CO</em>2 + 18H_2O)$

    • Recount (after doubling):

      • Left: C ($16$), H ($36$)

      • Right: C ($16$), H ($36$)

    • Balance O: Calculate total O on right: $(16 imes 2) + (18 imes 1) = 32 + 18 = 50$. We need $50$ O on the left. Since $(O_2)$ is diatomic, $50/2 = 25$.

    • Final Balanced Equation: $(2C<em>8H</em>{18}(l) + 25O<em>2(g) ightarrow 16CO</em>2(g) + 18H_2O(g))$

• Balancing Polyatomic Ions:

  • If a polyatomic ion (e.g., phosphate $(PO_4^{3-})$) is present on both sides of the equation and unchanged, balance it as a single unit or