Compounds and Chemical Equations

Forming Compounds and Energy Changes

Imagine baking a cake: the raw ingredients look completely different from the final sponge. Similarly, chemical reactions cause substances to combine and form entirely new materials. An element consists of only one type of atom.

When elements react, they form a compound, which is a pure substance made of two or more elements chemically combined in fixed proportions. Compounds have physical and chemical properties that are distinctly different from their constituent elements. Furthermore, unlike mixtures (which are not chemically combined), compounds can only be separated back into elements through chemical reactions, not by physical means like filtration.

The formation of a compound from its elements happens through a specific sequence of events involving energy changes and atomic rearrangement:

• Step 1: Breaking old bonds (Energy taken in): Existing chemical bonds in the reactant elements must be broken. This requires energy to be taken in from the surroundings, which is an endothermic reaction process.

• Step 2: Atoms rearrange to achieve stability: The freed atoms interact to achieve a stable, full outer shell of electrons. This happens through either Electron Transfer (atoms lose or gain electrons to form ionic bonds) or Electron Sharing (atoms share electron pairs to form covalent bonds).

• Step 3: Making new bonds (Energy released): As the atoms form these new bonds to create the final compound, energy is released to the surroundings, which is an exothermic reaction process.

The overall net energy change of the reaction depends on the balance between these steps:

• If more energy is released making bonds (Step 3) than is taken in breaking them (Step 1), the overall reaction is exothermic (temperature increases).
• If more energy is taken in breaking bonds (Step 1) than is released making them (Step 3), the overall reaction is endothermic (temperature decreases).

Worked Example: Calculating Energy Change

Calculate the overall energy change for the formation of hydrogen chloride: $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$ (Given bond energies: $\text{H}-\text{H} = 436\text{ kJ/mol}$, $\text{Cl}-\text{Cl} = 243\text{ kJ/mol}$, $\text{H}-\text{Cl} = 432\text{ kJ/mol}$)

Step 1: Write the formula for energy change.

$$\text{Total Energy Change} = \text{Energy in (breaking bonds)} - \text{Energy out (making bonds)}$$

Step 2: Calculate energy taken in to break bonds in the reactants.

• $\text{Energy in} = 436\text{ kJ/mol} + 243\text{ kJ/mol} = 679\text{ kJ/mol}$

Step 3: Calculate energy released forming bonds in the products.

• $\text{Energy out} = 2 \times 432\text{ kJ/mol} = 864\text{ kJ/mol}$

Step 4: Calculate the net change.

• $\text{Total Energy Change} = 679 - 864 = -185\text{ kJ/mol}$ (The negative sign proves this is an exothermic reaction).

Word and Balanced Symbol Equations

Chemical equations are like precise recipes, showing exactly what goes into a reaction and what comes out. The starting materials are called the reactants (written on the left), and the new substances formed are the products (written on the right).

A word equation represents a chemical reaction using the full chemical names.

Step-by-Step Guide to Constructing Word Equations:

• Step 1: Identify reactants: Determine the starting materials for the reaction.

• Step 2: Write reactants: Write their full chemical names on the left-hand side. If there is more than one reactant, separate them with a plus sign ($+$).

• Step 3: Draw the arrow: Draw an arrow ($\rightarrow$) pointing from left to right, which translates to "reacts to form".

• Step 4: Identify products: Determine the new substances formed by the reaction.

• Step 5: Write products: Write their full chemical names on the right-hand side. If there is more than one product, separate them with a plus sign ($+$).

For example: $\text{Magnesium} + \text{Oxygen} \rightarrow \text{Magnesium oxide}$

A balanced symbol equation uses chemical formulas to show the exact number of atoms of each element. This relies on the Law of Conservation of Mass, which states that no atoms are lost or made during a chemical reaction. Therefore, the total mass of the products must perfectly equal the total mass of the reactants, and the total number of atoms for each element must be equal on both sides of the arrow.

Non-enclosed Systems: In a non-enclosed system (such as an open flask), the mass may appear to change during a reaction:

• If a gas is a reactant, the mass appears to increase as the gas enters from the air and combines with the solids/liquids in the flask.
• If a gas is a product, the mass appears to decrease as the gas escapes into the air. The Law of Conservation of Mass still applies; the "missing" or "extra" mass is simply the gas moving in or out of the container!

Step-by-Step Guide to Balancing Equations: Balance the reaction between Aluminium and Copper(II) Oxide: $\text{Al} + \text{CuO} \rightarrow \text{Al}_2\text{O}_3 + \text{Cu}$

• Step 1: Count Atoms: Left side has $1\text{Al}, 1\text{Cu}, 1\text{O}$. Right side has $2\text{Al}, 1\text{Cu}, 3\text{O}$.

• Step 2: Balance Oxygen: The right side has 3 oxygen atoms. Place a coefficient of 3 in front of the $\text{CuO}$ on the left.

  • $\text{Al} + 3\text{CuO} \rightarrow \text{Al}_2\text{O}_3 + \text{Cu}$

• Step 3: Balance Aluminium: The right side has 2 aluminium atoms. Place a coefficient of 2 in front of the $\text{Al}$ on the left.

  • $2\text{Al} + 3\text{CuO} \rightarrow \text{Al}_2\text{O}_3 + \text{Cu}$

• Step 4: Balance Copper: The left side now has 3 copper atoms. Place a coefficient of 3 in front of the $\text{Cu}$ on the right.

  • $2\text{Al} + 3\text{CuO} \rightarrow \text{Al}_2\text{O}_3 + 3\text{Cu}$

• Step 5: Final Check: Both sides now have $2\text{Al}$, $3\text{Cu}$, and $3\text{O}$. The equation is balanced.

Always remember that seven diatomic non-metals must be written as pairs in symbol equations: $\text{H}_2, \text{N}_2, \text{O}_2, \text{F}_2, \text{Cl}_2, \text{Br}_2, \text{I}_2$. Equations should also include state symbols: $\text{(s)}$ for solid, $\text{(l)}$ for liquid, $\text{(g)}$ for gas, and $\text{(aq)}$ for aqueous (dissolved in water).

Ionic Equations and Spectator Ions (Higher Tier)

In a busy stadium, the spectators just watch the game while the players do the actual work. In chemistry, an ionic equation shows only the specific atoms and ions actively participating in a chemical change.

It strips away the spectator ions — ions that remain in the exact same state and charge on both the reactant and product sides. Ionic equations are most commonly used for neutralisation, displacement, and forming an insoluble precipitate.

Step-by-Step Mechanism for Ionic Equations: Find the ionic equation for: $\text{AgNO}_3\text{(aq)} + \text{NaCl}\text{(aq)} \rightarrow \text{AgCl}\text{(s)} + \text{NaNO}_3\text{(aq)}$

• Step 1: Write the full balanced symbol equation with state symbols (as above).

• Step 2: Split ONLY the aqueous $\text{(aq)}$ ionic compounds into their separate ions. Do NOT split $\text{(s)}$, $\text{(l)}$, or $\text{(g)}$ substances.

  • $\text{Ag}^+\text{(aq)} + \text{NO}_3^-\text{(aq)} + \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl}\text{(s)} + \text{Na}^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$

• Step 3: Cancel out the spectator ions that appear identically on both sides ($\text{Na}^+$ and $\text{NO}_3^-$).

• Step 4: Rewrite the final ionic equation.

  • $\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl}\text{(s)}$

Half Equations and Electron Transfer (Higher Tier)

A half equation focuses on the exact transfer of electrons (written as $\text{e}^-$) for a single element during a redox reaction.

To write them correctly, you must use the "OIL RIG" principle. Oxidation is the loss of electrons (electrons are written as products on the right). Reduction is the gain of electrons (electrons are written as reactants on the left).

Step-by-Step Mechanism for Half Equations: Write the half equation for the oxidation of Bromide ions.

• Step 1: Identify species: Reactant is $\text{Br}^-$ (ion), product is $\text{Br}_2$ (diatomic molecule).

• Step 2: Balance the atoms: You need two bromide ions to make one bromine molecule.

  • $2\text{Br}^- \rightarrow \text{Br}_2$

• Step 3: Balance the charges: The left side has a total charge of $2-$. The right side has a charge of $0$. Add $2\text{e}^-$ to the right side to balance it.

  • $2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-$

Balancing Ionic Equations with Different Charges (Higher Tier)

When metals displace ions with different electrical charges (e.g., $3+$ and $2+$), the final ionic equation must be perfectly balanced for BOTH the number of atoms and the total electrical charge.

The number of electrons lost by the oxidised metal must exactly equal the number of electrons gained by the reduced metal ion.

Worked Example: Balancing Displacement Charges

Balance the ionic equation for Aluminium displacing Copper(II) ions: $\text{Al}\text{(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Al}^{3+}\text{(aq)} + \text{Cu}\text{(s)}$

Step 1: Write the separate half equations.

• Oxidation: $\text{Al} \rightarrow \text{Al}^{3+} + 3\text{e}^-$
• Reduction: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$

Step 2: Find the Lowest Common Multiple (LCM) of the transferred electrons. The LCM of 3 and 2 is 6. You must transfer 6 electrons in total.

Step 3: Multiply both equations to reach 6 electrons.

• Multiply Aluminium by 2: $2\text{Al} \rightarrow 2\text{Al}^{3+} + 6\text{e}^-$
• Multiply Copper by 3: $3\text{Cu}^{2+} + 6\text{e}^- \rightarrow 3\text{Cu}$

Step 4: Combine the equations and cancel the $6\text{e}^-$ from both sides.

• $2\text{Al}\text{(s)} + 3\text{Cu}^{2+}\text{(aq)} \rightarrow 2\text{Al}^{3+}\text{(aq)} + 3\text{Cu}\text{(s)}$ (Final check: Both sides have $6+$ total charge, 2 Aluminium atoms, and 3 Copper atoms).