Monitoring and controlling chemical reactions Revision Notes

Understanding Theoretical Mass

The theoretical mass of a product is the maximum amount of product that can be formed from a given amount of reactant, assuming the reaction goes to completion and there are no losses. It is calculated using stoichiometry from a balanced chemical equation. This value represents the ideal yield and serves as a benchmark against which the actual experimental yield can be compared. If a reaction has multiple reactants, the theoretical yield is determined by the limiting reactant.

Key Steps for Calculation

To calculate the theoretical mass of a product:

1. Ensure the chemical equation is balanced. This gives the mole ratios.
2. Calculate the relative formula mass (Mr) for the given reactant and the desired product. Use the relative atomic masses (Ar) provided in the periodic table (e.g., C=12, O=16, H=1).
3. Convert the given mass of reactant into moles using the formula: moles = mass / Mr.
4. Use the molar ratio from the balanced equation to find the moles of the product that can be formed.
5. Convert the moles of product back into mass using the formula: mass = moles × Mr. This is your theoretical mass.

Example Calculation Walkthrough

Consider the reaction: 2Na + Cl₂ → 2NaCl. If you start with 2.3 g of sodium (Na), what is the theoretical mass of sodium chloride (NaCl) produced?

1. Balanced equation: 2Na + Cl₂ → 2NaCl (already balanced).
2. Relative formula masses: Na = 23, Cl = 35.5. Mr(Na) = 23 g/mol. Mr(NaCl) = 23 + 35.5 = 58.5 g/mol.
3. Moles of reactant (Na): Moles Na = 2.3 g / 23 g/mol = 0.1 mol.
4. Molar ratio: From the equation, 2 moles of Na produce 2 moles of NaCl. So, the ratio is 1:1. Therefore, moles NaCl = 0.1 mol.
5. Theoretical mass of product (NaCl): Mass NaCl = 0.1 mol × 58.5 g/mol = 5.85 g. The theoretical mass of NaCl is 5.85 g.

Understanding Molar Gas Volume (Higher)

The molar gas volume is the volume occupied by one mole of any gas at a specific temperature and pressure. At room temperature and pressure (RTP), which is typically taken as 20°C and 1 atmosphere (atm) or 101 kPa, the molar gas volume is approximately 24 dm³ per mole. This is a crucial constant for calculations involving gases in chemical reactions. This value means that whether you have one mole of hydrogen gas (H₂), oxygen gas (O₂), or carbon dioxide gas (CO₂), they will all occupy the same volume of 24 dm³ under RTP conditions. This principle allows for direct comparison of volumes of reacting gases based on their molar ratios from balanced equations.

Using Balanced Equations to Find Moles (Higher)

Before calculating gas volumes, you must have a balanced chemical equation for the reaction. The coefficients in a balanced equation represent the molar ratios of reactants and products. For example, in the reaction 2H₂(g) + O₂(g) → 2H₂O(g), the ratio of hydrogen gas to oxygen gas to water vapour is 2:1:2 moles. If you know the number of moles of one substance, you can use these ratios to find the number of moles of any other gaseous substance in the reaction. This step is fundamental to linking the stoichiometry of the reaction to the volumes of gases involved.

Calculating Gas Volume from Moles (Higher)

Once the number of moles of a gas is determined using the balanced chemical equation, the volume of the gas at RTP can be calculated using the molar gas volume constant. The formula is: Volume of Gas (dm³) = Moles of Gas × 24 dm³/mol. For instance, if you have 0.5 moles of carbon dioxide, its volume at RTP would be 0.5 mol × 24 dm³/mol = 12 dm³. It is essential to ensure that the volume unit is dm³ when using 24 dm³/mol. If the required units are cm³, convert the final answer (1 dm³ = 1000 cm³). Pay attention to the specified conditions of the reaction; this constant applies specifically to room temperature and pressure.

Calculating Moles from Gas Volume (Higher)

Conversely, if you are given the volume of a gas at RTP and need to find the number of moles, you can rearrange the formula: Moles of Gas = Volume of Gas (dm³) / 24 dm³/mol. For example, if 6 dm³ of nitrogen gas is produced, the number of moles would be 6 dm³ / 24 dm³/mol = 0.25 moles. This is particularly useful when experimental data provides gas volumes, and you need to work backwards to determine reactant amounts or verify product yields in terms of moles. Always convert volumes to dm³ if they are initially given in cm³ before using the 24 dm³/mol constant.

What is Titration?

Titration is a laboratory technique used to find the unknown concentration of a solution. It involves a controlled reaction between a solution of known concentration (titrant) and a solution of unknown concentration (titrand). The goal is to reach the equivalence point, where the reactants have completely reacted in their stoichiometric ratio. This is often an acid-base neutralisation reaction.

Key Apparatus and Their Uses

The main pieces of equipment for titration include a burette, pipette, conical flask, and a white tile. A burette is used to deliver variable, precise volumes of the titrant. A pipette measures a fixed, accurate volume of the titrand. The conical flask holds the titrand and indicator, allowing for swirling without spillage. A white tile under the flask helps to clearly observe the indicator's colour change.

Performing a Titration

First, rinse the burette with the titrant and the pipette with the titrand. Use the pipette to transfer a known volume of the titrand into a conical flask, adding a few drops of an appropriate indicator. Fill the burette with the titrant to the zero mark. Slowly add the titrant from the burette to the flask, swirling constantly. Stop immediately when the indicator changes colour permanently, marking the endpoint. Record the volume of titrant used. Repeat the experiment several times to obtain concordant results (within 0.1 cm³).

Calculating Concentration

Once concordant results are obtained, calculate the average volume of titrant used. Using this average volume, the known concentration of the titrant, the known volume of the titrand, and the balanced chemical equation, the unknown concentration of the titrand can be determined. For acid-base reactions, this involves using the formula: (Concentration of acid × Volume of acid) / (Moles of acid coefficient) = (Concentration of base × Volume of base) / (Moles of base coefficient), or mole calculations.

What are Reversible Reactions?

A reversible reaction is a chemical reaction where the products can react together to reform the original reactants. This is indicated by a double arrow (⇌) in the chemical equation, signifying that the reaction proceeds in both the forward and reverse directions simultaneously. For example, A + B ⇌ C + D. The forward reaction creates C and D from A and B, while the reverse reaction converts C and D back into A and B. At equilibrium, the rates of the forward and reverse reactions are equal, meaning the concentrations of reactants and products remain constant, although the reactions are still occurring.

How Conditions Affect Reversible Reactions

The position of equilibrium in a reversible reaction can be shifted by changing external conditions. These conditions include temperature, pressure, and the concentration of reactants or products. For instance, increasing the temperature will favour the endothermic direction of the reaction (the one that absorbs heat), while decreasing the temperature will favour the exothermic direction (the one that releases heat). For gaseous reactions, increasing the pressure will favour the side of the reaction that produces fewer moles of gas, reducing the overall pressure. Conversely, decreasing pressure will favour the side with more moles of gas. Changing the concentration of a reactant or product will also shift the equilibrium to counteract that change.

Examples of Reversible Reactions

A common example of a reversible reaction is the Haber process for ammonia synthesis: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Another example is the dehydration of copper(II) sulfate: CuSO₄·5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(g). In this reaction, hydrated copper(II) sulfate is blue, and when heated, it turns into white anhydrous copper(II) sulfate and water vapour. Adding water to anhydrous copper(II) sulfate reverses the reaction, turning it blue again. This principle is used in desiccating agents and chemical tests for water.