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AQA A-Level Chemistry Notes

1.2.5 Balanced Equations and Associated Calculations

Writing and Balancing Chemical Equations

Understanding Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants (substances that start a reaction) on the left side and the products (substances formed from the reaction) on the right. These equations must be balanced to obey the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.

Steps to Balancing Equations

  1. Identify the Reactants and Products: Begin by writing the correct chemical formulas for all the reactants and products involved in the reaction.
  2. Balance Atoms One Element at a Time: Adjust the coefficients - the numbers placed before the chemical formulas - to ensure that the number of atoms for each element is the same on both sides of the equation. Remember, altering subscripts in a formula changes the substance and is not allowed.
  3. Balance Polyatomic Ions as a Unit: If a polyatomic ion appears unchanged on both sides of the equation, balance it as a whole.
  4. Check for Balance: Confirm that all elements are balanced and that the equation adheres to the Law of Conservation of Mass.

Balancing Ionic Equations

In ionic equations, we represent ions in aqueous solutions separately.

  • Full Ionic Equations: These show all the ions present in the reaction mixture.
  • Net Ionic Equations: These are simplified versions that exclude spectator ions (ions that do not participate in the reaction) to highlight the actual chemical change.

Calculating Percentage Atom Economy

Definition and Importance

Atom economy is a measure that reflects the efficiency of a chemical reaction in terms of how well reactants are converted into useful products. It is a concept of great importance in green chemistry, aiming to minimize waste and maximize efficiency.

Calculation Method

The formula for percentage atom economy is:

[ \text{Atom Economy} = \left( \frac{\text{Molecular Mass of Desired Product}}{\text{Sum of Molecular Masses of All Reactants}} \right) \times 100 ]

Significance

A higher atom economy is preferable as it indicates a more environmentally friendly and resource-efficient reaction. It is particularly significant in industrial chemistry where the scale of reactions can lead to substantial waste if atom economy is low.

Stoichiometric Calculations

Using Balanced Equations

Balanced chemical equations provide a stoichiometric ratio, indicating the proportional amounts of reactants and products. This ratio is fundamental for calculating the quantities of substances involved in chemical reactions.

Calculating Masses and Volumes

  • Mass Calculations: Utilize the molar ratio from the balanced equation to find the mass of reactants or products required or produced in a reaction.
  • Gas Volumes: When dealing with gases, use the molar volume (24 dm3 at room temperature and pressure) to calculate the volume of gases involved in the reaction.

Concentrations and Solutions

  • When considering reactions in solution, concentration (expressed in mol dm–3) becomes a key factor. The balanced equation can help determine how the concentrations of different reactants and products relate to each other.

Percentage Yields and Atom Economies

Percentage Yield

  • Percentage yield is a measure of the efficiency of a reaction, indicating what proportion of the theoretical yield (as predicted by stoichiometry) was actually obtained in the experiment.
  • It's calculated using the formula:

( \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 )

  • Understanding and calculating percentage yield is important in both laboratory and industrial contexts, as it helps in assessing the practical efficiency of reactions.

Atom Economy in Context

  • Atom economy is not just a theoretical concept; it has practical implications in the context of sustainable chemistry. It encourages the development of processes that maximize the formation of useful products while minimizing waste.
  • This concept is integral to assessing the environmental and economic impact of chemical processes, particularly in industrial chemistry.

Practical Applications

Real-World Scenarios

  • The skills in balancing equations and performing stoichiometric calculations are not confined to the classroom. They are extensively used in chemical industries for scaling up reactions from the lab to production scale, determining the cost-effectiveness of processes, and assessing their environmental impact.
  • In research and development, these calculations are crucial for developing new products and processes.

Skills Development

  • Mastery of these concepts and calculationsis essential for chemists and chemical engineers. It equips them with the knowledge to design more efficient, cost-effective, and environmentally friendly chemical processes.
  • These skills also enhance a chemist's ability to critically evaluate and optimize existing chemical processes for better outcomes.

Understanding balanced chemical equations and the related calculations is fundamental to the study of chemistry. These concepts form the basis for much of the quantitative analysis in chemical reactions and are essential for any student aspiring to excel in A level Chemistry. This knowledge not only aids in academic success but also prepares students for practical applications in various fields of chemistry.

FAQ

Stoichiometric calculations, in their basic form, assume ideal conditions where reactions go to completion, reactants are pure, and there are no side reactions. However, in real-world scenarios, these conditions are seldom met. Factors such as reaction conditions (temperature, pressure, catalysts) and the purity of reactants significantly impact the outcome of a reaction. For instance, impurities in reactants can reduce the yield, and non-ideal conditions can shift the equilibrium of reversible reactions, affecting the amount of product formed. Moreover, side reactions can consume some reactants, leading to lower yields of the desired product. To account for these real-world factors, chemists use the concept of percentage yield, which compares the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry. This adjustment provides a more realistic view of the reaction’s efficiency. In industrial processes, understanding these variations is crucial for process optimisation, cost estimation, and environmental impact assessment. Chemists must often perform additional calculations or experiments to determine the actual efficiency and yields under specific conditions.

Balancing equations for combustion reactions, particularly those involving hydrocarbons (compounds consisting of hydrogen and carbon), follows a systematic approach. First, write the unbalanced equation with the hydrocarbon and oxygen on the reactant side, and carbon dioxide and water on the product side. Start by balancing the carbon atoms, followed by hydrogen atoms, and finally, adjust the oxygen atoms. An essential tip is to balance the oxygen atoms last since they are in both CO₂ and H₂O. For instance, in the combustion of methane (CH₄), the unbalanced equation is CH₄ + O₂ → CO₂ + H₂O. Balance the carbons (1 C in CH₄ and 1 C in CO₂), then hydrogens (4 H in CH₄ and 2 H in H₂O, so put a 2 before H₂O), and finally oxygens (2 O in CO₂ and 4 O in 2H₂O, totalling 6 O atoms, thus 3 O₂ molecules). The balanced equation becomes CH₄ + 2O₂ → CO₂ + 2H₂O. This systematic approach ensures accuracy, especially in complex hydrocarbon combustion reactions.

The stoichiometric coefficient in a balanced chemical equation represents the relative number of moles of each reactant and product involved in the reaction. These coefficients are essential for understanding the proportions in which different substances react and are formed. For instance, in the equation 2H₂ + O₂ → 2H₂O, the coefficients 2, 1, and 2 indicate that two moles of hydrogen react with one mole of oxygen to produce two moles of water. In reaction calculations, these coefficients are used to determine the molar ratios of reactants and products, which are crucial for calculating the amounts of substances involved. For example, if you know the amount of one reactant, you can use the stoichiometric coefficients to find the amount of another reactant required or the amount of product that will be formed. This concept is fundamental in stoichiometry and is applied in a wide range of calculations including determining the limiting reactant, calculating theoretical yields, and even in more complex scenarios such as gas law calculations where volumes are directly proportional to moles.

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, thus determining the amount of product formed. The excess reactant, on the other hand, is the substance that remains after the reaction has completed. When performing stoichiometric calculations, it is crucial to identify the limiting reactant as it dictates the maximum amount of product that can be formed. For example, if a reaction involves two reactants A and B in a 1:1 ratio, and there is more of reactant B present than required, reactant A becomes the limiting reactant. The amount of product formed is calculated based on the moles of reactant A. In practical scenarios, this concept is essential for optimising the efficiency and cost-effectiveness of chemical processes. For instance, in industrial settings, identifying and using the correct amount of limiting reactant can significantly reduce waste and increase yield, impacting both economic and environmental aspects of the process.

Stoichiometry plays a crucial role in environmental chemistry, especially in the field of pollution control. It is used to understand and quantify the reactions involved in the formation and breakdown of pollutants. For instance, stoichiometry is essential in calculating the amounts of reactants needed to neutralize pollutants in wastewater treatment or to reduce emissions in industrial processes. In the case of combustion reactions, stoichiometric calculations help in determining the optimal fuel-to-air ratio to minimise the production of pollutants like nitrogen oxides (NOx) and sulphur oxides (SOx). Additionally, these calculations are instrumental in designing processes for the removal of pollutants from the environment, such as in the scrubbing of flue gases to remove acidic gases. By understanding the exact stoichiometry of reactions involved in the formation and mitigation of pollutants, chemists can design more efficient and effective methods for pollution control, contributing significantly to environmental protection and sustainability. This application of stoichiometry demonstrates its importance not just in theoretical chemistry but also in practical, real-world scenarios where it can have a significant impact on environmental health.

Practice Questions

Given the unbalanced chemical equation for the combustion of ethanol: ( C_2H_5OH(l) + O_2(g) \rightarrow CO_2(g) + H_2O(g) ) Balance the equation and calculate the volume of carbon dioxide produced at room temperature and pressure when 23 grams of ethanol (C2H5OH) are completely burnt. (Molar mass of ethanol = 46 g/mol, Molar volume of gas at RTP = 24 dm³/mol)

The balanced chemical equation is:

( C2H5OH(l) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) )

To find the volume of CO₂ produced, first, calculate the number of moles of ethanol:

( \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{23}{46} = 0.5 \text{ moles} )

Since 1 mole of ethanol produces 2 moles of CO₂, 0.5 moles of ethanol will produce 1 mole of CO₂.

Finally, calculate the volume of CO₂ at RTP:

( \text{Volume of CO₂} = \text{Number of moles} \times \text{Molar volume} = 1 \times 24 = 24 \text{ dm³} )

Thus, 23 grams of ethanol produce 24 dm³ of CO₂ at RTP.

Acetylene (C2H2) is used in welding. When 52 grams of acetylene is burnt in excess oxygen, calculate the percentage yield if 88 grams of CO2 is produced. (Molar mass of C2H2 = 26 g/mol, CO2 = 44 g/mol)

First, write and balance the equation for the combustion of acetylene:

( 2C2H2(g) + 5O2(g) \rightarrow 4CO2(g) + 2H2O(g) )

Calculate the moles of acetylene:

( \text{Number of moles of C}2\text{H}2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{52}{26} = 2 \text{ moles} )

From the balanced equation, 2 moles of C2H2 produce 4 moles of CO2.

The theoretical yield of CO2 is:

( 4 \text{ moles} \times 44 \text{ g/mol} = 176 \text{ grams} )The percentage yield is:( \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{88}{176} \right) \times 100 = 50\% )

Therefore, the percentage yield of CO2 in this reaction is 50%.

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