Reaction Yield (HSC SSCE Chemistry): Revision Notes

Reaction Yield

Introduction

When chemical reactions occur, they rarely produce 100% of the theoretical amount of product. Understanding and maximising reaction yield is essential in industrial chemistry, where producing the maximum amount of product efficiently determines whether a process is economically viable. This note explores what yield means, how to calculate it, and the strategies used in industry to optimise product formation.

What is yield?

Yield describes how much product is actually formed in a chemical reaction. When performing stoichiometric calculations, we typically assume all reactants convert completely to products. However, in practice this rarely happens.

Theoretical yield is the maximum amount of product that could be formed if the reaction proceeded with perfect efficiency. This is calculated using stoichiometry based on the limiting reactant.

Actual yield is the amount of product that is really obtained from the reaction. This is always less than or equal to the theoretical yield.

Percentage yield expresses the actual yield as a percentage of the theoretical yield, providing a measure of reaction efficiency:

$\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100\%$

Maximising yield

Industrial chemists design reaction conditions to produce as much product as possible. The goal is to shift the equilibrium position as far to the right (toward products) as possible, meaning a much higher proportion of products than reactants exists at equilibrium. Several strategies achieve this:

Removing the product

When product molecules are removed as they form, their concentration decreases. According to Le Chatelier's principle, this disturbance causes equilibrium to shift right, favouring the forward reaction and producing more product. Continuous removal of product is a key strategy in industrial processes.

Recycling unreacted reactants

Rather than wasting unreacted starting materials, industrial processes typically recycle them back into the reaction vessel. This approach serves two purposes: it reduces waste and increases reactant concentration. Higher reactant concentration favours the forward reaction, again shifting equilibrium to the right and increasing yield.

Controlling temperature

Temperature affects both the yield and the rate of reaction. For exothermic reactions (which release heat), lower temperatures favour the forward reaction and increase yield. However, lower temperatures also slow the reaction rate because fewer particles have sufficient activation energy for successful collisions.

For endothermic reactions (which absorb heat), higher temperatures favour the forward reaction and increase yield, but this requires energy input.

Adjusting pressure for gaseous systems

For reactions involving gases, pressure affects the equilibrium position. According to Le Chatelier's principle, increasing pressure favours the side of the equation with fewer gas molecules. If the forward reaction produces fewer gas molecules than the reactants, higher pressure increases yield.

Higher pressure also increases the reaction rate by increasing concentration and therefore collision frequency between reactant particles.

However, extremely high pressures are expensive to generate and maintain, and pose safety risks due to potential equipment failure. Industrial processes therefore use moderate pressures that balance yield, rate, cost, and safety considerations.

Using catalysts

Catalysts increase the rate of reaction by providing an alternative reaction pathway with lower activation energy. This means more particles possess sufficient energy for successful collisions, allowing the reaction to proceed faster.

Enzymes are biological catalysts (proteins) that can be used in industrial processes. They are particularly useful when the temperature or pressure required for an adequate reaction rate would be uneconomical, or when yield would be too low at high temperatures or pressures.

Balancing yield and rate

In an ideal situation, a reaction would have both high yield and high rate, producing large amounts of product quickly and maximising profit while minimising waste. Unfortunately, the conditions that favour high yield often slow the reaction rate, and vice versa. Industrial chemists must find a compromise that produces sufficient product in reasonable time at acceptable cost.

Ensuring product purity

The purity of the required product is also critical. Unwanted by-products must be removed either before or after the final product is collected. Different grades of purity are acceptable for different uses:

• Technical grade: suitable for non-critical laboratory tasks like rinsing and dissolving
• Synthesis reagents: suitable for organic synthesis
• For analytical purposes: suitable for quantitative analysis, research, and quality control
• Pharmacopoeia grade: meets purity requirements for pharmaceutical production

Industrial case study: The contact process

The contact process produces sulfuric acid ($\text{H}_2\text{SO}_4$), one of the most important industrial chemicals worldwide. Sulfuric acid is used in metal production, agriculture, and manufacturing of paints, fertilisers, car batteries, detergents, dyes, and fibres.

The process involves four main steps:

Step 1: $\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g)$

or: $4\text{FeS}_2(s) + 11\text{O}_2(g) \rightarrow 2\text{Fe}_2\text{O}_3(s) + 8\text{SO}_2(g)$

Step 2: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$

Step 3: $3\text{SO}_3(g) + \text{H}_2\text{SO}_4(l) \rightarrow \text{H}_2\text{S}_2\text{O}_7(l)$ (oleum)

Step 4: $\text{H}_2\text{S}_2\text{O}_7(l) + \text{H}_2\text{O}(l) \rightarrow 2\text{H}_2\text{SO}_4(l)$

Excess reagent in the contact process

Air is used as the source of oxygen in Step 1. Air is cheap and readily available, making it an economical choice. An excess of air is deliberately used, providing enough oxygen for both Step 1 and Step 2. This makes the process more efficient since the reactants for Step 2 ($\text{SO}_2$ and $\text{O}_2$) are already mixed together.

The stoichiometric ratio for Step 2 is $2\text{SO}_2:1\text{O}_2$. However, the amount of air is controlled to give a 1:1 ratio of sulfur dioxide to oxygen, providing an excess of oxygen. This excess has two benefits:

• Increased reaction rate: Higher oxygen concentration increases collision frequency between reactant particles
• Increased yield: According to Le Chatelier's principle, increasing the concentration of a reactant (oxygen) shifts equilibrium to the right, increasing the yield of sulfur trioxide

Temperature in the contact process

Step 2 is an exothermic reaction with $\Delta H = -196\text{ kJ}\cdot\text{mol}^{-1}$. Temperature presents a dilemma:

• Increasing temperature increases the rate (more particles have sufficient energy for successful collisions) but decreases yield (favours the endothermic reverse reaction)
• Decreasing temperature increases yield (favours the exothermic forward reaction) but decreases rate

Pressure in the contact process

The pressure used is relatively low at 1–2 atm. Higher pressure would provide benefits:

• Increased rate: Higher pressure increases reactant concentration and collision frequency
• Increased yield: Since there are more gas molecules in the reactants (3 molecules) than products (2 molecules), increased pressure would shift equilibrium right, increasing sulfur trioxide yield

However, the cost of producing and maintaining very high pressure safely outweighs these benefits. Therefore, a lower, more economical pressure is used.

Catalyst in the contact process

Vanadium oxide ($\text{V}_2\text{O}_5$) is used as a catalyst in Step 2. Although this does not affect the yield of sulfur trioxide, it increases the rate at which it is produced. This allows the subsequent steps to proceed faster, making the overall process more efficient and economically viable.

Steps 3 and 4 explanation

Industrial case study: The Haber process

The Haber process produces ammonia from nitrogen and hydrogen gases. Developed in 1915, it remains the main method for producing nitrogen-based fertilisers. An industrial plant can produce around 1000 tonnes of ammonia per day.

The equation for the reaction is:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92.4\text{ kJ}\cdot\text{mol}^{-1}$

The Haber process exemplifies how industrial conditions represent a compromise between maximising yield and rate whilst remaining cost-effective.

Temperature compromise in the Haber process

This is an exothermic reaction, so temperature affects yield and rate in opposite ways:

• Increasing temperature increases the rate (more particles have sufficient energy for successful collisions) but decreases yield (favours the endothermic reverse reaction, reducing ammonia formation)
• Decreasing temperature increases yield (favours the exothermic forward reaction, producing more ammonia) but decreases rate

Pressure compromise in the Haber process

The equation shows that the forward reaction produces fewer gas molecules (2 molecules of $\text{NH}_3$) than the reactants (4 molecules total: 1 $\text{N}_2$ + 3 $\text{H}_2$). Therefore, pressure strongly influences this reaction:

• Increasing pressure increases the rate (higher concentration leads to more collisions) and increases yield (favours the side with fewer gas molecules, shifting equilibrium right)

Catalysts in the Haber process

An iron/iron oxide catalyst is used, with small amounts of potassium oxide and aluminium oxide. This provides an alternative reaction pathway with lower activation energy, meaning more particles have sufficient energy for successful collisions. This increases the reaction rate without affecting yield.

The catalyst allows the reaction to proceed at an acceptable rate despite the compromise temperature and pressure being used, making the process economically viable.

Calculating theoretical yield using limiting reactants

To calculate the theoretical yield, we must identify which reactant limits the amount of product that can be formed. The limiting reactant (or limiting reagent) is the reactant that runs out first, preventing any more product from being made. Other reactants present in larger amounts are excess reactants (or excess reagents).

Method for identifying the limiting reactant

There are several methods to determine which reactant is limiting. One effective approach compares the stoichiometric ratio (SR) from the balanced equation with the actual ratio (AR) of moles present:

1. Calculate the number of moles of each reactant
2. Determine the stoichiometric ratio (SR) using coefficients from the balanced equation
3. Determine the actual ratio (AR) using the calculated moles
4. Compare the ratios: If AR > SR, the reactant in the denominator (bottom) is limiting. If SR > AR, the reactant in the numerator (top) is limiting
5. Use moles of the limiting reactant to calculate moles of product
6. Convert moles of product to mass, volume, or concentration as required

Calculating percentage yield

The actual yield is usually expressed as a percentage of the theoretical yield. A percentage yield of 90% means only 90% of the theoretically possible amount of product was actually produced.

To calculate percentage yield:

1. Use stoichiometry to calculate the theoretical amount of product that could be formed from the reactants
2. Note the actual amount of product obtained
3. Apply the formula:

$\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100\%$

Determining purity

Whilst percentage yield measures the quantity of product obtained, purity measures the quality of the final product. Achieving 100% purity is almost impossible in chemical synthesis, so products must be analysed for impurities and percentage purity determined.

Grades of purity

Different purity levels are acceptable depending on the intended use:

• Technical grade: used for non-critical laboratory tasks such as rinsing, dissolving, and as raw materials
• Synthesis reagents: suitable for organic synthesis reactions
• For analytical purposes: suitable for quantitative analysis, research, and quality control
• Pharmacopoeia grade: meets purity requirements for pharmaceutical production (human consumption)

Calculating per cent purity

Per cent purity is calculated using:

$\text{Per cent purity} = \frac{\text{Mass of useful product}}{\text{Total mass of sample}} \times 100\%$