THEORETICAL PERFORMANCE: WHAT IT CONSISTS OF AND EXAMPLES - CHEMISTRY

The theoretical yield of a chemical reaction is the maximum quantity that can be obtained from a product assuming the complete transformation of the reactants. When for kinetic, thermodynamic or experimental reasons one of the reactants partially reacts, the resulting yield is less than theoretical.

This concept allows you to compare the gap between chemical reactions written on paper (chemical equations) and reality. Some may look very simple, but experimentally complex and low-yielding; while others can be extensive but simple and high-performance when carried out.

Source: Pxhere

All chemical reactions and amounts of reagents have a theoretical yield. Thanks to this, a degree of the effectiveness of the process variables and the hits can be established; the higher the yield (and the shorter the time), the better the conditions chosen for the reaction.

Thus, for a given reaction, a temperature range, stirring speed, time, etc. can be chosen and an optimal performance can be carried out. The purpose of such efforts is to approximate the theoretical yield to the actual yield.

What is the theoretical yield?

The theoretical yield is the amount of product obtained from a reaction assuming a conversion of 100%; that is, all of the limiting reagent must be consumed.

So, every synthesis should ideally give an experimental or real yield equal to 100%. Although this does not occur, there are reactions with high yields (> 90%)

It is expressed in percentages, and to calculate it you must first resort to the chemical equation of the reaction. From the stoichiometry, it is determined for a certain quantity of limiting reagent how much product originates. After this, the quantity of product obtained (actual yield) is compared with that of the theoretical value determined:

% Yield = (Actual Yield / Theoretical Yield) ∙ 100%

This% yield allows estimating how efficient the reaction has been under the selected conditions. Their values vary drastically depending on the type of reaction. For example, for some reactions a 50% yield (half the theoretical yield) can be considered a successful reaction.

But what are the units of such performance? The mass of the reactants, that is, their number of grams or moles. Therefore, to determine the yield of a reaction, the grams or moles that can theoretically be obtained must be known.

The above can be clarified with a simple example.

Examples

Example 1

Consider the following chemical reaction:

A + B => C

1gA + 3gB => 4gC

The chemical equation has only 1 stoichiometric coefficients for species A, B and C. As they are hypothetical species, their molecular or atomic masses are unknown, but the mass proportion in which they react is available; that is, for each gram of A, 3 g of B react to give 4 g of C (conservation of mass).

Therefore, the theoretical yield for this reaction is 4 g of C when 1g of A reacts with 3g of B.

What would be the theoretical yield if we have 9g of A? To calculate it, just use the conversion factor that relates A and C:

(9g A) ∙ (4g C / 1g A) = 36g C

Note that now the theoretical yield is 36 g C instead of 4 g C, since there is more reagent A.

Two methods: two returns

For the above reaction there are two methods to produce C. Assuming both start with 9g of A, each has its own actual yield. The classical method makes it possible to obtain 23 g of C in a period of 1 hour; while using the modern method, 29 g of C can be obtained in half an hour.

What is the% yield for each of the methods? Knowing that the theoretical yield is 36 g of C, the general formula is applied:

% yield (classical method) = (23g C / 36g C) ∙ 100%

63.8%

% yield (modern method) = (29g C / 36g C) ∙ 100%

80.5%

Logically, the modern method by originating more grams of C from the 9 grams of A (plus the 27 grams of B) has a yield of 80.5%, higher than the yield of 63.8% of the classical method.

Which of the two methods to choose? At first glance, the modern method seems more viable than the classical method; However, the economic aspect and the possible environmental impacts of each one come into play in the decision.

Example 2

Consider exothermic and promising reaction as an energy source:

H ₂ + O ₂ => H ₂ O

Note that as in the previous example, the stoichiometric coefficients of H ₂ and O ₂ are 1. If you have 70g of H ₂ mixed with 150g of O ₂, what will be the theoretical yield of the reaction? What is the yield if 10 and 90g of H ₂ O are obtained?

Here it is uncertain how many grams of H ₂ or O ₂ react; therefore, the moles of each species must be determined this time:

Moles of H ₂ = (70g) ∙ (mol H ₂ / 2g)

35 moles

Moles of O ₂ = (150g) ∙ (mol O ₂ / 32g)

4.69 moles

The limiting reagent is oxygen, because 1mol of H ₂ reacts with 1mol of O ₂; and since there are 4.69 moles of O ₂, then 4.69 moles of H _{2 will} react. Likewise, the moles of H ₂ O formed will be equal to 4.69. Therefore, the theoretical yield is 4.69 moles or 84.42g of H ₂ O (multiplying the moles by the molecular mass of water).

Lack of oxygen and excess impurities

If 10g of H ₂ O are produced, the yield will be:

% yield = (10g H ₂ O / 84.42g H ₂ O) ∙ 100%

11.84%

Which is low because a huge volume of hydrogen was mixed with very little oxygen.

And if, on the other hand, 90g H ₂ O are produced, the yield will now be:

% yield = (90g H ₂ O / 84.42g H ₂ O) ∙ 100%

106.60%

No performance can be higher than theoretical, so anything above 100% is an anomaly. However, it can be due to the following causes:

-The product accumulated other products caused by side or secondary reactions.

-The product became contaminated during or at the end of the reaction.

For the case of the reaction in this example, the first cause is unlikely, since there is no other product besides water. The second cause, if 90g of water were actually obtained under such conditions, indicates that there was an entry of other gaseous compounds (such as CO ₂ and N ₂) that were erroneously weighed together with the water.

References

Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning, p 97.
Helmenstine, Todd. (2018, February 15). How to Calculate Theoretical Yield of a Chemical Reaction. Recovered from: thoughtco.com
Chieh C. (June 13, 2017). Theoretical and Actual Yields. Chemistry LibreTexts. Recovered from: chem.libretexts.org
Khan Academy. (2018). Limiting reagents and percent yield. Recovered from: khanacademy.org
Introductory Chemistry. (sf). Yields. Recovered from: saylordotorg.github.io
Introductory course in general chemistry. (sf). Limiting reagent and performance. University of Valladolid. Recovered from: eis.uva.es

THEORETICAL PERFORMANCE: WHAT IT CONSISTS OF AND EXAMPLES - CHEMISTRY - 2025