The theoretical performance of a chemical reaction is the maximum quantity that can be obtained from a product assuming the complete transformation of the reactants. When one of the reactants partially reacts for kinetic, thermodynamic or experimental reasons, the resulting yield is less than theoretical..
This concept makes it possible 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.
All chemical reactions and amounts of reagents have a theoretical yield. Thanks to this, a degree of 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..
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The theoretical yield is the amount of product obtained from a reaction assuming a conversion of 100%; that is, all the limiting reagent must be consumed.
So, every synthesis should ideally give an experimental or real yield equal to 100%. Although this does not happen, 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 stoichiometry, it is determined for a certain amount of limiting reagent how much product originates. After this, the quantity of product obtained (real 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 yield of 50% (half the theoretical yield) can be considered as 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 be theoretically obtained must be known..
The above can be clarified with a simple example.
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 4g C, since there is more reagent A.
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 allows 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..
Consider the exothermic and promising reaction as an energy source:
Htwo + ORtwo => HtwoOR
Note that as in the previous example, the stoichiometric coefficients of Htwo metwo are 1. We have 70g of Htwo mixed with 150g of Otwo, What will be the theoretical yield of the reaction? What is the yield if 10 and 90g of HtwoOR?
Here it is uncertain how many grams of Htwo or ortwo they react; therefore, the moles of each species must be determined this time:
Moles of Htwo= (70g) ∙ (mol Htwo/2 g)
35 moles
Moles of Otwo= (150g) ∙ (mol Otwo/ 32g)
4.69 moles
The limiting reagent is oxygen, since 1mol of Htwo reacts with 1mol Otwo; and having 4.69 moles of Otwo, then 4.69 moles of H will reacttwo. Likewise, the moles of HtwoOr formed is going to be equal to 4.69. Therefore, the theoretical yield is 4.69 moles or 84.42g of HtwoO (multiplying the moles by the molecular mass of water).
If 10g of HtwoOr, the performance will be:
% yield = (10g HtwoO / 84.42g HtwoO) ∙ 100%
11.84%
Which is low because a huge volume of hydrogen mixed with very little oxygen.
And if, on the other hand, 90g H are producedtwoOr, the performance will now be:
% yield = (90g HtwoO / 84.42g HtwoO) ∙ 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 was 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 COtwo and Ntwo) that were wrongly weighed together with the water.
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