PREDICT RENDEMENT OF PRODUCT A REACTION

    In chemistry, the chemical yield, the yield of the reaction, or only the rendement refers to the amount of reaction product produced in the chemical reaction. Absolute rendement can be written as weight in grams or in moles (molar yield). The relative yield used as a calculation of the effectiveness of the procedure is calculated by dividing the amount of product obtained in moles by the theoretical yield in moles

 I. Estimated Formation of Deposition Based on Ksp Price

            The solubility product is generally denoted by Qc and the method of calculating Ksp is the product of the concentration of dissolved ions in a soluble salt. The difference is that the Qc value shows the solubility product in the unconfirmed state whether the solution is unsaturated, precisely saturated or saturated (precipitated) while Ksp is the product of the concentration of ions decomposed in a soluble salt at the maximum saturated). To determine the state of the solution, we must know the value of Ksp first.

If the value of Qc <Ksp then the solution is not saturated.

If the value of Qc = Ksp then the solution is exactly saturated.

If the value of Qc> Ksp then the solution is saturated (characterized by the formation of the precipitate because the solvent is no longer able to dissolve the solute which in this case is the soluble salt).

Problems example:

500 mL of Pb (NO 3) 2 10-3 M solution mixed with 1 liter of 10-2 M NaI solution. If Ksp PbI2 = 6 is known. 10-9, Determine whether sediment is formed or not?  

Answer:

Mol Pb2 + = V. M

                 = 0.5 liters × 10-3 M

= 5. 10-4 mol   Mol I-

= V. M = 1.0 liter × 10-2 M

= 1. 10-2 mol  

Concentration after mixing:

[Pb2 +] = mol Pb2 + / total volume

= 5. 10-4mol / 1.5L

= 3.33. 10-4 M [I-]

= total mole / total volume

= 1. 10-2 mol / 1.5 L

= 6.67. 10-3M  

Qc = [Pb2 +] [I-]

= (3.33.10-4) (6.67.10-3)

= 1.5.10-8 M

Price Qc> Ksp then PbI2 deposition occurs.

II. Predicting Chemical Reaction Balance Products

            Some rules that apply in determining the value of chemical equilibrium constant when the equilibrium reaction is manipulated (modified) include:

1. If the reaction can be expressed in the sum of two or more reactions, the equilibrium constant value of the whole reaction is the product of the equilibrium constant equilibrium of each reaction.

A + B <===> C + D  Kc '

C + D <===> E + F  Kc ''

A + B <===> E + F  Kc = Kc 'x Kc' '

2. If the reaction is written in reverse of the original reaction, the equilibrium constant value becomes the inverse of the original equilibrium constant value.

A + B <===> C + D  Kc '= [C] [D] / [A] [B]

C + D <===> A + B  Kc = [A] [B] / [C] [D] = 1 / Kc ' 3.

If an equilibrium reaction is multiplied by a factor n, the equilibrium constants value becomes the original equilibrium constant value raised by a factor of n.

A + B <===> C + D Kc '= [C] [D] / [A] [B]

2 A + 2 B <===> 2 C + 2 D Kc = [C] 2 [D] 2 / [A] 2 [B] 2 = {[C] [D] / [A] [B] 2 = (Kc ') 2

               One useful use of chemical equilibrium constant is to predict the direction of the reaction. To study the tendency of the reaction direction, the quantity Qc is used, ie the product of the initial concentration of the product is divided by the multiplication of the initial concentration of the reactant, each of which is removed by the reaction coefficient. If the value of Qc is compared to the value of Kc, there are three possible relationships that occur, among others:

A. Qc <Kc, Reversible reaction system excess reactants and product deficiencies. To achieve equilibrium, a number of reactants are converted into products. As a result, the reaction tends toward the product (to the right).

B. Qc = Kc, The system is in an equilibrium state. The reaction rate, both toward reactants and products, is the same.

C. Qc> Kc, Reversible reaction system of excess product and lack of reactants. To achieve equilibrium, some products are converted into reactants. As a result, the reaction tends toward the reactant (to the left).

Problems example:

Consider the following reaction: H2 (g) + I2 (g) <===> 2HI (g) Kc = 49.5 (4400C). For example in a closed container there is H2 0.1 mol; I2 0.2 mol; And HI 0.1 mol. Has the equilibrium been achieved?

Solution:

Calculate quota of reaction Q.

Q = [HI] 2 / [H2] [I2] = [0.1 mol] 2 / [0.1 mol] [0.2 mol] = 0.5

Since Q <Kc then the reaction is not balanced. The reaction will proceed to the right until equilibrium is obtained.

 III. Calculating Percentage of Results
When you do Stoichiometric Calculations to determine how many products are produced by the reaction, this is called "Theoretical Results". The expected number will be generated if everything goes right in the reaction but "Science" is like "alive" and not always right and something can happen is not as it should be And when it does, the results can be reduced, this is called "The Real Results ". So what could be wrong in chemical reactions to reduce the results?               
The reaction is not always complete and the dynamic balance is achieved which makes the reaction stop before all reactants are used.
Some products or reactants may be lost during filtering or moving from one container to another.The reactants may also be impure. Quantity can also be wrong measured. Sometimes the actual results can be greater than the theoretical results. This is usually due to the weight of an improper reactor or a mixed product,Percentage of Results = (Actual Results / Results in theory) x 100 
To calculate Percentage of Results from a reaction the following steps should be followed:
Step 1: Calculate the results theoretically. Theoretical results are what is expected from stoichiometric calculations using chemical equations.Step 2: Determine the actual result. This will usually come from doing experiments and measuring the amount of each product. But here you will get is quantity, like mass, volume or mole.Step 3: Percentage of Results = (Actual Results / Results in theory) x 100
Example 1:
Two other students did a burning reaction by burning 1.0kg of charcoal according to C (s) + O2 (g)> CO2 (g) reaction. They get 1050 liters of carbon dioxide on LTP.What Percentage of Results from CO2 (g)(RAM C = 12.0 and 1 mol of gas is 24.5 liters in LTP)
Step 1: The Results TheoryReacted mol C = mass / RMM C = 1000 / 12.0 = 83.3 mol.Mol Theory of CO2 (g) produced = Mol C x CO2 / C ratio = 83.3 x 1/1 = 83.3 mol.Volume Theory of CO2 generated on LTP = Mol CO2 x 24.5 = 83.3 x 24.5 = 2.041 liter
Step 2: Actual yield 1050 liters CO2 (g)
Step 3: Percentage of Results = (Actual Results / Theory Results) x 100 = (1050/2041) x 100 = 51.4%