The Determination of an Equilibrium Constant
I will be determining the equilibrium constant - Kc; of ethanoic acid reacting
with ethanol producing an equilibrium to form ethyl ethanoate and
water.
CH3COOH(aq) + C2H5OH(aq) ó CH3COOC2H5(aq) + H2O(l)
Following the method as detailed, I conducted experiment 4 and these
results were obtained:
Titration Trial
Volume of Sodium Hydroxide Neutralised (cm3)
1
7.65
2
7.75
3
7.80
4
7.70
5
7.75
μ
7.75
To calculate Kc, the concentrations of each reactant must be
calculated from the point of equilibrium at which the titration was
taken.
Therefore, using 7.75cm3 as the average titre,
Moles = 0.2M x (7.75cm3 / 1000 cm3) = 1.55 x 10-3.
However, this represents the total amount of acid in the system which
included ethanoic acid but also the acid catalyst: hydrochloric acid.
Therefore this must be taken away from the number of moles of acid to
calculate number of moles of ethanoic acid.
Since 25 cm3 of 1.0M HCl was included in the initial 250 cm3 mixture,
Moles = 1 x (25 / 250) = 0.1 moles.
But 1cm3 samples were taken at a time, therefore:
Moles = 0.1 x (1 cm3 / 1000 cm3) = 1 x 10-4.
Taking this away from the total amount of acid = 1.55 x 10-3 - 0.1 x
10-3 = 1.45 x 10-3 (moles of ethanoic acid).
Considering the stoichiometry of the equation, there is equal amounts
of ethanol as there is ethanoic acid, therefore there is 1.45 x 10-3
moles of ethanol at equilibrium also. This is because the total amount
of acid reacted is equal to the amount of acid, even though it is a
weak acid. Because it reacts with a strong base, it means that the
acid dissociates completely.
Now to calculate the number of moles of products;
The number of moles is calculated from multiplying volume by density
to calculate mass (which is provided on the enclosed sheet), and then
We then took 1ml of the 0.1% solution from test tube 2 using the glucose pipette and added it to test tube 3, we then used the H2O pipette and added 9ml of H2O into test tube 3 creating 10ml of 0.01% solution.
For example, a balanced chemical equation of a certain reaction specifies that an equal number of moles of two substances A and B is required. If there are more moles of B than of A, then A is the limiting reactant because it is completely consumed when the reaction stops and there is an excess of B left over. Increasing the amount of A until there are more moles of A than of B, however, will cause B to become the limiting reactant because the complete consumption of B, not A, forces the reaction to cease. Purpose
acid*1 mol s. Acid/ 138.1g s acid*1 mol aspirin/1 mol s. acid * 180.2 g aspirin/1 mol aspirin = 3.9145 aspirin
neutralize 35ml of our base. Once we weighed out the KHP we then dissolved it
the acid was at 14 C the magnesium took 141 seconds to react and 27 C
activation energy needed to be activated. I have chosen to do concentration of acid because surface area is difficult to measure, pressure is hard to do in a school lab, and temperature could be dangerous e.g. if acid boils it could spit.
Investigating the Rate of Reaction Between Marble Chips and Acid Introduction This is a test to demonstrate the reaction rate between marble chips (CaCO) and hydrochloric acid (HCl). Hopefully we will be able to prove that the concentration of the acid is directly proportional to the reaction rate. Aim To discover if the concentration of acid is directly proportional to the rate of reaction, by monitoring the amount of gas given off as the reaction takes place. Prediction We believe that the concentration of the acid will be directly proportional to the rate of reaction. We believe this due to the following theory: [IMAGE]The collision theory: This theory states that for a reaction to occur the reactant particles need to react with sufficient energy.
strong acid or base does not necessarily yield a drastic jump in pH. The acid
= 3 ´ E(C-H) + 1 ´ E(C-O) + 1 ´ E(O-H) + 1.5 ´ E(O=O)
* Concentration - I will try my best to use all the acid from the
= = pH 1 2 3 Average Rate of Reaction (cm3/s): 0 - 0. 3 0 0 0 0 0.000 5 0 0 0 0
One common thread sewn throughout all my high school classes is balance and equilibrium. The first math equation most children learn is 2+2=4. The simplicity was lost on me until my sophomore year honors chemistry and pre calculus classes. In honors chemistry we learned how equations always move towards their equilibrium; in pre calc we studied sine and cosine equations and how their curves were symmetric indefinitely. With each STEM course, I realized equilibrium was the common denominator uniting them: “=”holds more significance than I previously thought. The significance of balance and equilibrium continued throughout my Calculus and AP Physics classes where I learned Newton’s laws allow for order and balance in the universe. Although Newton discovered these laws and turned them into quantifiable equations, these
g. of KI in 10 mL of water. Add the KI solution dropwise to the test
ranging from 50 cm³ of acid and no water, to 12.5 cm³ of acid and 37.5