Chem-Lab-Report - kinetics hydrolysis lab report PDF

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Kinetics of the Hydrolysis of Methyl Methanoate Aim: To prove that the rate constant of methyl methanoate with respect to the ester will be first order within the acid-catalyses hydrolysis reaction and to determine the value for the rate constant. Introduction: Esters, such as methyl methanoate can be hydrolysed in an aqueous solution using an acid catalyst. The acid catalyst will allow for the donation of a H+ ion, allowing for the formation of methanol and methanoic acid once the ester undergoes hydrolysis within water. The equation for the hydrolysis of Methyl Methanoate is as followed; H +¿ions HCOOH +C H 3 OH → ←

HCOO H 3 + H 2 O ¿

Upon the ester being added into the aqueous solution, the equilibrium will begin to shif towards the right-hand side, the forward reaction being prompted resulting in an increase in products, as stated within Le Chatelier’s Principle, where if a change is made to any reaction at equilibrium, the system will adapt to change and undo this change. Within a normal first-rate reaction equation, the rate constant is equal to; Rate =k [ A ]

1

The rate of reaction depends on the ester concentration, and that the water and acid volumes are large in excess but that their concentrations are assumed constant through hydrolysis, only the esters concentration variating throughout. Integrating this expression allows us to find the first-rate order for the hydrolysis of methyl methanote, shown below; ln [ Ester ]t=−kt + ln [ Ester ] 0

Where [Ester]0 equal to the initial concentration of the ester before hydrolysis process and [Ester]t equal to the concentration of the ester at a given time, allowing us to calculate the rate of reaction at any specific time throughout the hydrolysis process. We can calculate any specific reaction rate at any given time during this hydrolysis reaction by using the equation; ln ( T ∞−T t ) =−kt +ln (T ∞ −T 0)

T∞ is the titre result which represents the completion of the hydrolysis reaction where no more methyl methanoate molecules are broken down into methanol and methanoic acid. The concentration of the ester will remain constant. The equation above is within the form of; y=mc + c

We can confirm that the reaction is a first order if there is a negative gradient when comparing the time of reaction ( −kt ) against the concentration of methyl methanoate at each interval ( ln (T ∞ −T t )¿ . The gradient of the line will give us the value for the rate constant k. Experimental Procedure: 100cm3 of 0.5M hydrochloric acid was added to a 250cm3 conical flask. The flask was clamped into a water bath at 30 C o . The conical flask was lef clamped into the water bath for 5 minutes to allow to come to the same temperature, with a thermometer used to check the temperature. While it was still clamped into the water bath, 5cm3 of methyl methanoate was added. A stop watch was immediately started as the ester was added and was swirled for five minutes.

Afer the five minutes, 5cm3 sample was pipetted out of the conical flask and transferred to a sperate conical flask containing ice, quenching of the equation to occur stopping hydrolysis. A burette was filled with 0.2M Sodium Hydroxide (NaOH) and the methyl methanoate solution was titrated against this with the addition of phenolphthalein as an indicator, the end of the titration indicated by a clear  Pink colour change. This was repeated every five minutes until the results of the concentration of the ester began to plateau between three interval titres. This marked the end of the hydrolysis reaction of the ester and occurred around 25-35 minutes into the reaction.

Results Time (cm3)

Titre (cm3)

5 10 15 20 25 30 35

16.3 24.7 24.1 26.4 27.4 27.6 28

T ∞−T 1 (cm3) 11.7 3.3 3.9 1.6 0.6 0.4 0

Ln( T ∞−T 1 ¿ (cm3 ) 2.46 1.14 1.36 0.42 -0.51 -0.92 --

The results above were transformed and made into the graph of Ln( T ∞−T 1 ) against time below allowing us to find the gradient of the line equal to the negative of the rate constant.

Integrated First Rate Reaction of Hydrolosis of Methyl Methanoate 3 2.5 f(x) = − 0.13 x + 2.94 R² = 0.94

Ln(T∞-T1) (cm3)

2 1.5 1 0.5 0

0

5

10

15

20

25

30

35

-0.5 -1 -1.5

Time (Minutes)

This equation resulted in the equation of the line equal to be y = -0.1302x + 2.9373, using Y −Y 1 m= 2 .This will result in that the gradient of the line will be equal to -0.13, X 2− X 1 therefore the rate constant (k) would be equal to 0.13 (as (-ve) k x (-ve) k = (+ve) k) Conclusion Based on the present work, the following was concluded; the rate constant of the hydrolysis reaction was equal to 0.13, and the negative gradient allowed us to prove that this reaction is first order. The actual value for the rate constant for this reaction was 0.1, and an accurate result would lie within the boundaries of 0.07 and 0.13. As the result of this rate constant was equal to 0.13 it can be concluded that this experiment was accurate, meeting all the aims of the experiment...