conclusion report

ConclusionAs a conclusion, the experiment was a success and all the objectives have been achieved. From the experiment conducted, we can determine the volume of NaOH titrated. From the result obtained, we can calculate the rate constant and the rate of reaction. The rate constant, k is 0.0859 min-1. From the graph plotted, it can be seen that the reaction is a first order reaction. Based on this experiment, the reacting material is Ethyl Acetate and Sodium Hydroxide. So, the conclusion is the experiment was successful in requiring the rate of reaction and the value of rate constant, k.Recommendation

1. Before conducting the experiment, ensure that the equipment was clean and was set up properly.

2. During titration, the final colour of the mixture must be same for all the 3 titration for each samples.

3. Titration with NaOH must done properly which in this case, the titration was done with burette was slightly opened to prevent excessive amount of NaOH dropped into the flask.

4. Observation must be efficiently done while titrating NaOH. The titration must be stopped as soon as the sample mixture turned colour from colourless to light pink.

References

1. Fogler, H.S. (2006). Elements of Chemical Reaction Engineering. 4th Edition, New Jersey: Prentice Hall. 2. Retrieved from http://sites.tufts.edu/andrewrosen/files/2013/09/reactor_design_guide1.pdf Retrieved from http://ocw.mit.edu/courses/chemical-engineering/10-37-chemical-and-biological-reaction-engineering-spring-2007/lecture-notes/lec05_02212007_g.pdf 3. Appendices(A) Sample : t = 5 min

(B) Volume of titrating NaOH(mL) = 13.3 mL = 0.0133 L

(C) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,stdWith NaOH in sample (mL) = 0.1 mol/L x 11.5 mL/0.25 mol = 5.32 mL

(D) Volume of HCl reacted with NaOH = VHCl (C)In sample (mL) =10 mL 5.32 mL= 4.68 mL = 0.00468 L

(E) Mole of HCl reacted with NaOH in = CHCl,stdx (D) = 0.25 mol/L x 0.00468 mL = 1.17x10-3mol

(F) Mole of NaOH unreacted in sample = (E) = 1.17x10-3mol

(G) Concentration of NaOH unreacted = (E)Vs= 1.17x10-3mol 0.05L= 0.0234mol/L

(H) Steadt state fraction conversion = 1 CA CA0= 1 (0.0234mol/L)/0.05(mol/L) = 0.532

(I) Concentration of NaOH reacted = CNaOH,0 (G)With Ethyl Acetate(mol/L) = 0.05 mol/L 0.0234mol/L = 0.0266mol/L

(J) Mole of NaOH reacted with = (I) x VsEthyl Acetate in sample(mol) = 0.0266mol/L X 0.05 L= 1.33 x 10-3mol

(K) Concentration of ethyl acetate = (J)/VsReacted with NaOH(mol/L) = 1.33 x 10-3mol mol/ 0.05 L = 0.0266mol/L

(L) Concentratin of Ethyl Acetate = CEA,0 (K)Unreacted (mol/L) = 0.1 mol/L 0.0266mol/L = 0.0734mol/L

Sample : t = 10 min

(A) Volume of titrating NaOH(mL) = 13.9 mL = 0.0139 L

(B) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,stdWith NaOH in sample (mL) = 0.1 mol/L x 13.9 mL/0.25 mol = 5.56 mL

(C) Volume of HCl reacted with NaOH = VHCl (C)In sample (mL) =10 mL 5.56 mL= 4.44 mL = 0.00444 L

(D) Mole of HCl reacted with NaOH in = CHCl,stdx (D)= 0.25 mol/L x 0.00444 mL= 1.11x10-3mol

(E) Mole of NaOH unreacted in sample = (E) = 1.11x10-3mol

(F) Concentration of NaOH unreacted = (E)Vs= 1.11x10-3mol 0.05L= 0.0222mol/L

(G) Steadt state fraction conversion = 1 CA CA0= 1 (0.0222mol/L)/0.05(mol/L)= 0.556

(H) Concentration of NaOH reacted = CNaOH,0 (G)With Ethyl Acetate(mol/L) = 0.05 mol/L 0.0222mol/L= 0.0278mol/L

(I) Mole of NaOH reacted with = (I) x VsEthyl Acetate in sample(mol) = 0.0278mol/L X 0.05 L= 1.39 x 10-3mol

(J) Concentration of ethyl acetate = (J)/VsReacted with NaOH(mol/L) = 1.39 x 10-3mol mol/ 0.05 L = 0.0278mol/L

(K) Concentratin of Ethyl Acetate = CEA,0 (K)Unreacted (mol/L) = 0.1 mol/L 0.0278mol/L = 0.0722mol/L

(A) Sample : t = 15 min


(C) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,stdWith NaOH in sample (mL) = 0.1 mol/L x 14.3 mL/0.25 mol= 5.72 mL

(D) Volume of HCl reacted with NaOH = VHCl (C)In sample (mL) =10 mL 5.72 mL = 4.28 mL = 0.00428 L

(E) Mole of HCl reacted with NaOH in = CHCl,stdx (D)= 0.25 mol/L x 0.00428 mL = 1.07x10-3mol


(G) Concentration of NaOH unreacted = (E)Vs= 1.07x10-3mol 0.05L = 0.0214mol/L

(H) Steadt state fraction conversion = 1 CA CA0= 1 (0.0214mol/L)/0.05(mol/L)= 0.572



(K) Concentration of ethyl acetate = (J)/VsReacted with NaOH(mol/L) = 1.43 x 10-3mol mol/ 0.05 L = 0.0286mol/L

(L) Concentratin of Ethyl Acetate = CEA,0 (K)Unreacted (mol/L) = 0.1 mol/L 0.0286mol/L = 0.0714mol/L



(C) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,stdWith NaOH in sample (mL) = 0.1 mol/L x 14.4 mL/0.25 mol = 5.76 mL


(E) Mole of HCl reacted with NaOH in = CHCl,stdx (D)= 0.25 mol/L x 0.00424 mL= 1.06x10-3mol


(G) Concentration of NaOH unreacted = (E)Vs= 1.06x10-3mol 0.05L= 0.0212mol/L




(K) Concentration of ethyl acetate = (J)/VsReacted with NaOH(mol/L)= 1.44 x 10-3mol mol/ 0.05 L = 0.0288mol/L

(L) Concentratin of Ethyl Acetate = CEA,0 (K)Unreacted (mol/L) = 0.1 mol/L 0.0288mol/L= 0.0712mol/L



(C) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,stdWith NaOH in sample (mL)= 0.1 mol/L x 14.4 mL/0.25 mol = 5.76 mL


(E) Mole of HCl reacted with NaOH in = CHCl,stdx (D)= 0.25 mol/L x 0.00424 mL= 1.06x10-3mol


(G) Concentration of NaOH unreacted = (E)Vs= 1.06x10-3mol 0.05L= 0.0212 mol/L


(I) Concentration of NaOH reacted = CNaOH,0 (G)With Ethyl Acetate(mol/L) = 0.05 mol/L 0.0212 mol/L = 0.0288 mol/L

(J) Mole of NaOH reacted with = (I) x VsEthyl Acetate in sample(mol) = 0.0288 mol/L X 0.05 L= 1.44 x 10-3mol

(K) Concentration of ethyl acetate = (J)/VsReacted with NaOH(mol/L) = 1.44 x 10-3mol mol/ 0.05 L = 0.0288 mol/L

(L) Concentratin of Ethyl Acetate = CEA,0 (K)Unreacted (mol/L) = 0.1 mol/L 0.0288 mol/L = 0.0712 mol/L

Continuous Stirred Tank Reactor Unit (Model SOLTEQ, BP100)

Physical appearance of samples after undergoes titration process.

conclusion report

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