Assignment 1: State the thermodynamic aspects of the process of recovering a pure metal from an aqueous solution containing Zn2+ and Cu2+ metal ions (assuming the remaining ion is only SO42), including: Draw process diagrams Electrode selection Show the reactions on the electrodes Propose suitable DC voltage for the process (you can state other assumptions).Assignment 2: A reaction, speeded up by one of its products, is called selfcatalyzed. The classic example of selfcatalysis is a reaction of acetone iodination in an acid medium, described by the total scheme: The reaction consists of two stages. In its first stage, ketone transforms into enol (slowed stage):Assignment 3: Reaction A + 3B C has first order for each reactant with the rate constant k2=0.15M1.min1. Plot the concentration of A, B, and C as a function of reaction time until 90% of B is converted. Given: A0=0.5M, B0=0.6M. This project is such a fantastic occasion for us to solve those tricky questions together, which helps us strengthen our teamwork skills and prepare for a significant upcoming final test. Therefore, we sincerely thank Prof. Long so much for creating this activity.
VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY HO CHI MINH UNIVERSITY OF TECHNOLOGY FACULTY OF CHEMICAL ENGINEERING Ho Chi Minh city, May 1th 2022 HCMUT-Physical Chemistry CC01-Group TABLE OF CONTENTS QUESTION .3 QUESTION .4 QUESTION .6 EVIDENCE OF TEAMWORK ACTIVITY COMMENTS ABOUT THE PROJECT This project is such a fantastic occasion for us to solve those tricky questions together, which helps us strengthen our teamwork skills and prepare for a significant upcoming final test Therefore, we sincerely thank Prof Long so much for creating this activity Page HCMUT-Physical Chemistry Assignment 1: State the thermodynamic aspects of the process of recovering a pure metal from an aqueous solution containing Zn 2+ and Cu2+ metal ions (assuming the remaining ion is only SO42-), including: - Draw process diagrams - Electrode selection - Show the reactions on the electrodes - Propose suitable DC voltage for the process (you can state other assumptions) Solution We have a process diagram as shown: Electrode selection - It contains two compartments One consists of a zinc strip dipping into a zinc sulfate (ZnSO4) solution called the anode, and the other of a copper strip dipping into a copper sulfate (CuSO4) solution called the cathode A conducting bridge links both, it precludes the mixing of the two electrolyte solutions, but it permits the current to pass from one to the other Moreover, both are in mutual electric contact through a metallic thread One galvanometer and an electric motor are inserted into the circuit Finally, a millivoltmeter is connected in parallel between the two lines - Two kinds of electrical currents exist in the cell, called ionic current, it also exists in the bridge Page HCMUT-Physical Chemistry 23 The reactions on the electrodes In anode: Zn(s) Zn2+ + (w) + 2e- (metal) E0 = -0.76V In cathode: Cu2+ + (w) + 2e- (metal) Cu(s) E0 = 0.34V Overall: Zn + Cu2+ Cu + Zn2+ Suitable DC voltage for the process E0cell = E0cathode – E0anode = 0.34 – (-0.76) = 1.1V So DC voltage for the process should be around 1.1V Assignment 2: A reaction, speeded up by one of its products, is called self-catalyzed The classic example of self-catalysis is a reaction of acetone iodination in an acid medium, described by the total scheme: The reaction consists of two stages In its first stage, ketone transforms into enol (slowed stage): Further, enol reacts with iodide to form final products As the reaction passes, hydronium content increases, which speeds up the reaction Thus, the rate of the reaction is defined by the rate of acetone enolization: Where CA is the current concentration of acetone, and k is a rate constant of the ratedetermining step If initial concentrations of acetone and hydronium are marked as C A0 and CB0, then kinetic equations of self-catalyzed reaction could be written in the form: (x: the extent of the reaction = CA0- CA) Page HCMUT-Physical Chemistry Given: [CH3COCH3]0 = 0.2 M ; [I2]0 = 0.1 M [H+]0 = 0.05 M ; ; k = 0.15 (M-1.h-1) Plot the Concentration of CH3COCH2I as a function of time How long does it take to convert 80% of I 2? Solution CH3COCH3 I CH 3COCH 2I HI Initial CAo Final o A x C x From the information on the problem, we have: dx (t) =k (C A 0−x ( t ) )×(C B + x (t ) ) dt Convert it to the form x=f (t ): dx (t ) =kdt (C A 0−x ( t )) ×( C B 0+ x ( t )) We integrate both sides: x t dx (t) ∫ (C −x ( t ))×(C + x ( t ) )=∫ kdt( 1) 0 A0 B0 Suppose that: A B = + = A ( C B 0+ x ( t )) +B ¿ ¿ ( C A 0−x ( t ) ) × ( C B + x ( t ) ) ( C A 0−x ( t ) ) ( C B + x ( t ) ) ¿ A C B 0+ B C A + x (t )( A−B) A−B=0 → → A=B= C B +C A (C A 0−x ( t ))×(C B 0+ x ( t )) A C B 0+ B C A 0=1 { From the result of A and B, we convert (1): x ∫ dx ( t ) ( C A 0−x ( t ) ) × ( C B + x (t ) ) x ¿∫ → C B +C A + C B 0+C A ( C A 0−x ( t ) ) ( C B + x ( t ) ) dx ( t )= C B 0+ C A [ x x ] dx ( t ) dx ( t ) −ln ( C A 0−x ( t ) ) x +ln ( C ∫ C −x ( t ) +∫ C + x ( t ) =¿ C +C ) ( B0 ) B0 A0 ( A0 [ C A0 (C B0+ x ) ln =kt (2) C B +C A C B ( C A 0−x ) [ ] C B 0=¿ ¿ C A 0=[ C H COC H ]0 =0.2 ( M ) k =0.15( M −1 h−1 ) Replace C B 0, C A 0, k into (2) we have: Page HCMUT-Physical Chemistry 0.2 ( 0.05+ x ) ln 0.05+ 0.2 0.05 ( 0.2−x ) 0.2 ( 0.05+ x ) ln =0.15t → =t 0.05+0.2 0.15 0.05 ( 0.2− x ) [ [ ] ] Suppose: 0.2 ( 0.05+ x ) ln 0.05+ 0.2 0.05 ( 0.2−x ) =f ( x ) 0.15 [ ] Time (h) Using Casio calculator to solve equation f ( x )=tthen using excel to plot the diagram: 50 45 40 35 30 25 20 15 10 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 [CH3COCH2I] (mol/L) To reduce 80% of I2, we need: x=C A 0−C A =C(I ¿¿2)0−C t= C A 0(C B0+ x ) ln C B +C A C B ( C A 0−x ) [ k ] (I ¿¿2)=0.8× 0.1=0.08 ¿ ¿ 0.2 ( 0.05+0.08 ) ln 0.05+ 0.2 0.05 ( 0.2−0.08 ) = =39.102( h) 0.15 [ ] Assignment 3: Reaction A + 3B C has first order for each reactant with the rate constant k2=0.15M-1.min-1 Plot the concentration of A, B, and C as a function of reaction time until 90% of B is converted Given: [A]0=0.5M, [B]0=0.6M Equation: A t=0: [A]0 Reacted: x t=t: [A]0-x + [A]0=0.5M; [B]0=0.6M ; 3B C [B]0 3x [B]0-3x x x k2 0.15M 1.min Page HCMUT-Physical Chemistry 90% B 90% 0.6 0.54 M 90% of [B]0 is converted at t=t: Let x be the concentration of specie A reacted at the time t 3x is the concentration of specie B converted at a time t 3x 0.54 x 0.18 [A] A x A B [B] B B 3x B 3( A [A]) The expression of rate law: R dA 1 1 k[ A][B ] d [A] dt d [A] dt dt k[ A][B ] 0.15 [A] [B]0 [A] [A] 1 1 ∫ d [A] ∫ d [A] 0.15 [A] (0.6 (0.5 [A]) 0.15 [A] [B]0 [A] [A] 20 200 ∫ d [A] d [A] ∫ [A] (5 (2[A] 1) 2) 3 1 [A] [A] 2 5 Solving ∫ d [A] [A] (5 (2[A] 1) 2) du 1 10 Substituteu 5(2[A] 1) 2 10 d [A] du , use : d [A] 10 [A] u 1 ∫ du ∫ du u (u 3) u ( 1) u dv u2 Substitute v 1 du dv : u du u ln 1 1 (2[A] 1) ∫ dv ln(v ) v 3 200 ln 1 200 (ln | 10 [A] | ln [A] ) (2[ A] 1) 200 d [A] C C ∫ [A] (5 (2[A] 1) 2) 27 27 1 ∫0.15 [A] (0.6 (0.5 [A]) d[A] ∫dt 200 (ln | 10 [A] | ln [A] ) 27 C t 0 [A] 0.5M 200 (ln | 10 0.5 | ln 0.5 ) C t t0 0 C 10.2688 27 200 (ln | 10 [A] | ln [A] ) 1 ∫ d [A] 10.2688 t 0.15 [A] (0.6 (0.5 [A]) 27 Page HCMUT-Physical Chemistry 200 (ln | 10 [A] | ln [A] ) 10.2688 t 27 The equation for [A]: When 90% of [B]0 is converted: [A]t [A]0 x 0.5 0.18 0.32 M t 13.75 Similar for [B] and [C] equation: dB 1 3k[ A][B] d [B] dt dt 3k [ A][B ] 1 d [B] dt [B] [B] 0.15 [B] [A]0 R 1 ∫ [B]0 [B] 0.15 [B] [A]0 200 ln 10 [ B] ln [B ] 1 ∫ [B] [B] 0.15 [B] [A]0 d [B] ∫dt 1 d [B] ∫dt 0.6 [B] 0.15 [B] 0.5 d [B] ∫ C t 27 t 0 [B] 0.6 M C 23.8435 The equation for B: 200 ln 10 [ B] ln [B] 27 dC k[ A][B ] d [C] dt dt k[ A][B ] d [C] dt 0.15 0.5 [C ] 0.6 3[C ] 23.8435 t R 200 ln [C ] ln 2[C ] 27 t 0 [C ] 0 M C 0 The equation for C: Concentration (M) ∫0.15 0.5 [C ] 0.6 3[C ] d [C] ∫dt C t 200 ln [C ] ln 2[C ] 27 t Plot the concentration of A, B, C following the equations: Concentration of A Concentration of B Concentration of C t (min) Figure: Concentration of A, B, C as a function of reaction time Page