Problems 4-6.docx

CHEMICAL KINETICS

a. The surface area of reactant 1. Which is wider surface area (with the same weight), marble in pieces or powder? Answer: Marble in powder , according with our results that when we use the pieces of marble, the reaction time is 421 second. While we use the powder of marble with the same weight, the reaction time is 120 second. 2. What are the effects of surface area of marble toward the rate reaction? Answer: The greater the surface area of the marble causes the touch will be faster so the reaction rate is faster too. Its because the smaller the size of the particle will cause the possibility of greater collisions. b. Concentration of reactants 1. What are the effect of the concentration of HCl to the rate of reaction and Na2S2O3 Answer: 5 ml of Na2S2O3 0.15 M was mixed with 5 ml of HCL 3 M recorded time required since addition of 5 ml HCL 3 M solution until sign (X) no longer seen, the time required was 11 s. The solution becomes turbid and endothermic reactions occur. 5 ml of Na2S2O3 0.15 M mixed with 5 ml of HCL 1,8 M, the time required since the addition of 5 ml of HCL 1,8 M solution until the sign (X) is no longer visible, the time required is 13 s. The solution becomes turbid. 5 ml of Na2S2O3 0.15 M mixed with 5 ml of HCL 0,6 M, the time required since the addition of 5 ml of HCL 0,6 M solution until the sign (X) is no longer visible, the time required is 18 s. The solution becomes turbid. So from the three experiments can be known the effect of concentration on time: The greater the concentration of HCl, the reaction will be faster too. 2. What are the influences of Na2S2O3 concentration to the rate of reaction? Answer: 5 ml of HCl 3 M mixed with 25 ml of Na2S2O3 0,12 M, the time required since the addition of 25 ml of Na2S2O3 0,12 M solution until the sign (X) is no longer visible, the time required is 14 s. The solution becomes turbid. 5 ml of HCl 3 M mixed with 25 ml of Na2S2O3 0,09 M, the time required since the addition of 25 ml of Na2S2O3 0,09 M solution until the sign (X) is no longer visible, the time required is 21 s. The solution becomes turbid. 5 ml of HCl 3 M mixed with 25 ml of Na2S2O3 0,06 M, the time required since the addition of 25 ml of Na2S2O3 0,06 M solution until the sign (X) is no longer visible, the time required is 30 s. The solution becomes turbid. 5 ml of HCl 3 M mixed with 25 ml of Na2S2O3 0,03 M, the time required since the addition of 25 ml of Na2S2O3 0,03 M solution until the sign (X) is no longer visible, the time required is 60 s. The solution becomes turbid.

So from the five experiments can be known the influence of concentration on time: The greater the concentration of Na2S2O3, the reaction will be faster too. 3. Draw a graph showed the relationship of 1/time and sodium thiosulfate concentration! Answer: 1 0.9 0.8

1/t (10-1)

0.7 0.6 0.5 0.4 0.3 0.2 0.1

0 1

2

3

4

5

6

Concentration of Na2S2O3 (M)

c. Temperature 1. What are the effect of temperature toward rate reaaction on the reaction between potassium permanganate with oxalic acid and sulphuric acid? Answer: when temperature in a reaction that take place is raised (in the practice is heating), it causes particles to move more actively so the collisions ar more frequent causing greater the reaction rate. 2. Write down the occured reaction Answer: 5H2C2O4 + 9H2SO4 + 6KMnO4 → 6MnSO4 + 10CO2 + 14 H2O + 3K2SO4 d. Catalyst 1. FeCl3 are containing two kinds of ions, cations, and anions, which ion works as a catalyst? Answer: Fe+3 2. In this reaction, H2O2 decomposes to form H2O and O2. Whether the catalyst is also changing as aresult of a reaction? Answer: No, catalyst is just as accelerator of reaction and does not participate react e. Autocatalyst 1. Why is the rate of reaction on the first droplets of aa solution of KMnO4 in the first tube is slower when compared with the second tube?

Answer: Because in the second tube there are two catalysts MnSO4 and KMnO4. MnO4 act as autocatalyst. 2. What is the effect of Manganese (II) sulfate in the reaction which takes place on the tube? Answer: as autocatalyst causes the reaction to take place quickly than thr first tube.

ELECTROCHEMISTRY

a. Redox titration 1) Write down all of the occured reaction in this experiment. Answer: Reaction of metal and metal nitrat solution 1. 2Al(s) + 3Pb(NO3)2(aq) → 2Al(NO3)2(aq) + 3Pb(s) 2Al(s) + 3Zn(NO3)2(aq) → 2Al(NO3)2(aq) + 3Zn(s) 2Al(s) + 3AgNO3(aq) → 2Al(NO3)2(aq) + 3Ag(s) 2Al(s) + NaNO3(aq) ↛ 2. Cu(s) + Pb(NO3)2(aq) ↛ Cu(s) + Zn(NO3)2 (aq) ↛ Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s) Cu(s) + NaNO3(aq) ↛ 3. Fe (s) + Pb(NO3)2(aq) → Fe(NO3)2(aq) + Pb(s) Fe(s) + Zn(NO3)2 (aq) ↛ Fe (s) + 2AgNO3 (aq) → Fe(NO3)2(aq) + 2Pb(s) Fe(s) + NaNO3(aq) ↛ 4. Cu (s) + HCl(aq) ↛ 2Fe(s) + 6HCl (aq) → 2FeCl3(aq) + 3H2(g) Mg(s) + 2HCl (aq) → MgCl2(aq) + H2(g) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Pb(s) + HCl(aq) → PbCl2(aq) + H2(g) 2) Sort of seven metals (Al, Pb, Zn, Ag, Na, Cu, Fe) by dimishing the power reduction and give explanation Answer: The power reduction of seven metals is Na>Al>Zn>Fe>Pb>Cu>Ag. According to the position of the left metal in the voltaic cell the characteristics of the reducer are stronger and the reduction potential is smaller. 3) Where is the position of Hydrogen in the series above Answer: Position of Hydrogen between Pb and Cu b. Electrolysis of KI solution. 1) What is the function of the salt bridge?

Answer: The function of the salt bridge is to neutralize the anion and cation excess in the solution and to close the circuit so that the reaction can take place continuously. 2) Explain why the reaction can take place spontaneously in terms of free energy?

Answer: A reaction can be said to be spontaneous when it qualifies for thermodynamics, ie, its free energy Gibbs (DGo) is equal to zero. The DGo value can be determined from the standard cell potential by the formula DGo = - n F Eocell Thus, it can be deduced if a cell has a positive Eocell, then DGo will be negative and its reaction is spontaneous. c. Electrolysis of CuSO4 1) Write equations for all reactions that occur in electrode! Answer: 1. Using C electrode occured there is gas bubbles in the anode and in the cathode there is a golden-brown plating.

Reaction: CuSO4(aq) Anode Cathode

→

Cu 2+ + SO42-

: 2 H2O(l) → 4 H+ + O2(g) + 4e: 2Cu2+ (aq) + 4e- → 2Cu(s) 2H2O(l) + 2Cu2+→ 4 H+ + O2(g) + 2Cu(s)

Eo = – 1,23 V Eo = + 0,34 V + Eo = – 0,89 V (not

spontaneous) 2. Using Fe electrode occured there is gas bubbles and the color of solution changes to be brownish black in the anode and in the cathode there is a goldenbrown plating. CuSO4(aq) → Cu 2+ + SO42Cathode Anode

: :

Cu2+ + 2e- → Cu Fe → Fe2+ + 2e Cu2+ + Fe → Cu + Fe2+

Eo = + 0,34 V Eo = + 0,41 V + Eo = + 0,75 V

(spontaneous) 2) What is the function of PP? Answer: The usefulness of Phenolphthalein (PP) is as indicator or to prove the basic solution in the presence of I2 on the anode because phenolphthalein has a pH range of 8.00 to 10.0 . 3) What is the function of CHCl3 ? Answer: The usefulness of CCl3 is as an polar organic solvent. 4) In the electrolysis using C electrode, what happens at each electrode? Write the equation Answer:

As the reaction: CuSO4(aq) Anode Cathode

→

Cu 2+ + SO42-

: 2 H2O(l) → 4 H+ + O2(g) + 4e: 2Cu2+ (aq) + 4e- → 2Cu(s) 2H2O(l) + 2Cu2+→ 4 H+ + O2(g) + 2Cu(s)

Eo = – 1,23 V Eo = + 0,34 V + Eo = – 0,89 V (not

spontaneous) 5) In the electrolysis using Fe electrode, what happens at each electrode? Write the equation Answer: Reaction: CuSO4(aq) → Cu 2+ + SO42Cathode : Anode : (spontaneous)

Cu2+ + 2e- → Cu Fe → Fe2+ + 2e Cu2+ + Fe → Cu + Fe2+

Eo = + 0,34 V Eo = + 0,41 V + Eo = + 0,75 V

ELEMENTAL CHEMISTRY

a. Preparation of halogen and its properties 1) Write all the occured reactions Answer : The reactions for this experiment are : 1. Chlorine + concentrated HCl Ca(OCl)2 + 4HCl  CaCl2 + 2Cl2 + 2H2O 2. Iodine in water I2 + H2O  HI + HOI 3. a. Chlorine water + CHCl3 + KBr 𝐶𝐻𝐶𝑙3

Cl2 + 2KBr → 2KCl + Br2 b. Chlorine water + CHCl3 + KI 𝐶𝐻𝐶𝑙3

Cl2 + 2 KI → 2 KCl + I2 4. Iodine water + CHCl3 +NaCl I2 + NaCl  5. a. NaCl + AgNO3  NaNO3 + AgCl AgCl(s) + NH3(aq)  [Ag(NH3)2]+ + Clb. KI + AgNO3  KNO3 + AgI AgI(s) + NH (aq)  [Ag(NH3)2]+ + Ic. KBr + AgNO3  KNO3 + AgBr AgBr(s) + NH3 (aq)  2) Sort the reactivity of halogen elements based on your experiments Answer : Cl > Br > I b. Complex ion formation 1) NH3 solution containing NH3, NH+ and OH-. Which one from complex ions with Cu2+ ion ? give an explanation. Answer : NH3 because NH3 is donor of electron. The reaction is : Cu2+ + 4NH3  [Cu(NH3)4]2+ 2) HCl solution containing H+ and Cl- ions. Which ions form complexes with Cu+ ion ? Give an explanation. Answer : Cl- , because liganol is an anion or a neutral molecule. Ligan Cl- is weakness than H2O. The reaction is : Cu2+ + 4Cl-  [Cu(Cl)42-] 3) Write down all the reactions in this experiment Answer : The reactions for this experiment are :

1. CuSO4.5H2O crystal heated CuSO4(s) + 5H2O(aq) CuSO4.5H2O(s) 2. a. CuSO4 + 4NH3  Cu(NH3)4SO4 b. CuSO4 + NaOH  [Cu(OH)2SO4]4- + 4Na+ c. CuSO4 + HCl  [CuCl2SO4]4- + 4H+ d. CuSO4 + 2HCl  CuCl2 + H2SO4 e. H2SO4 + 2NH4Cl  (NH4)2SO4 + 2HCl c. Typical properties of transition elements 1) Write the all the occured reaction in this experiment Answer : Iron powder + H2SO4 : Fe(s) + H2SO4(aq)  FeSO4(aq) + H2(g) Filtrate 1 :  2FeSO4 + K4[Fe(CN)6]  3K2SO4 + Fe3[Fe(CN)6]2  3FeSO4 + 2K3[Fe(CN)6]  3K2SO4 + Fe3[Fe(CN)6]2  FeSO4 + 2NaOH  Na2SO4 + Fe(OH)2 Iron powder + HNO3 : Fe(s) + 2HNO3(aq)  Fe(NO3)2(aq) + H2(g) Filtrate 2 :  2Fe(NO3)2 + K4[Fe(CN)6]  4KNO3 + Fe[Fe(CN)6]  3Fe(NO3)2 + 2K3[Fe(CN)6]  6KNO3 + Fe3[Fe(CN)6]  Fe(NO3)2 + NaOH  2NaNO3 + Fe(OH)2 2) Write the half reactions for all redox reactions that occur in this experiment. Answer : Oxidation : Mn2+ + 2 H2O  MnO2 + 2 e- + 4 H+ x3 Reduction : MnO4- + 3 e- + 4 H+  MnO2 + 2 H2O x2 Oxidation : 3 Mn2+ + 6 H2O  3 MnO2 + 6 e- + 12 H+ Reduction : 2 MnO4- + 6 e- + 8 H+  2 MnO2 + 4 H2O 2+ 3 Mn + 2 MnO4 + 2 H2O  5 MnO2 + 4 H+