Question 1 (a)

(ii) explain why the reaction isßé@represented by a net-ionic equation. The net-ionic Pb12(s) from Pb2+(aq) and 1-(aq) 1 point is earned for a valid explanation. non-reacting species K+(aq) and N03-(aq).

Question 2 (a)

(a) Identify a Brønsted-Lowry con •u ate acid-base which is the base. CH3CH2COOH and CH3CH2COO and H O acid H30 acid OR base base air in the reaction. Clearly label which is the acid and 1 point is earned for writing (or naming) either of the Brønsted-Lowry conjugate acid-base pairs with a clear indication of which is the acid and which is the base.

Question 3 (c)

The response should show at least one K+ ion moving toward the Cu compartment on the left and at least one N03- ion moving in the pposite direction. salt 1 point is earned for correct representation of both K + and N03- ions. (Including free electrons loses this point.) 1 point is earned for correctly indicating the direction of movement of both ions.

Anode loses mass N03- Na4 brldge Cl.l++ wire Cathode gains mass An or the Zn(s) Cu(s)

![Copper (cathode) Voltmeter + 1.10 V 0.76 v Salt bridge +0.34 V cu2•+

  • cu(s) Zinc (anode) +0.76 V ](./media/image199.png)

Question 3 (e)

(ii) Calculate the value of AGO for the reaction. Include units with your answer. AGO -nFEO AG 2 mol e- mol 96,485C 0.48 J c mol e- - - 93,000 J/mol„n - - 93 k.J/m01 1 point is earned for the correct number of electrons. 1 point is earned for the correct answer with unit.

Nernst Equation • Hydro-electrometallurgical processes often involve electrochemical reactions. For electrochemical reaction AG = -nFE, AGO = -nFEO, therefore Nernst Equation In which E = potential for reduction-oxidation reaction EO -standard potential for reduction-oxidation reaction n = number of electron involved in the electrochemical reaction, F = Faraday constant = 96485 Coulomb/mole of electron Spontaneous process E > 0 AG < O

Question 4 (c)

(c) After 20 minutes some C02(g) was injected into the container, initially raising the pressure to 1.5 atm. Would the final pressure inside the container be less than, greater than, or equal to 1.04 atm? Explain your reasoning. The final pressure would be equal to 1.04 atm. Equilibrium was reached in both experiments; the equilibrium pressure 1 point is earned for the correct at this temperature is 1.04 atm. As the reaction shifts toward answer with justification. the reactant, the amount of C02(g) in the container will ecrease until the pressure returns to 1.04 a

Question 5 (b)

1:28 / 1:28

Charge Formal Cl Valence Electrons NonBonding Val Electrons Bondinq Electrons 6/2 2/2

Question 6 (b)

CH3 Propene Vinyl Chloride (chloroethene) (b) The boiling point of liquid propene (226 K) is lower than the boiling point of liquid vinyl chloride (260 K). Account for this difference in terms of the types and strengths of intermolecular forces present in each liquid. Both substances have dipole-dipole interactions and London dispersion forces (or propene is essentially nonpolar with only LDFs while vinyl chloride has both LDFs and dipole-dipole forces). Propene contains a CH3 group, but vinyl chloride contains a Cl atom. Vinyl chloride thus has a larger electron cloud, is more polarizable, and has a larger dipole moment. Thus intermolecular attractions are stronger in vinyl chloride, which results in it having the higher boiling point. 1 point is earned for a discussion of intermolecular forces and for a comparison of their relative strengths.

Question 7 (a)

![Trial Number 1 2 3 4 Initial P (torr) cis-2-butene 300. 600. 300.

  1. 2.00 2.00 4.00 2.00 350. 350. 350. 365 t1/2 (s) 100. 100. 100.

  2. (a) The reaction is first order. Explain how the data in the table are consistent with a first-order reaction. For a first-order reaction, the half-life is independent of reactant concentration (or pressure) at constant T, as shown in trials 1, 2, and 3. 1 point is earned for a correct explanation. ](./media/image206.png)

    Order Rate Law Rate = Rate = kl Concentration - Time Equation = kot 2.303 Half Life 2k o o .aga Graphical

    0.20 0.15 mol 1st half life 0.10 2nd 0.05 half life 0.00 Time, sec \[A\] vs time for a O order reaction alf life ecrease with decreasin concentration. 3td half life 20 40 60 80

    \[A\] vs time for a 1st order reaction 0.20 engtho hålf life I constan 4 st half life half life half life 0.00 Time, sec.

    \[A\] vs time for a 2nd order reaction 0.20 Length of half life increases ith decreasing concentration. mol half half life half life 100 200 Time, sec. 300 400

    Half-Life The half-life ofa reaction, h n, is the time required for the concentration ofa reac- tant to reach half its initial value, \[A h 2 — Half-life is a convenient way to de- scribe how fast a reaction occurs, especially if it is a first-order process. A fast reaction has a short half-life. We can determine the half-life of a first-order reaction by substituting A = —kt1/2 ITC —kt1/2 Time A Figure 14.10 Comparison of first-order and zero-order reactions for the disappearance of reactant A with time. for CA\] tand t1/2 for t in Equation 14.12: In — t1/2 = \[14.15\] k 0.693 k From Equation 14.15, we see that t1/2 for a first-order rate law does not depend on the initial concentration of any reactant. Consequently, the half-life remains constant throughout the reaction. If, for example, the concentration of a reactant is 0.120 M at some instant in the reac- tion, it will be M) = 0.060 M after one half-life. After one more half-life passes, the concentration will drop to 0.030 M, and so on. Equa- tion 14.15 also indicates that, for a first-order reaction, we can calculate t1/2 if we know k and calculate k if we know t1/2. The change in concentration over time for the first-order rearrange- ment of gaseous methyl isonitrile at 199 oc is graphed in Figure 14.11. Because the concentration of this gas is directly proportional to its pres- sure during the reaction, we have chosen to plot pressure rather than concentration in this graph. The first half-life occurs at 13,600 s (3.78 h). At a time 13,600 s later, the methyl isonitrile pressure (and therefore, 150 75 h/2 37.5 10,000 20,000 Time (s) 30,000 concentration) has decreased to half of one-half, or one-fourth, of the initial value. In a first-order reaction, the concentration of the reactant decreases by one-half in each of a series of regularly spaced time intervals, each interval equal to ti/2. A Figure 14.11 Kinetic data for the rearrangement of methyl isonitrile to acetonitrile at 199 'C, showing the half-life of the reaction.

    The half-life for second-order and other reactions depends on reactant concentra- tions and therefore changes as the reaction progresses. We obtained Equation 14.15 for the half-life for a first-order reaction by substituting \[A h 2 = for \[A)tand ti/2 for t in Equation 14.12. We find the half-life of a second-oråer reaction by making the same substitutions into Equation 14.14: \[Alo — \[A\] 0 1 1 — kt1/2 + 2 C A 10 1 = kt1/2 1 \[14.17\] In this case, the half-life depends on the initial concentration of reactant—the lower the initial concentration, the longer the half-life.

Question 7 (c)

(c) Is the initial rate of the reaction in trial 1 greater than, less than, or equal to the initial rate in trial 2 ? Justify your answer. The initial rate in trial 1 is less than that in trial 2 because rate k \[cis-2-butene\] or rate kP cis-2-butene (with reference to values from both frials). OR because the initial concentration of cis-2-butene in trial 1 is less than that in trial 2 and k is constant. 1 point is earned for the correct answer with justification.

Question 7 (d)

(d) The half-life of the reaction in trial 4 is less than the half-life in trial 1. Explain why, in terms of activation energy. The temperature is higher in trial 4, meaning that the KEavg of the molecules is greater. Consequently, in this trial a greater fraction of collisions have sufficient energy to overcome the activation energy barrier, thus the rate is greater. 1 point is earned for a correct answer with justification.

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