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

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

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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