Review Quiz Part II

Question 1

Which metal will dissolve in hydrochloric acid?

Fe

For a metal to dissolve in HCl, the standard potential for the reaction must be positive. Because the H⁺/H₂ potential is zero, E^o_rxn = 0 - E^o_metal.

Question 2

The standard potential of the cell

Pb(s)|PbSO₄(s)|SO₄^2-(aq)||Pb²⁺(aq)|Pb(s)

5.9 x 10⁷

logK = nE^o_rxn/0.0592.

Question 3

Consider the following cell:

Ni(s)|Ni²⁺(aq, 0.200 M)||Cl^-(aq, 0.0200 M)|Cl₂(g, 0.500 atm)|Pt

+1.70 V

E = E^o - (0.0592/2)log([Cl^-]²[Ni²⁺]/P Cl₂)

Question 4

Consider the following cell:

Ag(s)|Ag⁺(aq, 0.100 M)||Ag⁺(aq, 1.00 M)|Ag(s)

+0.0592 V

The equation is (anode on left, cathode on right in cell diagram)
Ag⁺(1.00 M) ----> Ag⁺(0.100 M) and n = 1.
In this case E^o = 0. Finally, use the Nernst equation.

Question 5

Consider the following cell:

Zn(s)|Zn²⁺(aq, 0.200 M)||H⁺(aq, ?)|H₂(g, 1.00 atm)|Pt

9.1 x 10^-3 M

The equation for the reaction is
2H⁺ + Zn ----> H₂ + Zn²⁺
Apply the Nernst equation. Note, because you are taking exponentials, carry all decimal places in your calculation.

Question 6

Consider the following cell:

Pt|H₂(g, 1 atm)|H⁺(aq, ? M)||Ag⁺(aq, 1.0 M)|Ag(s)

8.8 x 10^-5 M

The equation for the reaction is
Ag⁺ + 1/2H₂ ----> Ag + H⁺
Apply the Nernst equation.

Question 7

When a lead-acid battery discharges,

sulfuric acid is consumed.

The cell reactions show that sulfuric acid is used up during discharge. See your text.

Question 8

The products of the electrolysis of CuF₂(aq) are

Cu(s), O₂(g), and H⁺(aq).

Fluoride is too difficult to oxidize (E^o = +2.87 V) with respect to water (E^o = +1.23 V or +0.81 V at pH 7) and the latter is preferentially oxidized to give oxygen gas and H⁺ ions.

Question 9

If 612 C of charge is passed through a solution of Cu(NO₃)₂(aq), calculate the number of moles of copper deposited.

0.00317 mol

mol e^- = C/F = 612 C/(96485 C.mol^-1) = 6.34 x 10^-3 mol e^-. But 2mol e^- are required for each mol Cu(s) produced.

Question 10

If the standard free energy change for the reaction

2H₂(g) + O₂(g) --> 2H₂O(l)

+1.23 V

The half reactions are

O₂ + 4H⁺ + 4e^- ----> 2H₂O
4H⁺ + 4e^- ----> 2H₂

Question 11

Consider the following cell:

Pb(s)|PbSO₄(s)|SO₄^2-(aq, 0.60 M)||H₂(g, 192.5 kPa)|H⁺(aq, 0.70 M)|Pt

E = 0.36 - 0.01285ln[1.90/{(0.70)²(0.60)}]

The half reactions are

2H⁺(aq) + 2e^- ----> H₂(g)
PbSO₄(s) + 2e^- ----> Pb(s) + SO₄^2-(aq)

Question 12

The rate of formation of NO₂(g) in the reaction

2N₂O₅(g) --> 4NO₂(g) + O₂(g)

2.89 (mol N₂O₅).L^-1.s^-1

5.78 (mol NO₂).L^-1.s^-1 x (2 mol N₂O₅)/(4 mol NO₂)
= 2.89 (mol N₂O₅).L^-1.s^-1

Question 13

Given:

2O₃(g) --> 3O₂(g) rate = k[O₃]²[O₂]^-1

1

Add the superscripts in the rate law.

Question 14

If the rate of a reaction increases by a factor of 64 when the concentration of reactant increases by a factor of 4, the order of the reaction with respect to this reactant is

3

rate = k[concentration]^order. Therefore, (4)^order = 64 and order = 3.

Question 15

For the reaction

2A + B --> products

rate = k[A]²

When the concentration of B is varied ([A] constant), the rate does not change. When the concentration of A is doubled and tripled ([B] constant), the rate increases by a factor of 4 and 9, respectively. (2)ⁿ = 4 and n = 2.

Question 16

A first-order reaction has a rate constant of 0.00300 s^-1. The time required for 55% reaction is

266 s

ln[A]_t = ln[A]_o - kt where [A]_t = 1 - 0.55 = 0.45 and [A]_o = 1.

Question 17

For a first-order reaction, after 230 s, 10% of the reactants remain. Calculate the rate constant for the reaction.

0.0100 s^-1

ln[A]_t = ln[A]_o - kt where [A]_t = 0.1, [A]_o = 1 and t = 230 s.

Question 18

Consider the following reaction:

2N₂O₅(g) ----> 4NO₂(g) + O₂(g), rate = k[N₂O₅]

400 s

ln[A]_t = ln[A]_o - kt where [A]_t = 1/8, [A]_o = 1 and k = 0.00520 s^-1.

Question 19

The HBr synthesis is thought to involve the following reactions:

Br₂ --> 2Br^* (1)

2 and 3

Chain propagation steps involve the reaction of chain carriers or radicals with reactant molecules to produce another chain carrier or radical.

Question 20

For a second-order reaction, a straight line is obtained from a plot of

1/[A]_t vs t

For a second-order reaction, 1/[A]_t = 1/[A]_o + kt.

Question 21

Given: SO₂Cl₂(g) --> SO₂(g) + Cl₂(g) rate = k[SO₂Cl₂]

If 70% of the SO₂Cl₂ has reacted after 80 s, calculate the rate constant.

0.015 s^-1

ln[A]_t = ln[A]_o - kt where [A]_t = 0.3, [A]_o = 1 and t = 80 s.

Question 22

For the reaction

cyclopropane --> propene

2.8 x 10^-4 s^-1

For a first-order reaction, the slope of a plot of ln[concentration] vs time is -k.

Question 23

Consider the dimerization reaction below:

2A-->A₂ rate = k[A]²

4.2 x 10^-4 M^-1.s^-1

The rate law indicates a second-order reaction for which
1/[A]_t = 1[A]_o +kt where [A]_t = 0.80 (40% of A remains), [A]_o = 2.0 and t = 30 x 60 s.

Question 24

For the reaction

C₂H₅Br(aq) + OH^-(aq) -->C₂H₅OH(aq) + Br^-(aq)

89.0 kJ/mol

The slope of a plot of lnk versus 1/T is -E_a/R.

Question 25

An elementary process has an activation energy of 40 kJ/mol. If the activation energy for the reverse reaction is 20 kJ/mol, what is the enthalpy change for the reaction?

20 kJ/mol

Because the activation energy of the reverse reaction is less than the activation energy for the forward reaction, the reaction is endothermic in the forward direction. The enthalpy change is 40 - 20 = 20 kJ/mol. See figure 18.35 (a) in the text.

Question 26

A reaction that has a very high activation energy

has a rate that is very sensitive to temperature.

The slope of a plot of lnk vs 1/T is directly proportional to E_a, the activation energy. Therefore, a large activation energy means a large slope and a large change in rate constants with temperature.

Question 27

Given:

O(g) + O₃(g) --> 2O₂(g)

rate = k[O][O₃]

Because this is an elementary reaction, the rate law is the product of the concentration of each species times the rate constant for the reaction.

Question 28

The reaction

NO₂(g) + CO(g) --> CO₂(g) + NO(g)

NO₃(g)

A reaction intermediate is a species that appears in the mechanism but not in the equation for the overall process. The species NO₃(g) does not appear in the overall equation but does appear in the mechanism.