Chem 1022 Solutions

7/31/2019 Chem 1022 Solutions

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SMART CHEM 1022 Review Guided Solutions

Atomic Properties

1. For which of the following would dipole-induced dipole be the strongest type of intermolecularforce between the two compounds?

Answer a

a.)

dipole-induced dipole, dispersion

b.)

ion-dipole, dispersion

c.)

H bonding, dipole-dipole, dispersion

d.)

ion-induced dipole, dispersion

e.)

dispersion

The only pair that has a dipole-induced dipole is methanol and chlorine.

Solutions


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1. C6H12O6 is dissolved in 1.00L of H2O whose initial vapor pressure is 23.8 torr. After addition of thesugar, the vapor pressure of the solution drops to 21.4 torr. Using this information, calculate

both the mole fraction of the solute and the mass, in grams, used.

Using our handy-dandy periodic table, or just a love of organic chemistry, we can determine that

the molar mass of C6H12O6 (or glucose) is the sum of the molar masses for each element present.This would be 12.01 for Carbon, 1.01 for Hydrogen, and 16.00 for Oxygen, so .

The handy vapor pressure equation ( ) tells us that the vaporpressure of a solvent is equal to the product of the mole fraction of the solvent and vapor

pressure of the solvent. .

By definition, the mole fraction in this situation would be . Somemay be familiar with the density of water (it is 1g/ml), and from that we could determine the

moles of water. The net equation to determine moles of glucose will be: . Multiplying this bythe molar mass of glucose gives us the mass, in grams, or 1121.37g

2. Which of the following liquids would have the lowest boiling point?a. NaCl (0.75 m in water)b. NaN03 (0.25 m in water)c. methanol (0.25 m in water)d. pure watere. the lowest boiling point cannot be determined

Answer dA solution boils at a higher temperature than a pure solvent so water has the lowest boiling

point because it is the only pure liquid.

3. Assuming ideal behavior, what is the freezing point of a 0.70m solution of glucose in water?(water has Kf=1.86 C/m, Kb=0.51C/m, 18.0 g/mol)

a. 1.30 Cb. 0.36 Cc. 0.00 Cd. -0.36 Ce. -1.30 C

Answer e

The change in freezing point only depends on the concentration of the solute and the addition

of solute causes freezing point depression, so

( )


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Rate Laws

1. An experiment is done to determine the rate law for a reaction which uses two reactants, A andB. Determine the rate law from the following experimental data:

My favorite way to solve these problems is to look at how changing the initial concentration

affects the initial rate. Remembering that the rate for this equation should be .If we decrease the concentrations of one of the reactants by half, then if it is first order the rate

should decrease by half. If it is second order, it would decrease by 2, and so forth. If you notice,

when decreasing the concentration of B by half, we change the rate from 80 to 640, which

would be

. From this, we can see that it is like that m is 3.

Keeping this in mind, by examining the change from experiment 2 to 3, while decreasing A by a

fourth, the rate decreases by a fourth, suggesting that n is equal to one. Thus, the overall rate

law is

2. The reaction 2N205(g) 02 (g) + 4NO2 (g) is first order in N205. At a particular temperature,the rate constant is 1.0 x 10-4 s-1. If initially the [N205] = 0.0040 M, what is the [N205] after20,000 seconds.

a. 0.0041Mb. 0.0026Mc. 0.0015Md. 0.00054Me. 0.00020M

Answer d

( )

( )

Experiment Initial Rate

1 4.00 4.00 640.0

2 4.00 2.00 80.0

3 1.00 2.00 20.0

4 1.00 1.00 2.5


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3. For a reaction in which A and B react to form C, what is the rate law for the reaction in which thefollowing initial raw data was obtained?

[A]0 (mol/L) [B]0 (mol/L) Initial Rate of Formation of C (mol/Ls)

0.2 0.4 2.00

0.2 0.8 4.00

0.2 1.6 8.00

0.4 0.4 8.00

0.8 0.4 32.00

a. rate = k [A]2[B]b. rate = k [A][B]3c. rate = k [A]2[B]2d. rate = k [A]2e. rate = k [A]2[B]3

Answer a

[

]

so b = 1

[]

so a = 2

Properties of solutions

1. 0.500 grams of a non electrolyte is dissolved in 25.0 mL of H2O which decreases the freezingpoint by 0.180C. Given that the Kfand Kb for H2O is -1.86 and 0.512

respectively, what is the

molar mass of the solute?

First things when handling questions like this, is to figure out what role each piece of

information plays in the question. For instance, the fact that it is a non-electrolyte tells us that it

is a non-ionic compound and will not disassociate upon addition to water. Since the question

states that the freezing point is changing, we can determine that we are only going to have to

use the equation for freezing point depression, and all things related. This equation, which

should be given to you on the exam, is .While the question gives us a Kb as well, this is a boiling point elevation constant, and irrelevant

to the question, and i is a constant referring to the number of moles the solute breaks up into


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for every mole added to solution (i.e. KCl dissolves into K+

and Cl-, so i=2 for that). Here we do

not have to worry about that. The only thing remaining is m, the molality of the solute. Using

our super-funzies density of water (which is super awesome, exciting, and memorable) 1g/ml,

we can get the kg of solvent and solve the equation for moles.

.

Using the mass provided, we can see that the molar mass is

Equilibrium

1. If the equilibrium for the reaction is , what is thepartial pressure of ClO2 from the reaction of 1.00 atm of O2 and Cl2?

The definition of a equilibrium constant is

. Now weget to put all of this into another favorite tool of mine, a RICE or ICE table. This stands for

Reaction, Initial, Change, and Equilibrium which is constructed as so:

Reaction Initial 1.00 1.00 0

Change -x -x +x

Equilibrium 1.00-x 1.00-x x

This makes our equation, at equilibrium . Unfortunately, there isno other way to solve this perfectly other than to make some fundamental assumptions.

Temporarily putting K in for the equilibrium, we can get the equation

, but since Kis very very small, we may assume that the concentration of products, x, is so very small, that 1-

x is approximately equal to 1, making our equation K=x2, and then .For those of you interested, it is possible to solve this exactly using Newtons Method (which

you may have learned in your calculus class. Both methods do give the same answer to the 5th

decimal place.

2. Calculate the equilibrium constant for the following reaction. 2A B + 2CInitially the concentration of A was 5.5M, B was 2.0 M and no C was present. At equilibrium theconcentration of A was 2.5 M.

a. 6.3b. 5.0c. 4.6d. 2.4e. 1.0


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[A] [B] [C]

initial concentration (mol/L) 5.5 2.0 0change -2x +x +2x

final concentration 2.5 3.5 3.0

Acid Base

1. If 0.050 L of 0.100M acetic acid, with a pKa of 4.76, is titrated with 0.015L of 0.100M NaOH, whatwill the resultant pH be?

Here is a fun little question, where we get to employ a super awesome equation known as the

Henderson Hasselbalch. This one goes in a little ditty I call puh is pika and a ha! or

. For this equation to work we need a pKa, which is given, and the amount of acid

and conjugate base.

Those of you familiar with acetic acid, or vinegar, may know that it is a weak acid while sodium

hydroxide is one of the strongest bases. Thus, we can assume that the base will react to

completion with the acetic acid, meaning we have to use some stoicheometry. In order to do

this, we need to convert both of our reactants into moles. Fortunately, the product of Molarity

and volume is moles. If we assume that all of the hydroxide deprotonates the acid, then theamount of HA or acid will be given by

. When you plug all ofthis in to the equation it goes something like

While this equation suggeststhe necessity of using concentrations, if the acid and base are in the same reaction flask, then

the two concentrations would use the same volume, canceling one another out, leaving only the

moles left.

2. What is the pH of 0.90 M NH3 solution? Kb=1.8x10-5a. 2.56b. 4.56c. 7.40d. 9.10e. 11.60


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The value of Kb is very small so we can assume that x is very small so the equation becomes-

Thermodynamics

1. Under what conditions is a reaction non-spontaneous at all temperatures?Something which I highly recommend learning is that spontaneous reactions occur when the

change in gibbs free energy is less than zero, meaning non-spontaneous reactions occur when

the change in Gibbs free energy is greater than zero. Setting up the equation as an inequality where then Playing around with this, it ispossible to see that so long as and , then the reaction will be non-spontaneous.

2. For the following reaction at equilibrium, which choice gives a change that will shift theequilibrium to the left? H= 30kJ

2NOBr (g) 2NO (g) + Br2 (g)

a. increase the volumeb. remove Br2c. remove NOBrd. remove NOe. two of these changes

Answer c

a. An increase in volume would decrease the pressure, so equilibrium would shift to theright.

b. Removal of Br2 would decrease Q so the reaction would have to shift to the right to goback to Keq.

c. Removal of NOBr would increase Q so the reaction would have to shift to the left to goback to Keq.d. Removal of NO would decrease Q so the reaction would have to shift to the right to go

back to Keq.

3. Which of the following results in a decrease in the entropy of the system?a. 02(g), 300K 02 (g), 400 Kb. H20 (l) H20 (g)


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c. 2N02(g) N204 (g)d. NH3(s) NH3 (l)e. 2H20 (g) 2H2 (g) + 02 (g)

Answer c

a. Heat lost by the surroundings is gained by the system so there is an increase in entropyof the system.

b. The entropy of the system increases as water absorbs heat and changes to gas.c. The entropy of the system decreases because the amount of gas decreases.d. The entropy of the system increases as solid NH3 absorbs heat and becomes liquid.e. The entropy of the system increases because the amount of gas increases.

4. Given N2H4 (l) + 2H20 (l) N2 (g) + 4H20 (g) H = -530kJ at 298KWhich of the following is true?

a. the reaction is spontaneous at high temperaturesb. the reaction is spontaneous at low temperaturesc. the reaction is spontaneous at all temperaturesd. the reaction is not spontaneous at any temperature

Answer c

The H < 0 so heat is released by the system and so the total entropy will be positive and the

reaction is spontaneous at all temperatures

Electrochemistry

1. A galvanic cell uses a hydrogen electrode and lithium. Given that the standard reductionpotential of lithium is -3.0401 and that of hydrogen is zero, what is the chemical potential of the

battery if the initial concentrations of the ions are 5mM each?

Looking at the periodic table, we can guess that Lithium, like Hydrogen, will be only capable of

forming a plus one charge state. Thus, the total reaction is going to be ,and the standard reduction of the reaction will be 3.0401V. Using the standard reduction

equation, , we can determine the reduction potential of the cell. Pluggingin our reaction, we get something closer to being a plug n chug:

. Our

remaining variable, n, corresponds to the number of electrons transferred in the reaction. As a

rule, it will always be the total amount of change in charge on each side, in this case it is 2. Thus,

our answer is .

2. An electrochemical cell involves the following half-reactions. Under standard conditions, whatspecies are produced at each electrode?

Ag+(aq) + e

- Ag(s) 0.80V


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Cu2+

(aq) + 2e-Cu(s) 0.34V

a. Ag+ (aq) is produced at the cathode and Cu(s) at the anodeb. Ag+ (aq) is produced at the anode and Cu (s) at the cathodec. Ag (s) is produced at the anode and Cu2+ (aq) at the cathoded. Ag (s) is produced at the cathode and Cu

2+

(aq) at the anodeAnswer d

The spontaneous reaction that occurs is:

2Ag+ + Cu(s) 2Ag(s) + Cu2+

Ecell = Ecathode(reduction) - Eanode(oxidation) = 0.80V-0.34V = 0.46V

The copper is oxidized at the anode and the silver is reduced at the cathode in the overall

reaction because Ecell must be positive for a spontaneous process.

3. For the reaction Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) the cell voltage is 1.10 V under standardconditions. What is the voltage if [Cu2+] =0.200M and [Zn2+]=0.060M?

a. 1.01 Vb. 1.12 Vc. 1.19 Vd. 1.26 Ve. 1.41 V

Answer b

Standard concentrations are: [Zn2+

] = [Cu2+

]= 1 M at 298.15K where

()

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