# Practice Problems for Exam 2

Practice problems for exam 2 are listed below.

## Topic: Solubility

1. #### Conversion of Concentration Units

An aqueous solution of nitric acid contains 10% HNO3 by mass. Determine the mole fraction of HNO3 and the molality and molarity of the solution.

X(HNO3) = 0.03
m(HNO3) = 1.75 m
M(HNO3) = 1.58 M

##### Explanation
Begin by assuming you have 100 g of solution. Thus if the solution is 10% by mass HNO3 then

moles HNO3 = (10 g)/(63 g/mol) = 0.158 mol
moles H2O = (90 g)/(18 g/mol) = 5.0 mol total moles = 0.158 mol HNO3 + 5.0 mol H2O = 5.158 mol

The mole fraction is

X(HNO3) = mol HNO3/total mol = 0.158 / 5.158 = 0.03

The molality is

m(HNO3) = mol solute/kg solvent = (0.158 mol HNO3)/(0.090 kg H2O) = 1.75 m

Assuming the solution density is 1 g/cm3, (100 g solution)/(1 g/cm3) = 100 cm3 = 100 mL and the molarity is

M(HNO3) = mol solute/L solution = (0.158 mol HNO3)/(0.100 L solution) = 1.58 M

2. #### Parts per million units

An ore contains 2.86 g of silver per ton of ore. What is the concentration of silver in ppm?
3.15 ppm Ag
##### Explanation
The mass of Ag is 2.86 g the total mass is 9.08 x 105 g (2000 lb * 454 g/lb).

ppm = (mass of component)/(total mass) * 106 = 2.86 g/(9.08x105)*106 = 3.15

3. #### Henry's Law

Henry's law constants of gases are used to calculate gas solubilities. A sealed flask containing water at 30o is filled with helium gas to a partial pressure of 1.5 atm. Determine the solubility of helium in water. The Henry's law constant for helium in water is 3.7 x 10 -4 M/atm.
SHe = 5.6 x 10 -4 M
##### Explanation

SHe = PHe * k
SHe = (1.5 atm)*(5.6 x 10-4 M/atm)
SHe = 5.6 x 10-4 M

4. #### Precipitation reactions

Consider the following statements regarding precipitation reactions. Determine whether each statement is true or false
• In a precipitation reaction, a solid is formed
• In an insoluble ionic compound, the attraction forces between oppositely charged ions are stronger than those in a soluble ionic compound.
• In an insoluble ionic compound, water molecules are repelled by the oppositely chared ions.
• Writing NaCl(aq) in a chemical reaction implies the presence of Na+ and Cl-
• True
• True
• False
• True
##### Explanation
• The precipitation reaction results in the formation of a solid product
• It is the relative strength of the inttermolecular interactions that will determine the solubility of an ionic compound.
• Water molecules are not repelled by oppositely chaged ions. The intermolecular interactions between the dipole nature of water and the ionic charges are not stronger than the ion-ion interactions in the insoluble compound.
• Since NaCl is a compound which includes an alkali metal, it is soluble in water and will lead to formation of Na+ and Cl-
5. #### Solubility and solution chemistry

Consider the following statements regarding solubility and solution chemistry. Determine whether each statement is true or false
• When NaCl(s) is dissolved in water, a chemical reaction occurs
• The separation of solute molecules when forming a solution is an endothermic process
• For dissolution to be spontaneous, the solvent-solute interactions must be attractive
• Water is miscible in gasoline, a non-polar substance
• Dissolution at constant temperature leads to an increase in randomness of the system
• Processes in which the energy content of the system decreases tend to occur spontaneously
• Like dissolves like
• False
• True
• True
• False
• True
• True
• True
##### Explanation
• THe formation of a solution is not a chemical reaction, just a mixing process (homogeneous mixing.
• The separation of solute molecules requires energy to overcome the intermolecular interactions. Therefore, the process is endothermic.
• The solvent-solute interaction must be comparable in magnitude to the energy needed to break the solvent-solvent and solute-solute intermolecular interactions.
• Non-polar liquids tend to be insoluble in poalr liquids.
• Whena solute is dispersed homogeneously in the solvent, the randomness, or entropy, of the system is increased.
• This statement is true. Some endothermic processed are spontaneous.
• Substances with similar intermolecular attractive forces tend to be soluble in one another.
6. #### Factors affecting solubility

Consider the following statements regarding the factors that affect solubility. Complete each statement with the word increases, decreased, or not change.
• The solubility of simple gases in water ... with increasing molecular mass
• The solubility of simple gases in water ... with decreasing polarity
• The solubility of a gas ... with an increase in partial pressure of the gas above the solution.
• The solubility of most solid solutes in wqater ... as the temperature of the solution decreases
• The solubility of gases in water ... with decreasing temperature
• increases
• decreases
• increases
• decreases
• increases
##### Explanation
• The dispersion forces will increase with molar mass. Since the intermolecular forces increase the solubility will increase.
• The weaker the attractions between solute and solvent molecules the lower the solubility
• The solubility will increase in dircet proportion to the partial pressure. This is related to Henry's Law.
• There are some expceptions, but in general an increase in temperature will increase the solubility of most solids.
• This is in contrast to solid solutes. As temperature is increased the solubility of gases in solution is decreased.

## Topic: Colligative Properties

1. #### Colligative properties

Determine whether each of the following statements pertaining to colligative properties is true or false.
• If two non-isotonic solutions are separated by a semipermeable membrane, the solute will flow from the low concentration to the high concentration.
• Rauolt's law gives better results for solvents in dilute solutions.
• At equal low molal concentrations, acetic acid and hydrochloric acid will have the same effect on the freezing point of water
• A water solution with a high osmotic pressure will boil at a lower temperature than a water solution with a lower osmotic pressure
• If an aqueous solution of CaCl2 boils at 102 o, it will freeze at -2 oC
• False
• True
• False
• False
• False
##### Explanation
• Solute does not flow through the semipermeable membrane. The membrane is only permeable to solvent. Solvent will flow from low concentration to high concentration.
• Raoult's law is best applied to solutions at low concentrations.
• Acetic acid is a weak acid, and only dissociates slightly. HCl is a strong acid and will (nearly) completely dissociate in water. Therefore, HCl will have a greater effect on the freezing point of water.
• A water solution with higher osmotic pressure has a higher molal concentration and therefore the boiling temperature will be raised more than for a water solution with a lower osmotic pressure.
• The freezing point depression and boiling point elavation constants are not of the same magnitude for a given solute.
2. #### Boiling Point of a syrup solution

A sweet syrup is made by adding 588 g of fructose (fruit sugar, MW 180.2 g/mol) to 1000 g of water. The strup is placed in a pressure cooker that operates with a safety value set to reliece the inside pressure if it exceeds 3.4 atm. Use the vapor pressure curve of pure water given below to estimatethe boiling point of the syrup in the pressure cooker. You can assume the solution is ideal and obeys Raoult's law.
140 o
##### Explanation
The number of moles of frustose and water are

(588 g)/(180.2 g/mol) = 3.26 mol
(1000 g)/(18 g/mol) = 55 mol

The mole fraction of fructose is

3.26 mol/(3.26 mol + 55 mol = 0.055

The vapor pressure of water is therefore reduced P = XPo. At 100 oC, P = (0.0945)(1 atm) = 0.945 atm. This will shoft the vapor pressure curve to lower pressures, see below and the new boiling point will increase to 140 o
3. #### Concentration effect on vapor pressure

List the following solutions in order of decreasing vapor pressure
• 0.2 mol MgCl in 1.0 kg of water
• 0.2 mol butiric acid in 1.0 kg of water
• 0.2 mol ethlyene glycol in 0.5 kg water
• 0.2 mol MgSO4 in 1 kg of water
• 0.2 mol butiric acid in 1.0 kg of water
• 0.2 mol MgSO4 in 1 kg of water
• 0.2 mol ethlyene glycol in 0.5 kg water
• 0.2 mol MgCl in 1.0 kg of water
##### Explanation
Colligative properties only depend on concentration. The higher concentration of solute will lead to a lower vapor pressure
• 0.2 mol butiric acid in 1.0 kg of water = 0.2 + x molal solution where x accounts for small dissociation of butiric acid which is a weak acid
• 0.2 mol MgSO4 in 1.0 kg of water = 0.4 - y molal solution where y accounts for small amount of MgSO4 that does not dissociate due to high concentration
• 0.2 mol ethylene glycol in 0.5 kg water = 0.4 molal solution since ethlyene gylcol does not dissociate in water
• 0.2 mol MgCL2 in 1.0 kg of water = 0.6 -z molal solution where z accounts for small amount of MgCl2 that does not dissociate due to high concentration.
4. #### Binary Solution and Vapor Pressure

At a temperature of 353 K, the vapor pressure of pure benzene is 720 torr and the vapor pressure of pure toluene is 300 torr. For a benzene mole fraction of 0.5, determine the total vapor pressure in equilibrium with the liquid. You can assume that the binary system behaves ideally.
510 torr
##### Explanation
Raoult's law can be used to describe the behavior of binary solution. The vapor pressure of each component depends linearly on the concentration.

Pbenzene = X Pobenzene

Where Pbenzene is the vapor pressure of benzene in solution, X is the mole fraction of benzene, and Pobenzene is the vapor pressure of pure benzene. The vapor pressure due to toluene can be calculated in an analogous manner.

Pbenzene = (0.5)(720 torr) = 360 torr
Ptoluene = (0.5)(300 torr) = 150 torr
Ptot = Pbenzene + Ptoluene = 510 torr

5. #### Osmosis of Sugar Solution

What is the osmotic pressure at 273 K of a 0.5 mM solution of calcium chloride assuming complete dissociation and ideal behavoir of the electrolyte?
0.033 atm or 25.5 torr
##### Explanation
The osmotic pressure is directly proportional to the molar concentration of individual particles.

P = MRT
M = 0.0005 m, R = 0.08206 Latm/Kmol, T = 273 K

Remember that CaCl2 will dissociate completely at low molal concentrations to one ion of Ca2+ and 2 ions of Cl-
6. #### Boiling Point Elevation

A solution of benzene is 20 mass percent glucose, M.W. = 180.1 g/mol. Calculate the boiling point of the solution given the boiling point of the pure solvent at 1.00 atm pressure is 80.1 oC. Assume ideal behavior of the solution. The molal boiling point elevation constant for benzene is 2.53 oC/m
83.6oC
##### Explanation
The change in boiling temperature is related to the molal concentration of particle in solution. To calculate molal concentration from mass percent, assume 100 g of solution. The mass of glucose is then 100 g * 0.2 = 20 g and the mass of benzene is 100 g - 20 g = 80 g.

molal concentration = [20 g solute/(180 g/mol)]/(0.080 kg solvent) = 1.39 m.

The molal boiling point elevation constant for benzene is 2.53 oC/m from Table 13.4 in the textbook

DT = Kbm = (2.53 oC/m)(1.39 m)
DT = 3.51 oC
Tb = 83.6 oC

7. #### Molar Mass from Colligative Properties

10.5 g of an unknown organic compound were dissolved in 200 g of carbon tetrachloride. The measured freezing point of the soluton was -32.7 oC. Determine the molar mass of the unknown compound
150 g/mol
##### Explanation
The depression of the freezing point will depend on the molal concentration, which can be used to determine the molar mass. The freezing point of pure carbon tetrachloride is -22.3 oC and Kf is 29.8 oC/m (see Table 13.4 in the textbook). Therefore,

D = -32.7 - (-22.3) = -10.4oC
m = (10.4oC)/(29.8oC/m) = 0.349 m
mol solute = (0.349 m)(0.200 kg solvent) = 0.07 mol solute
M.W. = (10.5 g)/(0.07 mol) = 150 g/mol

## Topic: Reaction Rates

1. #### Factors that affect reaction rates

Complete each of the following statements pertaining to the factors that affect chemical reaction rates. Fill in the blank space using the words: increase, decrease, or not change.
• If the reactant volume is increased, without changing reactant concentration, the rate of the reaction will ...
• When the temperature of the reaction vessel is increased, the reaction rate will usually ...
• If a solid reactant is subdivided into smaller particles, the reaction rate will ...
• If the reactant concentration is decreased, the reaction rate will usually ...
• If a catalyst is added to the reaction vessel, the reaction rate will ...
• Not change
• increase
• increase
• decrease
• increase
##### Explanation
• If the reactant concentration is unchanged, the rate will not change
• Increasing the temperature will increase the reaction rate. The rate constant k will increase according to the Arrhenius equation
• Subdividing a solid reactant increases surface area, which leads to an increase in reaction rate
• In most instances, the rate law shows a direct relation between rate and concentration. When the concentration is decreased the rate decreases as well
• A catalyst will speed the rate of chemical reaction by lowering the activation barrier
2. #### Stoichiometry and Rates

For the reaction below:

NO + H2 -> N2 + H2O

Balance the chemical reaction and determine the rate of produaction of H2O given the rate of destruction of H2 is 3.500 x 10-4 mol/s
3.500 x 10-4 mol/s
##### Explanation
The balanced reaction is:

2NO + 2H2 -> N2 + 2H2O

Since there is a 1:1 ratio in the stoichiometry between H2 and H2O the rate of production of H2O equals the rate of destruction of H2
3. #### Average Reaction Rates

The rate of disappearance of HCl was measured for the following reaction:

CH3OH (aq) + HCl (aq) -> CH3Cl (aq) + H2O (l)

From the following data calculate the average rate of the reaction between 54.0 and 107.0 minutes.
 Time (min) [HCl] (M) 0.0 1.85 54.0 1.58 107.0 1.36 215.0 1.02 430.0 0.580
4.15 x 10 -3 M/min
##### Explanation
The average rate is the change in concentration divided by the change in time.

Average rate = -D[HCl]/Dt = (1.58 M - 1.36 M)/(54.0 min - 107.0 min)
Average rate = 4.15 x 10-3 M/min

## Topic: Concentration and Rate

1. #### Reaction Orders

Consider the following statementts pertaining to the orders of chemical reactions. Determine whether each statement is true or false.
• THe overall reaction order is the sum of the reaction orders of all reactants
• M-1s-1 is the unit for the rate constant of an overall second order reaction
• The reaction order can be 0
• The reaction order of a reactant is determined experimentally
• The reaction order usually changes as the concentration of one of the reactants changes
• True
• True
• True
• True
• False
##### Explanation
• The overall reaction order is the sum of the orders with respect to each reactant in the rate law
• Since the rate law is R = k[A]2 the units of k must be M-1s-1
• The reaction order can be 0, negative, a fraction, etc.
• Stoichiometry cannot be used to determine reaction order, unless the reaction is an elementary reaction. Therefore experimental data are needed.
• Although the reaction rate changes with concentration the reaction order does not
2. #### Reaction Rate and pOH

A particular reaction was found to depend on the concentration of the hydroxyl ion [OH-]. The initial rates varied as a function of initial pOH as follows.
 pOH 2.699 2.398 2.097 Initial Rate (M/s) 1.600 x 10-2 6.400 x 10-2 2.560 x 10-1
Predict the initial reaction rate when pOH is 2.155
1.96 x 10-1
##### Explanation
pOH is a measure of [OH-] and pOH = -log[OH-] and the concentration can be added to the table
 pOH 2.699 2.398 2.097 [OH-] 2.00 x 10-3 M 4.00 x 10-3 M 8.00 x 10-3 M Initial Rate (M/s) 1.600 x 10-2 6.400 x 10-2 2.560 x 10-1
An increase by a factor of two in [OH-] leads to an increase in the reation rate by a factor of four. Therefore, the reaction is second order in [OH-]. The rate constant can be determined from one of the data points: R = k[OH-]2

1.600 x 10-2 M/s = k(2.00 x 10-3 M)2
k = 4000 M-1s-1

Using the same relation and the [OH-] for pOH = 2.155

R = k[OH-]2
R = 4000 M-1s-1(7 x 10-3)2
R = 1.96 x 10-1 M/s

3. #### Reaction Order from Experimental Data

The initial rates for the reaction A + B -> C + D at varying concentrations of A and B are given below:
 Exp init[A] (M) init[B] (M) Initial Rate (M/s) 1 3.50 4.00 5.900 x 10-6 2 3.50 20.0 1.475 x 10-4 3 3.50 100.0 3.688 x 10-3 4 17.5 4.00 7.375 x 10-4 5 87.5 4.00 9.219 x 10-2
Determine the order of the reaction for the reactants A and B and the overall reaction order.

3rd order in [A]
2nd order in [B]
5th order overall

##### Explanation
Comparing Exp 1 and 2

Rate = k[A]m[B]n 5.900 x 10-6 M/s = k[3.5 M]m[4.00 M]n
1.475 x 10-4 M/s = k[3.5 M]m[20.00 M]n

Taking the ratios of these two rates

0.04 = [4]n/[20]n
0.04 = 1/[5]n
25 = 5n
n = 2

The reaction is second order with respect to [B] Comparing Exp 1 and 4

Rate = k[A]m[B]n 5.900 x 10-6 M/s = k[3.5 M]m[4.00 M]n
7.375 x 10-4 M/s = k[17.5 M]m[4.00 M]n

Taking the ratios of these two rates

0.008 = [3.5]m/[17.5]m
0.008 = 1/[5]m
125 = 5m
n = 3

The reaction is third order with respect to [A] The reaction is 5th order overall
4. #### Reaction rate and equilibirum constant

For the equilibrium reaction A <=> B, various measurements were run using A only, and B only. From the data below, extract the equilibrium constant for the reaction.
 Exp [A] (M) Initial Rate (M/s) 1 4.5 8.900 x 10-5 2 9.00 3.560 x 10-4 3 18.00 1.424 x 10-3 Exp [B] (M) Initial Rate (M/s) 1 3.00 9.767 x 10-7 2 15 4.883 x 10-6 3 75.00 2.442 x 10-5
Keq = 13.5
##### Explanation
The equilibrium constant is

Keq = kf/kr

To find the rate constants for the forward and reverse reactions, need to determine reaction order. Comparing Exp 1 and 2 for A only

Rate = k[A]m
8.900 x 10-5 M/s = Kf[4.5 M]m
3.560 x 10-4 M/s = Kf[9.0 M]m

Taking the rations of these two rates:

0.25 = [4.5]m/[9.0]m
0.25 = [1]m/[1]m
m = 2, order with respect to [A]
kf = Rate/[A]2
kf = 8.900 x 10-5/[4.5]2
kf = 4.395 x 10-6

Comparing Exp 1 and 2 for B only

Rate = k[B]n
9.767 x 10-7 M/s = Kf[3.0 M]n
4.883 x 10-6 M/s = Kf[15.0 M]n

Taking the rations of these two rates:

0.200 = [3.0]n/[15.0]n
0.200 = 1/[5]n
n = 1, order with respect to [B]
kr = Rate/[B]
kr = 9.676 x 10-7/[3.0]2
kr = 3.256 x 10-7

Thus the over all reaction rate is

Keq = kf/kr
Keq = 4.395 x 10-6/3.256 x 10-7
Keq = 13.5

## Topic: Order of Reaction

1. #### Primer on First-Order Reactions

Which of the following statements regarding first order chemical reactions are true?
• A graph of reactant concentration versus the natural logarithum of time gives a straight line
• It is not necessary to know the initial reactant concentration to determine half-life
• The reaction rate may depend on the concentration of a single reactant raised to the first power
• The first-order nature of a reaction can be determine by experiment
• The concentration of the reactant decreases by one-half over a time interval equal to one half-life
• False
• True
• True
• True
• True
##### Explanation
• A graph of the natural log of the reactant concentration versus time will be a straight line for a first order reaction.
• The half-life of a first-order reaction i inversely proportional to the rate constant, and independent of concentration
• For a first-order reaction, Rate = k[A]1
• The only methods to determine rate orders is through experiment
• The half-life is defined as the average time required to reduce the reactant concentration by 1/2 of its initial value.
2. #### Primer on second order reactions

Which of the following statements, regarding second order chemical reactions, are true?
• A graph of natural logarithum of reactant concentration versus time gives an exponential decay curve
• It is not necessary to know the initial reactant concentration to determine the half-life
• The reaction rate may depend on the concentration of a single reactant raised to the first power.
• The second-order nature of a reaction can be determined simply by examining the stoichiometry of the chemical reaction.
• The concentration of the reactant decreases by one-half over a time interval equal to one half-life
• False
• False
• False
• False
• True
##### Explanation
• A graph of 1/[A] versus time will yield a straight-line curve
• The half-life of a second-order reaction does depend on initial concentration: t1/2 = 1/(k[Ao])
• For a second order reaction, Rate = k[A]2
• The reaction order must be determined from experimental data.
• The half-life is defined as the average time needed to decrease the initial concentration by a factor of 2
3. #### Half-life determination for first-order process

A reactant undergoes a dissociation reaction A -> 2B, which follows first-order kinetics. The initial concentration of reactant A is 0.40 M, and after 75 hr the concentration of A decreases to 0.082 M. Determine the half-life of the first-order process
t1/2 = 32.7 hr
##### Explanation
The half-life can be solved using:

ln[A]t = -kt + ln[A]0
ln[A]t = ln[0.082] = -2.50
ln[A]0 = -0.916
t = 75 hr
Therefore, k = 0.02112 hr-1 t1/2 = ln(2)/k
t1/2 = 0.693/(0.02112 hr-1)
t1/2 = 32.7 hr

4. #### Reactant Concentration for Second-Order Process

A reactant undergoes a combination reaction 2A -> B, which follows second-order kinetics with a rate constant of 0.715 M-1hr-1. The initial concentration of reactant A is 0.300 M. Calculate the concentration of A after 12.0 hr
[A] = 0.049 M
##### Explanation
The concentration of [A] at time t for a second order process is given as:

1/[A]t = kt + 1/[A]0
k = 0.715 M-1s-1
t = 12.0 hr
1/[A]0 = 1/(0.3 M) = 3.33 M-1
1/[A]t = 11.91 M-1
[A] = 0.084 M

5. #### Iodate Reaction Rate and pH

The iodate ion is reduced by sulfite according to the following reaction:

IO3-(aq) + 3SO32-(aq) -> I-(aq) + 3SO42-(aq)

The rate of reaction is found experimentally to be first order in IO3-, first order in SO32-, and first order in H+. By what factor will the rate of reaction change if the pH is lowered from 5.00 to 3.40 with all other reactant concentrations remaining constant?
The rate will increase by a factor of 40.
##### Explanation

Rate = k[H+][IO3-][SO32-]
For pH = 5.00, [H+] = 1 x 10-5M
For pH = 3.40, [H+] = 3.98 x 10-4M

If all other concentrations are the same:

Rate1 =k[1 x 10-5M][IO3-][SO32-]
Rate2 =k[3.98 x 10-4M][IO3-][SO32-]

Taking the ratio of these two relations:

Rate1/Rate2 = (1 x 10-5M)/(3.98 x 10-4M)
Rate1/Rate2 = 0.025

Therefore, Rate2 is faster than Rate 1 by a factor of 40
6. #### Rate Order Determination from Graphing

The data below were collected for the dissociation reaction A -> B + C. Plot these data to determine whether the reaction is first or second order.
 Exp Time [A] (mol) 1 0.0 2.50 2 28.4 1.895 3 56.8 1.436 4 85.2 1.088 5 113.6 0.825
The reaction is first order with respect to [A]
##### Explanation
The reaction order can be determined by plotting the [A] v. t in different ways. A linear curve will be obtained for a plot of ln[A] v. t if the reaction is first order. A linear curve will be obtained for a plot of 1/[A] v. t if the reaction is second order. Such plots for the data here are shown below

The reaction is first order with respect ot [A].

## Topic: Temperature and Rate

1. #### Arrhenius Equation

Consider the following statements, regarding the Arrhenius equation, and decided whether each is true or false.
• The collisional model can be used to describe the dependence of reaction rate on temperature.
• A chemical reaction will be initiated regardless of the orientation of the reactant molecules.
• The activation energy is equal to the energy released in a chemical reaction.
• The increase in reaction rate with increasing temperature has a linear behavior.
• Product formation occurs upon each molecular collision.
• True
• False
• False
• False
• False
##### Explanation
• The collisional model is based on Kinetic Molecular Theory and provides a means to describe reaction rates as a function of temperature
• Orientation is important for a chemical reaction to occur. In the Arrhenius expression, the molecular orientation is accounted for in the frequency factor
• The activation energy is the energy needed to overcome the activation barrier. It is the minimum energy need to initiate a chemical reaction
• The reaction rate increases exponentially with temperature.
• Product formation will not occur with every collision. There is a probability factor that governs product formation
2. #### Activation Energy Calculation

The rate constant increases by a factor of 2 when the temperature is increased from 80.0 oC to 115.0 oC for a given reaction. Estimate the Arrhenius activation energy
Ea = 22.5 kJ/mol
##### Explanation
The Arrhenius equation for a reaction at two different temperatures is given as:

ln(k1/k2) = Ea/R[(1/T2) - (1/T1)]
The ratio k1/k2 = 1/2
T2 = (115.0 + 273.15) = 388 K
T1 = (80.0 + 273.15) = 353 K
R = 8.314 J/K-mol

Ea = 22.5 kJ/mol
3. #### Reaction Rate at Different Temperature

The rearrangement of methyl isonitrile to acetonitrile has an activation energy of 25.0 kJ/mol and a rate constant of 8.10 x 10-4 s-1 at 80.0 oC. What is the value of the rate constant at 255.0 oC.
k(255 oC) = 1.36 x 10-2 s-1
##### Explanation
The Arrhenius equation for a reaction at two different temperatures is given as:

ln(k1/k2) = Ea/R[(1/T2) - (1/T1)]
k1 = 8.1 x 10-4 s-1
T2 = (255.0 + 273.15) = 528 K
T1 = (80.0 + 273.15) = 353 K
R = 8.314 J/K-mol

Ea = 25 x 103 J/mol k2 = 1.36 x 10-2 s-1
4. #### Arrhenius Curve and Mechanism Changes

An Arrhenius plot is shown below for the hydration of fumaric acid. This plot shows a break at a temperature of 20oC, indicating a change in the reaction mechanism. Indicate which statement best describes the relationship between the activation energies for the two sections of the Arrhenius plot.
• Ea(AB) is the same as Ea(BC)
• Ea(AB) is greater than Ea(BC)
• Ea(AB) is smaller than Ea(BC)

Ea(AB) is greater than Ea(BC)
##### Explanation
The slope of the relation ln K versus 1/T is equal to -Ea/R. The steeper the slope, the greater the value of Ea, since R is a constant.
5. #### Reaction Pathway and Kinetics

Consider the following statements, pertaining to the reaction coordinate energy diagram given below:

Determine whether each statement is true or false.
• For the reaction Y -> X, U is the activation energy
• For the reaction X -> Y, T is the amount of heat consumed
• For the reaction Y -> X, U+T is the amount of heat consumed
• For the reaction X -> Y, T is the activation energy.
• The reaction Y -> X has the same rate as X -> Y
• For the reaction X -> Y, W is the transition state
• False
• False
• False
• False
• False
• True
##### Explanation
• The activation energy for Y -> X is T+U, and represents the energy difference between reactants and the transition state
• X -> Y is an exothermic reaction, therefore, heat in the amount of T is released
• Y -> X is an endothermic process, but the heat consumed is T. The activation energy U is not consumed nor released in the reaction. It is an energy required to initiate the reaction.
• The activation energy for X -> Y is equal to U, again the difference in energy between reactants X and the transition state W.
• Since Y -> X has a higher activation barrier (U+T) than X -> Y (which is only U), the rate of Y -> X will be slower.
• W is the transition state for both X -> Y and Y -> X

## Topic: Reaction Mechanism

1. #### Basics of Elementary (Single-Step) Reactions

The following reaction is a single-step, bimolecular reaction:

H + I2 -> HI + I

Consider the following statements, pertaining to the elementary process above. Determine whether each statement is true or false.
• The rate law for a single-step reaction can be determined from the reaction molecularity
• The single-step nature of the reaction could be determined by experiment
• The rate law for the reaction is k[H][I]2
• The overall reaction order cannot be determined
• The reaction is bimolecular since it leads to the formation of two products
• True
• True
• False
• False
• False
##### Explanation
• If the reaction is an elementary (single-step) reaction, the the rate law follows from the molecularity
• All rate laws can be determined experimentally
• The rate law for the reaction is k[H][I2]. It is an overall second-order reaction since it is bimolecular in nature.
• The overall reaction order can be determined, since it is an elementary reaction. The basis of molecularity to determine the rate law comes from the collision model.
• The reaction is bimolecular since it involves the reaction of two species as reactants
2. #### Rate Laws for Multistep Mechanisms

Determine whether each statement, regarding rate laws of multistep mechanisms, is true or false.
• A multistep reaction which proceeds via two elementary steps will always have a total reaction order of two
• The slowest elementary step will limit the overall chemical reaction rate
• The rate-determining step governs the rate law for the overall chemical reaction
• Whenever a slow step preceeds a fast step, the fast step has no influence on the overall reaction rate
• Whenenver a fast step preceeds a slow step, the concentration of an intermediate can be determined by assuming all reactant is converted to product in the slow step
• The fastest elementary step in a multistep reaction is called the rate-determing step
• False
• True
• True
• True
• False
• False
##### Explanation
• The reaction order will depend on the molecularity of the rate-determining (slow) step.
• The slowest step will determine the rate law for the reaction
• The slowest step is called the rate-determining step, and determines the rate law for the overall reaction
• Since the slow step determines the rate law, and following steps, which are faster, have no influence on the rate law
• The concentration of the intermediate can be determined by assuming any preceeding reactions are in equilibrium
• The slowest elementary step will be the rate-determining step in the reaction
3. #### Relative rates of Elementary Steps

Exchange of an oxygen atom between SO3 and CO is believed to proceed via two elementary steps:

(1) 2SO3(g) -> SO4(g) + SO2(g)
(2) SO4(g) + CO(g) -> CO2(g) + SO3(g)

Complete the following:
• Write the balanced chemical equation for the overall reaction.
• Give the rate law for each elementary step.
• The experimental rate law is Rate = k[SO3]2. What can be said about the relative rates of the elementary steps (1) and (2)?
• SO3 + CO -> CO2 + SO2
• Rate(1) = k1[SO3]2
• Rate(2) = k2{k1/k-1}[CO]{[SO3]2/[SO2]}
• Rate (1) slower than Rate (2)
##### Explanation
The overall reaction is obtained by summing the first and second elementary reactions:

SO3 + CO -> CO2 + SO2

The rate laws for each step are based on the molecularity, since they are both elementary steps:

Step (1) is bimolecular, Rate (1) = k1[SO3]2
Step (2) is bimolecular, Rate (2) = k2[SO4][CO]

but SO4 is an intermediate, whose concentration is unknown, but can be determined if we assume Step (1) to be in equilibrium:

Rate (1) forward = Rate (1) reverse
k1[SO3]2 = k-1[SO4][SO2]
[SO4] = {k1/k-1}{[SO3]2/[SO2]}
therefore Rate(2) = k2{k1/k-1}[CO]{[SO3]2/[SO2]}

The substitution for intermediate concentrations is performed according to the equilibrium method. Since the experimental rate law matches that of the first step, the first step is the rate determining step. Therefore, it is the slowest step in the 2 step mechanism.
4. #### Catalysis using Cobalt Chloride

The following question relates to the catalysis of the hydrogen peroxide and potassium sodium tartrate reaction using cobalt chloride:

5H2O2(aq) + KNaC4H4O6(aq) -> CO2(g) + NaOH(aq) + KOH(aq) + 6H2O(l)

Determine whether each statement is true or false.
• The catalyst does not undergo a chemical change during the reaction.
• Adding cobalt chloride to the reaction increases the amount of heat released per unit mole of reactant consumed.
• The activation energy of the reaction is unaltered with addition of cobalt chloride.
• Cobalt chloride will increase the rate of the chemical reaction.
• The mechanism for the reactions is not changed upon addition of cobalt chloride.
• Cobalt chloride is not consumed in the reaction.
• False
• False
• False
• True
• False
• True
##### Explanation
• The catalyst will undergo a chemical change during the reaction, however, at the completion of the reaction it will go back to its original form. It does not undergo a net chemical change
• The enthalpy of the reaction is not changed by addition of a catalyst. The relative energies of the reactants and products remains the same
• As a general rule, the effect of a catalyst is to lower the overall activation energy for a chemical reaction
• Cobalt cloride will increase the rate of reaction by finding an alternate reaction pathway with a lower activation barrier
• Since the reaction pathway is altered, effectively a new reaction mechanism is considered when the catalyst is present
• A catalyst undergoes no net chemical change during the reaction
5. #### Catalyst Affect on reaction Rate

The activation energy of an uncatalyzed reaction is 95 kJ/mol. The addition of a catalyst lowers the activation energy to 55 kJ/mol. Assuming that the collisional factor remains the same, by what factor will the catalyst increase the rate of reaction at 125 oC
Rate will increase by factor of 1.8 x 105
##### Explanation
The Arrhenius equation is given as:

k = Ae-(Ea/RT)

for T = (125 +273.15) = 398 K, R=8.314 J/mol-K and assuming that A is constant:

k(cat.) = Ae-[55000J/({8.314 J/mol-K}{398 K})]
k(uncat.) = Ae-[95000J/({8.314 J/mol-K}{398 K})]

The ratio of the two relations will give the Rate increase.

k(cat.)/k(uncat.) = e-16.62/e-28.7
k(cat.)/k(uncat.) = 1.78 x 105

## Topic: Previous Exams (Spring 2010)

1. #### Partial Pressure

The partial pressure of ethanol over an ethanol/water solution is shown below as a function of the mole fraction of ethanol. The vapor pressure of pure ethanol is 350 torr and the vapor pressure of pure water is 100 torr.
1. What is the partial pressure of ethanol over the solution when the mole fraction of ethanol is 0.2?
2. What is the total pressure over the solution when the mole fraction of ethanol in the solution is 0.2.?
1. 70 torr
2. 150 torr
##### Explanation
1. Pethanol = Xethanol * Pethanolo
Pethanol = 0.2 * 350 torr = 70 torr
2. PH2O = XH2O * PH2Oo
PH2O = 0.8 * 100 torr = 80 torr
Ptotal = 70 torr + 80 torr = 150 torr
2. #### Colligative properties

The molal-freezing-point depression constant is 1.86 oC/m for water. A 1 L solution of NaCl and water has a freezing point of -2.12 oC. What mass of NaCl is dissolved in the solution? You may assume that the density of the solution is 1g/cm3
32.3 g
##### Explanation
ΔTf = Kf * m
m = ΔTf / Kf = 2.12 oC / (2 * 1.86 oC/m) = 0.57 m
m = (mass / MW) / kg solvent
0.57 m = (x g)(mol / 58.5 g) / (1 - x/1000)
0.57 m * (1 - x/1000) = x / 58.5 g
33.53 - 0.033x = x
x = 32.3 g
3. #### Reaction Mechanism

The following three elementary steps have been proposed for the gas phase reaction of chloroform (CH3Cl) and chlorine:

Cl2 (g) → 2 Cl (g)
Cl (g) + CHCl3 → HCl (g) + CCl3 (g)
Cl (g) + CCl3 (g) → CCl (g)

1. What are the intermediates in these reactions?
2. If the second step is rate limiting what is the rate law of the reaction?
1. Cl , CCl3
2. Rate = k[Cl2]1/2[CHCl3]
##### Explanation
1. Intermediates are those chemicals that are produced in one step and consumed in another
2. If the second step is rate limiting the rate law can be expressed as
Rate = k2[Cl][CHCl3]
An intermediate chemical can not enter into a rate expression, use step one to determine an expression for the Cl concentration
kf[Cl2] = kr[Cl]2
[Cl] = kf/kr*[Cl2]1/2
Rate = k2*kf/kr * [Cl2]1/2[CHCl3] = k[Cl2]1/2[CHCl3]
4. #### Reaction Order

For the following reaction,

SO2Cl2 (g) → SO2 (g) + Cl2 (g)

the pressure (in atm) of SO2Cl2 is plotted as a function of time on the graph below. The equation that describes the pressure of SO2Cl2 as a function of time is shown on the graph.
1. What is the order of the reaction with respect to SO2Cl2?
2. What is the value of the rate constant and what is the half-life?
3. What is the initial concentration?
4. If the initial concentration is doubled what is the resulting half-life?
1. 1st order
2. k = 2.2 * 10-5 s-1, t1/2 = 31506 s
3. 1 atm
4. 31506 s
##### Explanation
1. a plot of ln[A] versus time is linear, therefore first order
2. slope of line is the negative of rate constant, half-life = ln 2 / k
3. from intercept of line, ln[A]o = 8.87*10-5
4. half-life of a first-order reaction is independent of initial concentratin
5. #### Osmotic pressure

A large enzyme is dissolved in water. A solution containing 0.15 g of this enzyme in 210 mL of solution has an osmotic pressure of 0.953 torr at 25 oC. What is the molar mass of this enzyme?
13929 g
##### Explanation
π=MRT
M = π/RT = (0.953 torr)(1 atm / 760 torr) / (0.08206 Latm/Kmol * 298 K) = 5.13 * 10-5 mol/L
0.15 g * (mol / x g) = 5.13 * 10-5 mol/L * 0.210 L
x = 13929 g
6. #### Reaction rates

Consider the following statements related the factors that affect chemical reaction rates. Please complete the following statements with either; not change, increase, decrease
1. If the reactant volume is decreased, without changing reactant concentration, the rate of the reaction will __________
2. When the temperature of a reaction is decreased the reaction rate will usually ___________
3. If a solid reactant is divided into smaller particles the reaction rate will ___________
4. If the reactant concentration is decreased the reaction rate will usually ____________
5. If a catalyst is added to the reaction, the reaction rate will typically ____________
1. not change
2. decrease
3. increase
4. decrease
5. increase
##### Explanation
1. reaction rate depends on concentrations not on quantity
2. the Arrhenius equation provides an expression for the temperature dependence of the reaction rate
3. increasing the surface area of a solid reactant will increase the reaction rate
4. As long as the reaction order is greater than zero a decrease in concentration will decrease the reaction rate.
5. a catalyst lowers the activation energy of a reaction thus increasing the rate
7. #### Arrhenius equation

Consider the following statements related to the Arrhenius equation and determine if they are true or false.
1. All chemical reactions occur regardless of the orientation of the individual reactant molecules.
2. The activation energy is the energy released in a chemical reaction.
3. After the addition of a catalyst, the reaction rate decreases because the activation energy has also increased.
4. The reaction rate decreases with increasing temperature.
1. False
2. False
3. False
4. False
##### Explanation
1. molecules must be oriented correctly in order to react
2. the activation energy is the energy that must be overcome for a reaction to proceed
3. a catalyst decreases the activation energy and increases the reaction rate
4. increasing the temperature increases the reaction rate
8. #### Reaction Coordinate

The reaction coordinate diagram is shown below for the conversion of A into D. The overall reaction occurs in a series of three elementary steps.
1. How many transitions states are shown on the diagram?
2. How many intermediates are shown on the diagram?
3. If the overall reaction proceeds from left to right, which elementary step is the fastest?
4. If the overall reaction proceeds from left to right, which elementary step is the slowest?
5. On the diagram, indicate the activation energy of the slowest step.
6. Is the overall reaction endothermic or exothermic?
1. 3
2. 2
3. C -> D
4. B -> C
5. exothermic
9. #### Phase Diagram

The phase diagram of pure water is shown below.
1. How many triple points are located on the phase diagram?
2. Label the triple point, critical point, normal boiling point and normal freezing point of pure water on the phase diagram.
3. If some NaCl is dissolved in the water, what happens to the vapor pressure of the resulting solution.
4. What happens to the normal boiling point and normal freezing point for the NaCl solution?
1. 1
2. the vapor pressure of the solution is reduced
3. the normal boiling point is increased and the freezing point is lowered.

## Topic: Previous Exams (Spring 2011)

1. #### Reaction Mechanism

The destruction of ozone in the presence of NO is described by the following balanced chemical reaction

O3 (g) + O (g) + NO (g) → 2 O2 (g) + NO (g)

This reaction has been proposed to occur through two elementary reaction steps shown below:

Step1: O3 (g) + NO (g) → NO2 (g) + O2 (g)
Step2: NO2 (g) + O (g) → NO (g) + O2 (g)

1. What is the intermediate?
2. What is the catalyst?
3. What is the rate law if the first step is rate limiting step?
4. What is the rate law if the second step is the rate limiting step and you can assume that the first step is at equilibrium?
1. NO2
2. NO
3. Rate = k[O3][NO]
4. Rate = k[O3][NO][O]/[O2
##### Explanation
1. An intermediate is a species that is created in one step and consumed in a second step
2. A catalysts does not undergo a permanent chemical change
3. If the first step is slow the overall reaction rate can be determined from the rate of the first step.
4. If the second step is slow assume the first step is in equilibrium:
Ratef = Rater = kf[O3][NO]=Rater[NO2][O2]
[NO2] = (kf/kr)*[O3][NO]/[O2]
and use it to solve for the concentration of the intermediate
Rate = k2[NO2][O]
Rate = (k2kf/kr)*[NO2][O][O3][NO]/[O2]
Rate = k*[NO2][O][O3][NO]/[O2]
2. #### Boiling points

Label the following solutions in order of increasing boiling points. Use the numbers one through five with 1 to label the solution with the lowest boiling point and 5 to label the solution with the highest boiling point. You may or may not need Kf,H2O = 1.86 oC/m and Kb,H2O = 0.51 oC/ma.
1. 0.1 mol of glucose, C6H12O6, dissolved in 1 kg of water
2. 0.1 mol of magnesium chloride, MgCl2, dissolved in 1 kg of water
3. 0.2 mol of calcium chloride, CaCl2, dissolved in 1 kg of water
4. 0.1 mol of sodium chloride, NaCl, dissolved in 1 kg of water
5. 0.5 mol of ethanol, C2H6O, dissolved in 1 kg of water
Lowest Boiling Point a, d, b, e, c highest boiling point
##### Explanation
The boiling point elevation is defined by ΔTb=Kbm*(van't Hoff). The van't Hoff factors for the five solutions are a - 1, b - 3, c - 3, d - 2, e - 1.
3. #### Collisional Model

Consider the following statements concerning the collisional model. In the blanks please write either increase, decrease, or not change.
1. The reaction rate of a first order reaction will _______________ if the concentration is decreased.
2. The reaction rate will ________________¬¬¬¬ ¬¬¬¬¬¬¬¬¬¬¬ if the temperature is increased.
3. The reaction rate will ________________ if the activation energy is decreased.
4. If a solid reactant is divided into smaller particles the reaction rate will ______________.
5. If the reactant volume is increased without changing reactant concentration the rate of the reaction will ________________.
1. decrease
2. increase
3. increase
4. increase
5. not change
##### Explanation
1. Based on the rate law Rate = k[A]
2. Based on Arrhenius's Law k = A*e-Ea/RT
3. Based on Arrhenius's Law k = A*e-Ea/RT
4. Increasing the surface area of a solid reactant will increase the reaction rate
5. The reaction rate depends on concentration
4. #### Henry's Law

A Coke can is pressurized with 2 atm of CO2. Calculate the concentration of dissolved CO2 in the carbonated soft drink? The Henry’s Law constant for CO2 in water at this temperature is 0.031 mol / L atm.
0.062 mol/L
##### Explanation
Sg = kPg
Sg = (0.031 mol/Latm)(2 atm) = 0.062 mol/L
5. #### Colligative properties

Determine whether each of the following statements is true or false. Write the entire word TRUE or FALSE or else no credit will be given:
1. Colligative properties depend on the solute dissolved but not the quantity.
2. The flow of solvent between two solutions with different osmotic pressure through a rigid semi-permeable membrane is from the solution with a high concentration to the solution with the low concentration.
3. Raoult’s Law describes the reduction in vapor pressure caused by dissolving a solute within a solvent
4. In a mixture of two volatile liquids the vapor over the liquid is enriched in the more volatile of the two liquids.
5. A water solution with a high osmotic pressure will boil at a lower temperature than a water solution with a lower osmotic pressure.
1. False
2. False
3. True
4. True
5. False
##### Explanation
1. Colligative properties depend on the amount of solute but not its identity
2. The flow of solvent will be from the solution with low solute concentration to the solution with high solute concentration
3. Raoult's Law is PA = XAPAo
4. From Raoult's Law the composition of the vapor over a solution of two volatile components will be enriched in the more volatile component.
5. A high osmotic pressure indicates a high solute concentration and thus an elevated boiling point.
6. #### Arrhenius Equation

Answer the following questions refering to the picture above which shows measured reaction rate constants for the decomposition of N2O5 as a function of temperature with and without a catalyst.
1. Which reaction, 1 or 2, was carried out with a catalyst?
2. What is the activation energy of reaction 1?
3. What is the frequency constant for reaction 2?
1. 2
2. 500 kJ/mol
3. 3.0*10-17
##### Explanation
1. At a given temperature the reaction rate is higher for reaction 2. Or, the activation energy is lower for reaction 2 based on the gentler slope of the line.
2. Slope = -Ea/R
Ea = -slope * R = 60136 K * 8.3145 J/Kmol = 500000 J/mol -> 500 kJ/mol
3. ln A = -38.045
A = e-38.045 = 3.0*10-17
7. #### Reaction order

The decomposition of C2H5Cl:

C2H5Cl (g) → C2H4 (g) + HCl (g)

was carried out at two different temperatures. Answer the following questions based on the graph above which shows the concentration as a function of time for the reaction.
1. What is the reaction order?
2. What is the rate constant of reaction 1?
3. What is the half-life of reaction 1?
4. What is the initial concentration of reaction 1?
5. Which reaction was run at the higher temperature?
1. 1
2. k = 0.0462 s-1
3. 15 s
4. 0.1 M
5. 1
##### Explanation
1. A plot of ln[A] as a function of time is linear for a first order reaction
2. slope = -k
3. t1/2=ln 2 / k = 0.693 / 0.0462 s-1 = 15 s
4. ln[A]0 = -2.3026
[A]0 = e-2.3026 = 0.1 M
5. At a higher temperature the reaction proceeds faster so the drop in the concentration of A as a function of time is steeper
8. #### Reaction order and rate

The initial rate of the reaction for the decomposition of the gas NOBr is monitored in four different experiments.

2 NOBr (g) → 2 NO (g) + Br2 (g)

 Experiment number Initial concentration (M) Initial rate (M/s-1) 1 0.1 0.005 2 .2 .02 3 0.3 0.045
Based on the data presented below answer the following questions:
1. What is the reaction order?
2. What is the rate constant?
3. What is the half-life for experiment 3?
1. 2
2. 0.5 mol-1s-1
3. 6.6 s
##### Explanation
1. Rate2 = 0.02 = k[2NOBr]n
Rate1 = 0.005 = k[NOBr]n
4 = 2n n = 2
2. Rate = k[NOBr]2
k = Rate / [NOBr]2 = 0.005 M/s / (0.1 M)2 = 0.5 M-1s-1
3. t1/2 = 1/(k*[A]o) = 1 / (0.5 M-1s-1 * 0.3 M) = 6.6 s
9. #### Reaction coordinate

The production of HI given below

H2 (g) + I2 (g) → 2 HI (g)

is proposed to occur through two different elementary steps.

Step 1: I2 (g) → 2 I (g) (H = 0, Ea = 50 kJ/mol)
Step2: H2 (g) + 2 I (g) → 2 HI (g) (H = -100 kJ, Ea = 100 kJ/mol)

On the diagram given below, sketch the qualitative reaction coordinate diagram for the reaction. Label the location of the intermediates, transition states, activation energies for each step, and indicate which step is rate limiting.