Chemistry resources

Hydrophobic vs. Hydrophilic
Learning Goal: To understand the basic chemical properties that contribute to hydrophilicity and predict the hydrophilicity of different molecules based on these properties.
Chemists generally refer to solvents as hydrophilic (water-loving) or hydrophobic (water-hating) based on their ability to dissolve in water. The following factors favor hydrophilicity. They are listed in order of importance.
1) The presence of ions.
2) The presence of a dipole moment.
3) The presence of London dispersion forces.

The presence of ions
For example a salt such as magnesium sulfate, \rm MgSO_4, is very soluble in water because the water molecules will solvate the charged magnesium atom and the sulfate molecule very easily.

The presence of a dipole moment
A polar molecule will interact with water molecules which also have a dipole moment. Dipole moments are caused by the close proximity of electronegative atoms such as oxygen or sulfur to less electronegative atoms such as hydrogen or carbon. Acetone, \rm H_3C(CO)CH_3 and methanol, \rm HOCH_3, are readily soluble in water due to their dipole moments from the presence of the electronegative oxygen atom. On the other hand, a molecule like butane, \rm CH_3CH_2CH_2CH_3, which has no dipole moment, is hydrophobic and hence insoluble in water.

The presence of London dispersion forces
Substances that lack ions or a dipole moment are generally hydrophobic. However, when comparing two nonpolar gases, the one with the greater London forces will generally be the more hydrophilic of the two. In general, London forces become larger as the size of the atom or molecule increases. So \rm O_2 (MW=32) is more soluble in water than \rm N_2 (MW = 28).

Part A
Rank the following molecules in order from most soluble in water to least soluble in water.
methane: \rm CH_{4}, hexanol: \rm C_6H_{13}OH, table salt: \rm NaCl, propane: \rm C_{3}H_{8}
Hint A.1
Which molecule is the most hydrophilic?

Hint not displayed
Hint A.2
Which molecule is the least hydrophilic?

Hint not displayed
Rank from most to least soluble in water. To rank items as equivalent, overlap them.
ANSWER:


View

mass % to molarity
A 28.00%-by-mass solution of nitric acid, {\rm HNO}_3, in water has a density of 1.20 g/cm^3 at 20^\circ \rm C.
Part A
What is the molarity of {\rm HNO}_3 in this solution?
ANSWER:
[{\rm HNO}_3] = 5.33 \rm M

Molality
Part A
How many grams of water would you add to 1.00 \rm kg of 1.38 m {\rm CH}_3{\rm OH}(aq) to reduce the molality to 1.00 \rm m {\rm CH}_3{\rm OH}?
Express your answer using two significant figures.
ANSWER:
m = 360 \rm g

Recrystallization
One way to recrystallize a solute from a solution is to change the temperature. Another way is to evaporate solvent from the solution. A 330 -\,g sample of a saturated solution of {\rm KNO}_3(s) in water is prepared at 25.0^\circ \rm C.
Part A
If 50 g {\rm H}_2{\rm O} is evaporated from the solution at the same time as the temperature is reduced from 25.0 to 0.0^\circ \rm C, what mass of {\rm KNO}_3(s) will recrystallize? (Refer to Figure 13-8 in the textbook.)
Express your answer using two significant figures.
ANSWER:
m = 63 \rm g

Henry's Law
The solubility of a gas in a liquid increases with increasing pressure.

To understand the above statement, consider a familiar example: cola. In cola and other soft drinks, carbon dioxide gas remains dissolved in solution as long as the can or bottle remains pressurized. As soon as the lid is opened and pressure is released, the \rm CO_2 gas is much less soluble and escapes into the air.

The relationship between pressure and the solubility of a gas is expressed by Henry's law: C = kP, where C is concentration in M, k is the Henry's law constant in units of \rm mol/L\cdot atm, and P is the pressure in \rm atm.

Note: Since temperature also affects the solubility of a gas in an liquid, the Henry's law constant is specific to a partcular gas at a particular temperature.
The following table provides some information on carbon dioxide solubility in water.
C
(\rm mol/L) P
(\rm atm) k
(\rm mol/L\cdot atm) T
(\rm ^\circ C)
3.70×10−2 1.00 20.0
7.60×10−2 20.0
1.00 3.40×10−2 25.0
Part A
What is the Henry's law constant for \rm CO_2 at 20\rm ^{\circ}{C}?
Hint A.1
How to approach the problem

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ANSWER:
0.037 \rm mol/L\cdot atm
Part B
What pressure is required to achieve a \rm CO_2 concentration of 7.60×10−2 at 20\rm {}^{\circ}{C}?
Hint B.1
How to approach the problem

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ANSWER:
2.05 \rm atm
Part C
At 1 \rm atm, how many moles of \rm CO_2 are released by raising the temperature of 1 liter of water from 20\rm ^\circ{C} to 25\rm ^\circ{C}?
Hint C.1
Find the concentration

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Hint C.2
Determine the difference in concentration

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ANSWER:
0.003 \rm mol

Raoult's Law
Part A
Calculate \chi_{{\rm C}_6{\rm H}_6} in a benzene–toluene liquid solution that is in equilibrium at 25^\circ \rm C with a vapor phase that contains 62.0 mol\;\% {\rm C}_6{\rm H}_6. (At 25^\circ \rm C the vapor pressure of {\rm C}_6{\rm H}_6\;=\;95.1\;\rm mmHg, the vapor pressure of {\rm C}_7{\rm H}_8\;=\;28.4\;\rm mmHg.)
ANSWER:
\chi_{{\rm C}_6{\rm H}_6} = 0.328

Osmotic Pressure
The molecular mass of hemoglobin is 6.86\times10^4\rm u.
Part A
What mass of hemoglobin must be present per 100.0 \rm mL of a solution to exert an osmotic pressure of 7.10 mmHg at 25^\circ \rm C?
ANSWER:
m = 2.62 \rm g

Boiling Point Elevation and Freezing Point Depression for Organic Solutions
The temperature at which a solution freezes and boils depends on the freezing and boiling points of the pure solvent as well as on the molal concentration of particles (molecules and ions) in the solution. For nonvolatile solutes, the boiling point of the solution is higher than that of the pure solvent and the freezing point is lower. The change in the boiling for a solution, DeltaT_b, can be calculated as

\Delta T_{\rm b}=K_{\rm b}\cdot m
in which m is the molality of the solution and K_b is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, DeltaT_f, can be calculated in a similar manner:

\Delta T_{\rm f}=K_{\rm f}\cdot m

in which m is the molality of the solution and K_f is the molal freezing-point-depression constant for the solvent.
Part A
Cyclohexane has a freezing point of 6.50 ^\circ \rm C and a K_f of 20.0 ^\circ {\rm C}/m. What is the freezing point of a solution made by dissolving 0.540 g of biphenyl (\rm C_{12}H_{10}) in 25.0 g of cyclohexane?
Hint A.1
How to approach the problem

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Hint A.2
Determine the number of moles of solute

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Hint A.3
Determine the solvent mass in kilograms

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Hint A.4
Determine the molal concentration of the solution

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Hint A.5
Determine the freezing-point depression of the solution

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Express the temperature numerically in degrees Celsius.
ANSWER:
3.70 ^\circ \rm C
Part B
Paradichlorobenzene, \rm C_6H_4Cl_2, is a component of mothballs. A solution of 2.00 \rm g in 22.5 \rm g of cyclohexane boils at 82.39 ^\circ \rm C. The boiling point of pure cyclohexane is 80.70 ^\circ \rm C. Calculate K_b for cyclohexane.
Hint B.1
How to approach the problem

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Hint B.2
Determine the boiling-point elevation of the solution

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Hint B.3
Determine the molality of the solution

Hint not displayed
Express the constant numerically in degrees Celsius per molal.
ANSWER:
K_b = 2.80 ^\circ {\rm C}/m

van't Hoff Factors
Calculate the van't Hoff factors of the following weak electrolyte solutions.
Part A
4.8×10−2 m {\rm HCHO}_2 which begins to freeze at −9.84×10−2^\circ C
Express your answer using two significant figures.
ANSWER:
i = 1.1
Part B
0.120 M {\rm HNO}_2 which has a hydrogen ion (and nitrite ion) concentration of 6.89×10−3 M
ANSWER:
i = 1.06

Rates of Reaction
At 65^\circ \rm C, the half-life for the first-order decomposition of {\rm N}_2{\rm O}_5 is 2.38 \rm min.
{\rm N}_2{\rm O}_5\,(g)\;\to\;2\,{\rm NO}_2\,(g)\;+\;\frac{1}{2}\,{\rm O}_2\,(g)
1.50 g of {\rm N}_2{\rm O}_5 is introduced into an evacuated 14 -\,L flask at 65^\circ \rm C.
Part A
What is the initial partial pressure, in \rm mmHg, of {\rm N}_2{\rm O}_5(g)?
Express your answer using two significant figures.
ANSWER:
P =

Answer not displayed
\rm mmHg
Part B
What is the partial pressure, in \rm mmHg, of {\rm N}_2{\rm O}_5(g) after 2.38 \rm min?
Express your answer using two significant figures.
ANSWER:
P =

Answer not displayed
\rm mmHg
Part C
What is the total gas pressure, in \rm mmHg, after 2.38 \rm min?
Express your answer using two significant figures.
ANSWER:
P =

Answer not displayed
\rm mmHg

Reaction Order
For the reaction \rm A + B + C \rightarrow D + E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:
Trial [\rm A]
(M) [\rm B]
(M) [\rm C]
(M) Initial Rate
(M/{\rm s})
1 0.10 0.10 0.10 3.0 \times 10^{-5}
2 0.10 0.10 0.30 9.0 \times 10^{-5}
3 0.20 0.10 0.10 1.2 \times 10^{-4}
4 0.20 0.20 0.10 1.2 \times 10^{-4}
Part A
What is the reaction order with respect to \rm A?
Hint A.1
Compare trial 1 with trial 3

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Hint A.2
How to set up the math

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ANSWER:


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Part B
What is the reaction order with respect to \rm B?
Hint B.1
Compare trial 3 with trial 4

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Hint B.2
How to set up the math

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ANSWER:


Answer not displayed

Part C
What is the reaction order with respect to \rm C?
Hint C.1
Compare trial 1 with trial 2

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Hint C.2
How to set up the math

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ANSWER:


Answer not displayed

Part D

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Kinetics and Graphs
Learning Goal: To learn how graphs can be used to answer kinetics questions.
A student collected time (t) and concentration (\rm [A]) data at 295 \rm K for the reaction \rm 2A\rightarrow B. These time and concentration data are shown in the table to the right. The student then plotted graphs of \rm [A] versus t , \rm \ln{[A]} versus t , and \rm 1/[A] versus t .

t
(\rm s) \rm [A]
(M) \rm \ln{[A]} \rm 1/[A]
0.00 0.500 -0.693 2.00
20.0 0.389 -0.944 2.57
40.0 0.303 -1.19 3.30
60.0 0.236 -1.44 4.24
80.0 0.184 -1.69 5.43
Part A
What is the order of this reaction?
Hint A.1
How to approach the problem

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Hint A.2
Which graph shows a straight line?

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ANSWER:

0
1
2


Answer not displayed
Part B

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Transition will be visible after you complete previous item(s).
Part C

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Introduction to Integrated Rate Laws
Learning Goal: To understand how to use integrated rate laws to solve for concentration.
A car starts at mile marker 145 on a highway and drives at 55 \rm mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours?

This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145.

\rm 55~{mi /hr} \times 2~hr = 110~miles~traveled

\rm milemarker~{145} - {110}~miles = milemarker~{35}

If we were to write a formula for this calculation, we might express it as follows:

\rm milemarker = milemarker_0 - (speed \times time)

where \rm milemarker is the current milemarker and \rm milemarker_0 is the initial milemarker.

Similarly, the integrated rate law for a zero-order reaction is expressed as follows:

\rm [A] = [A]_0 - rate\times time

or

[{\rm A}] = [{\rm A}]_0 - kt

since

{\rm rate} = k [{\rm A}]_0 = k
A zero-order reaction proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, first-order reaction rates do change over time as the reactant concentration changes.

Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction.

The integrated rate law for a first-order reaction is expressed as follows:

[{\rm A}] = [{\rm A}]_0e^{-kt}

where k is the rate constant for this reaction.

The integrated rate law for a second-order reaction is expressed as follows:

{1 \over [{\rm A}]} = kt + {1 \over [{\rm A_0}]}

where k is the rate constant for this reaction.
Part A
The rate constant for a certain reaction is k = 6.00×10−3 s^{-1}. If the initial reactant concentration was 0.950 M, what will the concentration be after 10.0 minutes?
Hint A.1
How to approach the problem

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Hint A.2
Determine the reaction order

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ANSWER:


Answer not displayed
M
Part B
A zero-order reaction has a constant rate of 2.60×10−4 M/\rm s. If after 80.0 seconds the concentration has dropped to 6.50×10−2 M, what was the initial concentration?
Hint B.1
How to approach the problem

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Hint B.2
What is the rate constant?

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ANSWER:


Answer not displayed
M

Integrated Rate Laws
The integrated rate law is an equation that describes the concentration of a reactant over time (t). The integrated rate law comes in different forms depending on the reaction order. The rate laws below are given for a simple general reaction

\rm A\rightarrow\rm products

Order Integrated rate law Linear graph Slope
0 \rm [A]=-{\it kt}+[A]_0 \rm [A]\;vs.\; \it t -k
1 \rm \ln [A]=-{\it kt}+ln[A]_0 \rm \ln [A]\; vs.\; \it t -k
2 \rm \frac{1}{[A]}={\it kt}+\frac{1}{[A]_0} \rm \frac{1}{[A]}\; vs.\; \it t k
Part A
In a study of the decomposition of hydrogen iodide, \rm HI, via the reaction

\rm 2HI{(g)} \rightleftharpoons H_2{(g)}+I_2{(g)}

the following concentration-time data were collected:

Time \rm (min) \rm [HI]\;(\it M)
0 0.467
1 0.267
2 0.187
3 0.144
4 0.117
5 0.099
6 0.085
7 0.075

What is the order of the reaction?
Hint A.1
How to approach the problem

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Hint A.2
Determine if the reaction is zeroth order

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Hint A.3
Determine if the reaction is first order

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Hint A.4
Determine if the reaction is second order

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ANSWER:
The reaction is
zeroth
first
second
order.

Answer not displayed
Part B

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Ammonia on a hot tungsten wire
Ammonia decomposes on the surface of a hot tungsten wire. Following are the half-lives that were obtained at 1100^\circ \rm C for different initial concentrations of {\rm NH}_3: [{\rm NH}_3]_0 = 0.0031 \rm M, t_{1/2} = 7.6 \rm min; 0.0015 \rm M, 3.7 \rm min; 0.00068 \rm M, 1.7 \rm min.
Part A
For this decomposition reaction, what is the order of the reaction?
ANSWER:

zero order
first order
second order


Answer not displayed
Part B

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Mechanisms and Molecularity
Learning Goal: To understand how elementary steps make up a mechanism and how the rate law for an elementary step can be determined.

Very often, a reaction does not tell us the whole story. For instance, the reaction

\rm NO_2 + CO \rightarrow NO + CO_2

does not involve a collision between an \rm NO_2 molecule and a \rm CO molecule. In reality, this reaction occurs in two steps:

1. \rm 2NO_2 \rightarrow N_2O_4
2. \rm N_2O_4 + CO \rightarrow NO + CO_2 + NO_2

Each individual step is called an elementary step. Together, these elementary steps are called the reaction mechanism.

Overall, the resulting reaction is

\rm NO_2 + CO \rightarrow NO + CO_2

Notice that in the elementary steps \rm N_2O_4 appears both as a product and then as a reactant; therefore it cancels out of the final chemical equation. \rm N_2O_4 is called a reaction intermediate.

Molecularity is the proper term for "how the molecules collide" in a reaction. For example, step 1 is bimolecular because it involves the collision of two molecules. Step 2 is also bimolecular for the same reason. Unimolecular reactions involve only one molecule in the reactants. Though rare, collisions among three molecules can occur. Such a reaction would be called termolecular.
Order and rate law of a reaction

The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 is

{\rm rate} = k\rm[NO_2]^2

The exponent "2" is used because the reaction involves two \rm NO_2 molecules. The rate law for step 2 is

{\rm rate} = k{\rm[N_2O_4]^1[CO]^1} = k\rm [N_2O_4][CO]

because the reaction involves only one molecule of each reactant.
Analyzing a new reaction

Consider the following elementary steps that make up the mechanism of a certain reaction:

1. \rm 2A \rightarrow B + C
2. \rm B + D \rightarrow E + C

Part A
What is the overall reaction?
Hint A.1
How to approach the problem

Hint not displayed
Express your answer as a chemical equation.
ANSWER:


Answer not displayed

Part B
Which species is a reaction intermediate?
Hint B.1
How to approach the problem

Hint not displayed
ANSWER:

A
B
C
D
E


Answer not displayed
Part C
What is the rate law for step 1 of this reaction?
Hint C.1
How to approach the problem

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Hint C.2
Determine the molecularity

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Hint C.3
Find the order with respect to A

Hint not displayed
Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[{\rm A}][{\rm C}]^3 type k*[A]*[C]^3.
ANSWER:


Answer not displayed

Part D
What is the rate law for step 2 of this reaction?
Hint D.1
How to approach the problem

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Hint D.2
Determine the molecularity

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Hint D.3
Find the order with respect to B

Hint not displayed
Hint D.4
Find the order with respect to D

Hint not displayed
Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[{\rm A}][{\rm C}]^3 type k*[A]*[C]^3.
ANSWER:


Answer not displayed


Reaction Mechanisms
A reaction mechanism is defined as the sequence of reaction steps that define the pathway from reactants to products. Each step in a mechanism is an elementary reaction, which describes a single molecular event of usually one or two molecules interacting.

The rate law for an overall reaction is the rate law for the slowest step in the mechanism, which is directly related to the stoichiometric coefficients of the reactants.

The exception to this rule occurs when the slowest step contains intermediates. In these cases, the slowest step is usually preceded by an equilibrium step, which can be used to substitute for the intermediates in the overall rate law.
Part A
Consider the reaction

\rm 2XY_2 + Z_2 \rightleftharpoons 2XY_2Z
which has a rate law of

\rm rate={\it k}[XY_2][Z_2]
Select a possible mechanism for the reaction.
Hint A.1
How to approach the problem

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Hint A.2
Determine whether the mechanisms add up to the overall reaction

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Hint A.3
Determine the rate law for each mechanism

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ANSWER:


A \matrix{\rm Step\, 1\!: & \hfill \rm Z_2 & \rightarrow & \rm Z+Z & \rm (slow) \cr \rm Step\, 2\!: & \rm XY_2+Z & \rightarrow & \rm XY_2Z \hfill & \rm (fast) \cr \rm Step\, 3\!: & \rm XY_2+Z & \rightarrow & \rm XY_2Z \hfill & \rm (fast)}

B \matrix{\rm Step\, 1\!: & \rm XY_2+Z_2 & \rightarrow & \rm XY_2Z+Z & \rm (slow) \cr \rm Step\, 2\!: & \rm XY_2+Z & \rightarrow & \rm XY_2Z \hfill & \rm (fast)}

C \matrix{\rm Step\, 1\!: & \hfill \rm XY_2+Z_2 & \rightarrow & \rm XY_2Z_2 & \rm (slow) \cr \rm Step\, 2\!: & \rm XY_2Z_2 & \rightarrow & \rm XY_2Z +Z & \rm (fast)}

D \matrix{\rm Step\, 1\!: & \hfill \rm 2XY_2 & \rightleftharpoons & \rm X_2Y_4 \hfill & \rm (fast) \cr \rm Step\, 2\!: & \rm X_2Y_4+Z_2 & \rightarrow & \rm 2XY_2Z & \rm (slow)}

E \matrix{\rm Step\, 1\!: & \hfill \rm 2XY_2+Z_2 & \rightarrow & \rm 2XY_2Z & \rm (slow)}


Answer not displayed
Part B
What is the rate law for the following mechanism?

\matrix{\rm Step\, 1\!: & \hfill \rm A+B & \rightleftharpoons & \rm C \hfill & \rm (fast) \cr \rm Step\, 2\!: & \rm B+C & \rightarrow & \rm D & \rm (slow)}
Hint B.1
How to approach the problem

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Hint B.2
Determine the rate law for the slow step

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Hint B.3
Identify a substitution for [C]

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Express your answer in terms of k and the necessary concentrations (e.g., k[A]^3[D]).
ANSWER:
rate =

Answer not displayed


Enzyme kinetics
The following substrate concentration [{\rm S}] versus time data were obtained during an enzyme-catalyzed reaction: t=0\;\rm min, [{\rm S}]=1.00\;\rm M; 20 \rm min, 0.90 \rm M; 60 \rm min, 0.70 \rm M; 100 \rm min, 0.50 \rm M; 160 \rm min, 0.20 \rm M.
Part A
What is the order of this reaction with respect to \rm S in the concentration range studied?
ANSWER:

zero order
first order
second order


Answer not displayed

Activation Energy and Catalysts
Learning Goal: To help you understand how to interpret potential energy diagrams.
The graphs shown here are called potential energy diagrams. They show the potential energy of a system as it changes from reactants to products. The first graph shows the uncatalyzed reaction ; the second graph shows the catalyzed reaction .
Activation energy

The activation energy of a reaction is the difference in energy between the reactants and the activated complex. Activation energy is always a positive number.
Internal energy change

The internal energy change, DeltaE, of a reaction is the difference in energy between the reactants and the products. DeltaE may be positive or negative depending upon whether the reaction is endothermic or exothermic. To ensure that you always get the correct sign for DeltaE, use the following equation:

\Delta E = E_{\rm products} - E_{\rm reactants}
Enthalpy change

The enthalpy change, DeltaH, of a reaction is equal to the internal energy change, DeltaE, for a reaction as long as the system is at constant pressure. This is very often the case and should be assumed for any problem unless you are told otherwise.
Part A
What is the value of the activation energy of the uncatalyzed reaction?
Hint A.1
How to approach the problem

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Hint A.2
Determine the energy of the reactants

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Hint A.3
Determine the energy of the activated complex

Hint not displayed
Express your answer in kilojoules.
ANSWER:
activation energy =

Answer not displayed
\rm kJ
Part B
What is the value of the enthalpy change of the uncatalyzed reaction?
Hint B.1
How to approach the problem

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Hint B.2
Determine the energy of the reactants

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Hint B.3
Determine the energy of the products

Hint not displayed
Express your answer in kilojoules.
ANSWER:
DeltaH =

Answer not displayed
\rm kJ
Part C
What is the value of the activation energy of the uncatalyzed reaction in reverse?
Hint C.1
How to approach the problem

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Hint C.2
Determine the starting energy

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Hint C.3
Determine the energy of the activated complex

Hint not displayed
Express your answer in kilojoules.
ANSWER:
activation energy =

Answer not displayed
\rm kJ
Part D
What is the value of the enthalpy change of the uncatalyzed reaction in reverse?
Hint D.1
How to approach the problem

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Hint D.2
Determine the starting energy

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Hint D.3
Determine the ending energy

Hint not displayed
Express your answer in kilojoules.
ANSWER:
DeltaH =

Answer not displayed
\rm kJ
Part E
How does the presence of a catalyst affect the activation energy of a reaction?
Hint E.1
How to approach the problem

Hint not displayed
ANSWER:

A catalyst increases the activation energy of a reaction.
A catalyst decreases the activation energy of a reaction.
A catalyst does not affect the activation energy of a reaction.
It depends on whether you are talking about the forward or the reverse reaction.


Answer not displayed
Part F
How does the presence of a catalyst affect the enthalpy change of a reaction?
Hint F.1
How to approach the problem

Hint not displayed
ANSWER:

A catalyst increases the enthalpy change of a reaction.
A catalyst decreases the enthalpy change of a reaction.
A catalyst does not affect the enthalpy change of a reaction.
It depends on whether you are talking about the forward or the reverse reaction.


Answer not displayed