Search found 50 matches

by 805303639
Thu Mar 12, 2020 8:31 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 7.11 HW help
Replies: 2
Views: 56

Re: 7.11 HW help

The reaction profile displayed in the solution manual represents a more likely scenario than the one you described. The third step will not be endothermic (products having more potential energy) because doing so necessitates that an intermediate (reactant in step 3) be more thermodynamically stable ...
by 805303639
Thu Mar 12, 2020 8:18 am
Forum: General Rate Laws
Topic: Different Types of Rate Laws
Replies: 2
Views: 43

Re: Different Types of Rate Laws

Differential rate laws express the instantaneous rate of a reaction as a function of reactant concentrations. Integrated rate laws express the concentration of reactant remaining as a function of time. You'll have 2k as a constant if you solve for the rate of consumption/formation of a particular sp...
by 805303639
Thu Mar 12, 2020 8:00 am
Forum: Method of Initial Rates (To Determine n and k)
Topic: 3/2 overall order
Replies: 3
Views: 88

Re: 3/2 overall order

A reaction with an overall order of 3/2 requires that its slow, rate-determining elementary step has fractional stoichiometric coefficients. The rate law for a 3/2 order reaction would have a rate constant with units (L/mol)^1/2 * s^-1.
by 805303639
Thu Mar 12, 2020 7:45 am
Forum: General Rate Laws
Topic: Difference between unique rate and rate law
Replies: 3
Views: 46

Re: Difference between unique rate and rate law

To find the unique rate of a reaction, you take the rate (of any product or reactant) and divide by the stoichiometric coefficient of that chosen species Usually you'll first be asked to find the rate law, in which case you'll often use the method of initial rates. Recognizing that this rate law gi...
by 805303639
Thu Mar 12, 2020 7:28 am
Forum: First Order Reactions
Topic: 7B.17b
Replies: 2
Views: 38

Re: 7B.17b

In the first step, the solutions manual finds the concentration of A when the concentration of B has increased to 0.19 mol/L. The next step where 0.055(molA)/L = 0.37[A]0 is just a mathematical formalism expressing [A] in terms of [A]0. As you can see from the subsequent calculation steps, this seco...
by 805303639
Wed Mar 04, 2020 5:04 pm
Forum: Second Order Reactions
Topic: [A] v. Time
Replies: 16
Views: 208

Re: [A] v. Time

The graph of 1/[A] v Time for a second-order reaction yields a line with slope=k and y-intercept=1/[A]i. Hence the graph of [A] v Time for a second-order reaction would not be linear. The graph would be a rational function with a vertical asymptote at t= -1/(k[A]i).
by 805303639
Wed Mar 04, 2020 4:43 pm
Forum: General Rate Laws
Topic: rate law definition
Replies: 4
Views: 46

Re: rate law definition

Rate laws can be differential and take the form -d[A]/(a dt) or k[A]^n for the reaction aA --> P. Differential rate laws give the rate of the reaction with respect to the concentrations of reactants. Differential rate laws give insight into the reaction profile. For example, the rate law R = k [A]^2...
by 805303639
Wed Mar 04, 2020 4:30 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Initial Rates summarized
Replies: 3
Views: 49

Re: Initial Rates summarized

Consider the equation Rate = k [A]^n [B]^m. For a table of experiments containing data on the initial [A], [B], and rate for each experiment, use the method of initial rates to determine n, m, and k. Determine m and n by setting up ratios of Rate = k [A]^n [B]^m for different experiments such that e...
by 805303639
Wed Mar 04, 2020 1:40 pm
Forum: Zero Order Reactions
Topic: order of reactions
Replies: 4
Views: 53

Re: order of reactions

Use the method of initial rates to determine the reaction order with respect to each of the reactants. Then sum those orders to obtain the overall reaction order. Alternatively, one can determine the overall reaction order from the plots of [A] vs time, ln[A] vs time, and 1/[A] vs time (the integrat...
by 805303639
Wed Mar 04, 2020 1:24 pm
Forum: First Order Reactions
Topic: Graph
Replies: 9
Views: 186

Re: Graph

If the reaction is first order then the graph would be linear with a slope of -K. The same goes for zero and second order reactions. Their graphs should be linear. If the reaction is first order, then the plot of ln[A] vs time is linear and the slope equals -k. Similarly, for a zero order reaction,...
by 805303639
Thu Feb 27, 2020 9:12 pm
Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
Topic: Metal dissolution
Replies: 10
Views: 399

Re: Metal dissolution

Is there a way of knowing if the metal will be oxidized? Do certain metals oxidize easier than others? Which metals are least likely to oxidize and dissolve into solution? Refer to the ordered electrochemical series. Metals with half-reactions that appear later in the table (more negative E^0) are ...
by 805303639
Thu Feb 27, 2020 8:57 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: cell potential
Replies: 3
Views: 62

Re: cell potential

The maximum cell potential occurs at the instant when the anode and cathode are first connected and the flow of electrons begins. Beyond that point, the concentrations in the half-reactions change. The production of product decreases the cell potential per increasing Q in the Nernst equation: E= E^0...
by 805303639
Thu Feb 27, 2020 8:39 pm
Forum: Balancing Redox Reactions
Topic: Redox rxns- differences for solving?
Replies: 2
Views: 45

Re: Redox rxns- differences for solving?

I assume you are referring to the reactions posed in parts a and b of question 6L.7. Balancing these reactions requires some additional insight because the species involved in the redox half-reactions cancel in the full balanced equation. For the bronsted neutralization reaction, you need to add O2(...
by 805303639
Thu Feb 27, 2020 8:17 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: homework question 6L.9
Replies: 2
Views: 81

Re: homework question 6L.9

The balanced redox reaction should be: MnO4-(aq) + 5Fe^2+(aq) + 8 H+(aq) --> Mn^2+(aq) + 5Fe^3+(aq) + 4H2O(l). As for the galvanic cell, we know our cathode and our anode from the balanced half-reactions: a) MnO4-(aq) +8H+(aq)+5e- yields Mn2+(aq) +4H2O (cathode half reaction) 5[Fe2+ (aq) yields Fe3+...
by 805303639
Thu Feb 27, 2020 8:03 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: electrochemical series 6.45
Replies: 1
Views: 38

Re: electrochemical series 6.45

The question asks you to order the metals starting with the most strongly reducing metal. The "most strongly reducing metal" is the strongest reducing agent, which is the metal half-reaction with the most negative E^0 value (in this question, Al). You are right that the more positive the E...
by 805303639
Thu Feb 20, 2020 2:52 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Standard hydrogen electrode [ENDORSED]
Replies: 3
Views: 124

Re: Standard hydrogen electrode [ENDORSED]

Absolute potentials cannot be measured, so scientists need a reference electrode. The selection of hydrogen for this purpose is arbitrary. Early researchers of electrochemistry likely selected hydrogen given the relative ease of obtaining/constructing a hydrogen electrode/solution.
by 805303639
Thu Feb 20, 2020 2:37 pm
Forum: Balancing Redox Reactions
Topic: HW 6K.3
Replies: 2
Views: 56

Re: HW 6K.3

In S2O3 2-(aq), S has an oxidation number of +2. In SO4 2-(aq), S has an oxidation number of +6. After balancing the oxidation half-reaction, SO4 2-(aq) has a stoichiometric coefficient of 2. Since S2O3 2-(aq) has 2 S atoms, each with an oxidation number of +2, those S atoms together have an oxidati...
by 805303639
Thu Feb 20, 2020 2:23 pm
Forum: Balancing Redox Reactions
Topic: 6K.5 partA
Replies: 2
Views: 36

Re: 6K.5 partA

Although oxygen's oxidation number doesn't change from O3 to O2, it changes from O3 to BrO3 2- (aq). O has an oxidation number of -2 in BrO3 2-, so oxygen gains 2 electrons (reduced).
by 805303639
Thu Feb 20, 2020 2:14 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Salts used in Salt bridge
Replies: 2
Views: 53

Re: Salts used in Salt bridge

The salt used in a salt bridge should not form a precipitate with the ions in either of the half-reaction cells. Thus Cl- would actually not be a good salt when analyzing the redox potential for Ag+(aq) + e- --> Ag(s) because Cl- reacts with Ag+(aq). Provided K+ or Cl- does not form a precipitate wi...
by 805303639
Thu Feb 20, 2020 1:46 pm
Forum: Balancing Redox Reactions
Topic: Homework problem 6K.1
Replies: 5
Views: 88

Re: Homework problem 6K.1

Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7...
by 805303639
Wed Feb 12, 2020 10:32 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 4I.9
Replies: 3
Views: 64

Re: 4I.9

The ΔS of the irreversible process is the same as that of the reversible process because entropy is a state property. The pathway is irrelevant.
by 805303639
Wed Feb 12, 2020 10:29 pm
Forum: Calculating Work of Expansion
Topic: 4.7A
Replies: 1
Views: 56

Re: 4.7A

Even if you balanced the equation with whole-number coefficients, you are still only reacting 1.00 mol C6H6 (not 2.00 mol). Calculating the moles of gaseous reactants/products using 1.00 mol C6H6 will give the same delta n regardless of which balanced equation you use.
by 805303639
Wed Feb 12, 2020 10:00 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: HW 4F.17
Replies: 1
Views: 49

Re: HW 4F.17

Yes, you need to use delta S= n C ln(T2/T1). To calculate the standard entropy of vaporization of water at 85 C, you need to break the problem down into four steps: 1. Calculate the entropy change as water heats from 85 C to 100 C using delta S= n C ln(T2/T1). 2. Record the entropy of vaporization a...
by 805303639
Wed Feb 12, 2020 9:54 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: 4G.5
Replies: 2
Views: 46

Re: 4G.5

There are 15 ways to group two items from a set of 6 (6 choose 2). This is the total number of configurations of a pair of distinct atoms around an octahedral molecule, so #cis configurations + #trans configurations = 15. There are 3 trans configurations because there are 3 ways to group two atoms (...
by 805303639
Wed Feb 12, 2020 9:40 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Change in entropy for a monatomic ideal gas vs diatomic molecules
Replies: 3
Views: 136

Re: Change in entropy for a monatomic ideal gas vs diatomic molecules

1 mole of atoms making up diatomic molecules equals 0.5 moles of diatomic molecules. 1.0 mole of a monoatomic gas has a greater number of microstates than 0.5 moles of diatomic molecules because the monoatomic gas involves a greater number of particles.
by 805303639
Thu Feb 06, 2020 8:01 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 4F.15
Replies: 2
Views: 32

Re: 4F.15

Estimate the boiling point from (delta Svap) = (delta Hvap)/T where delta Svap= 85 J K^-1 and delta Hvap = 21.51 kJ/mol. The calculation yields a boiling point of 253 K. The actual boiling point is 248-249 K. Trouton's rule states that roughly the same increase in positional disorder occurs when any...
by 805303639
Thu Feb 06, 2020 3:34 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Cv or Cp
Replies: 5
Views: 364

Re: Cv or Cp

CristinaMorales1F wrote:Also, Cp = 5/2R and Cv=3/2R are only used for monoatomic ideal gases.


For linear molecules, use Cp = 7/2 R and Cv= 5/2 R to account for rotational kinetic energy along two axes.
For nonlinear molecules, use Cp = 4R and Cv = 3R to account for rotational kinetic energy along a third axis.
by 805303639
Thu Feb 06, 2020 3:25 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Heat vs Temp
Replies: 3
Views: 62

Re: Heat vs Temp

Heat is the transfer of energy resulting from a temperature difference between a system and its surroundings. At the molecular level, heat is the transfer of kinetic energy between particles via collisions.
by 805303639
Thu Feb 06, 2020 3:07 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: W
Replies: 2
Views: 37

Re: W

Emily Chirila 2E wrote:Degeneracy is the number of ways of achieving a given energy state. W=2^n where n= # of particles


W = 2^n for a two-state system. In general, W = m ^n where m is the # of possible states of a single molecule and n is the # of molecules
by 805303639
Thu Feb 06, 2020 3:04 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: micro states
Replies: 1
Views: 38

Re: micro states

For an octahedral molecule made of two atoms (A and B) attached to the central atom: 6 x A, 0 x B: 1 microstate because the molecule is symmetrical in all orientations 5 x A, 1 x B: 6 microstates because atom B can occupy each of the six positions attached to the central atom 4 x A, 2 x B: 15 micros...
by 805303639
Thu Jan 30, 2020 9:14 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: HW 4D.9
Replies: 2
Views: 48

Re: HW 4D.9

Yes, start by calculating the enthalpy change using the standard enthalpies of formation (use table 4D2 and the TNT formation enthalpy value provided in the problem). You should get -13168.48 kJ for the reaction. To calculate the number of liters of TNT that produces that enthalpy change, put on you...
by 805303639
Thu Jan 30, 2020 8:53 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Work done by expansion
Replies: 8
Views: 62

Re: Work done by expansion

For a reversible expansion, all changes in volume occur in tiny steps. We calculate the work using the definite integral from v1 to v2 of - P dV. To derive an equation for constant pressure conditions, set P as a constant. The definite integral evaluates to - P delta V, the work equation for constan...
by 805303639
Thu Jan 30, 2020 8:41 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 4A.13
Replies: 1
Views: 48

Re: 4A.13

Internal energy is a state property, so we need not consider the reaction's mechanism. This problem first asks you to calculate the calorimeter constant. Using qcal = Ccal delta T, you can plug in -3.50 kJ for qcal and 7.32 C for delta T. Solve and get Ccal = 0.478 kJ C^-1. Since we now know the hea...
by 805303639
Thu Jan 30, 2020 1:03 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Internal energy change, isothermal expansion
Replies: 1
Views: 23

Internal energy change, isothermal expansion

I'm confused about why the change in internal energy for the isothermal expansion of a gas is 0. I understand that the kinetic energy of the system isn't changing due to the constant temperature. But wouldn't the potential energy of the gas decrease because of reduced intermolecular interactions/rep...
by 805303639
Thu Jan 30, 2020 12:54 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Qv vs Qp
Replies: 7
Views: 51

Re: Qv vs Qp

Both of these are used to find enthalpy/delta H but also find change in internal energy which is q + w. Therefore, q can either be q p or q v depending on the given conditions. Enthalpy/delta H is equal to qp, the amount of heat released or absorbed at a constant pressure. Using qv implies that the...
by 805303639
Thu Jan 23, 2020 9:07 pm
Forum: Phase Changes & Related Calculations
Topic: Work, reversible path
Replies: 2
Views: 36

Work, reversible path

The textbook stresses that "the greatest work is done for a process that takes place reversibly." I'm confused as to why this is always the case.
by 805303639
Thu Jan 23, 2020 9:02 pm
Forum: Phase Changes & Related Calculations
Topic: 4A. 1 Identifying open and closed system
Replies: 11
Views: 270

Re: 4A. 1 Identifying open and closed system

An open system can exchange both matter and energy with the surroundings. A car engine is an open system because it exchanges heat and matter (carbon dioxide, carbon monoxide, water, and other byproducts) with its surroundings. An isolated system can exchange neither energy nor matter with its surro...
by 805303639
Thu Jan 23, 2020 8:45 pm
Forum: Phase Changes & Related Calculations
Topic: Enthalpy and Pressure
Replies: 4
Views: 51

Re: Enthalpy and Pressure

The equation deltaH = deltaE + delta(PV) describes the change in enthalpy of a system. We need not measure the enthalpy at constant pressure per se but doing so simplifies the equation to deltaH = q (after some algebra). As a result, we know the heat (q) released or absorbed in the reaction equals t...
by 805303639
Thu Jan 23, 2020 8:24 pm
Forum: Phase Changes & Related Calculations
Topic: Enthalpy of Glucose
Replies: 2
Views: 66

Re: Enthalpy of Glucose

Glucose is a great source of chemical potential energy. The oxidation of glucose releases 2880 kJ of energy/mol, 2823 kJ of which is available to do work. Although eukaryotic cells cannot harness all of this energy, cellular respiration is still ~34% efficient in converting the chemical energy of gl...
by 805303639
Thu Jan 23, 2020 8:03 pm
Forum: Phase Changes & Related Calculations
Topic: Internal Energy Change
Replies: 3
Views: 49

Re: Internal Energy Change

We measure the internal energy change instead of the absolute internal energy because the absolute internal energy is unknowable, encompassing the energies of all the atoms (nuclei and electrons) in the system. As Maya stated, delta U is useful because it quantifies work and heat flow through the sy...
by 805303639
Wed Jan 15, 2020 9:53 pm
Forum: Ideal Gases
Topic: approximate ideal gas law
Replies: 3
Views: 50

Re: approximate ideal gas law

The ideal gas law is a limiting law that performs well as P approaches 0. High pressures highlight its shortcomings. Van der Waals equation, another approximation of gas behavior, is more accurate because it accounts for the volume of the gas molecules and their intermolecular attractions. However, ...
by 805303639
Wed Jan 15, 2020 9:39 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Self Test 5I.3b
Replies: 4
Views: 41

Re: Self Test 5I.3b

Hi Gabby. Yes, I2 is not included in the equilibrium expression K= (PCl2)(PHI)^2/(PHCl)^2. Since we know the K value, the initial partial pressure of HCl, and the stoichiometric coefficients, we can use an ICE table to solve for the equilibrium partial pressures. Set up an ice table with the initial...
by 805303639
Wed Jan 15, 2020 9:15 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5.33
Replies: 3
Views: 70

Re: 5.33

Hi Sara. Reducing the volume increases the concentrations of both X2 and X (concentration = n/V). According to the reaction quotient Q = [X]^2/[X2], products increase relative to reactants as a result of the concentration increases. Q > K and the system favors the reverse reaction, yielding more X2 ...
by 805303639
Wed Jan 15, 2020 9:00 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Dilute solution cutoff
Replies: 2
Views: 27

Dilute solution cutoff

Lavelle explained that solvents/liquids are excluded from equilibrium expressions for dilute solutions because their concentrations remain essentially unchanged throughout the reaction. When would a solution not be considered "dilute" and no longer treated as a pure substance? Is there a s...
by 805303639
Wed Jan 15, 2020 8:46 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Module 4 Q2
Replies: 2
Views: 49

Re: Module 4 Q2

As Justin explained, compression affects the reaction by increasing the concentrations of gaseous reactants and products. The decrease in volume serves to multiply all concentrations by some factor. Since all concentrations increase by the same factor, the side with more moles experiences a greater ...
by 805303639
Thu Jan 09, 2020 8:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Dynamic equilibrium
Replies: 2
Views: 43

Re: Dynamic equilibrium

The rates of the forward and reverse reactions are equal at equilibrium; they are not 0. The reactions, therefore, do not stop.
by 805303639
Thu Jan 09, 2020 8:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Reaction quotient basis
Replies: 1
Views: 25

Reaction quotient basis

What is the reasoning behind raising the product/reactant concentrations to their stoichiometric coefficients in the reaction quotient/equilibrium expression?
by 805303639
Thu Jan 09, 2020 8:06 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: solids and liquids
Replies: 6
Views: 52

Re: solids and liquids

The equilibrium expression (Kc) does not include solid reactants/products because solids do not have concentrations. The expression neglects liquids/solvents because they are usually present in such excess that their concentrations are essentially unchanged.
by 805303639
Thu Jan 09, 2020 7:49 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: N2O4 <--> 2NO2, color?
Replies: 3
Views: 52

Re: N2O4 <--> 2NO2, color?

If you are interested, Professor Lavelle shows the temperature-dependent change in equilibrium for the 2NO2 <--> N2O4 at the end of the Chemical Equilibrium Part 4 video module (starting at 47 minutes). As temperature increases, the solution appears dark amber/brown because of an increase in NO2 con...
by 805303639
Thu Jan 09, 2020 3:37 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Pressure
Replies: 4
Views: 57

Re: Pressure

No, a higher pressure does not affect a reaction's equilibrium constant. The system does not respond to changes in pressure directly. Take the example of increasing the pressure by decreasing the volume, which changes the concentrations (n/V) of the reactants and products. It is this change in conce...

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