Search found 40 matches

by Sally Nason - 1K
Fri Mar 16, 2018 10:57 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Reducing Power
Replies: 5
Views: 187

Re: Reducing Power

Find the redox equations on the sheet for each of the elements given and the most negative value for E has the most reducing power.
by Sally Nason - 1K
Fri Mar 16, 2018 10:53 am
Forum: Method of Initial Rates (To Determine n and k)
Topic: Test 3 question 5
Replies: 3
Views: 250

Re: Test 3 question 5

You have to set the rate equal to k times the concentration of H2O (the moles of H2O (1.8/18.016) times 1/.750L) times the concentration of O2 squared (moles of O2 (1.8/32) times 1/.750L and square that). The equation should look like: 2.7 x 10^2 = k*[(1.8/18.016)*(1/.750)]*[(1.8/32)*(1/.750)] and t...
by Sally Nason - 1K
Fri Mar 16, 2018 10:48 am
Forum: Calculating Work of Expansion
Topic: Irreversible Expansion vs. Reversible Expansion
Replies: 2
Views: 286

Re: Irreversible Expansion vs. Reversible Expansion

Since you use the equation w = -P*deltaV for irreversible expansion, you would likely be given the pressure or the change in moles to use for substitution with PV = nRT. For reversible expansion, we use the equation w = -nRT*ln(V2/V1) so you would likely be told it is isothermal and the moles would ...
by Sally Nason - 1K
Mon Mar 12, 2018 5:43 pm
Forum: Calculating Work of Expansion
Topic: different types of work
Replies: 4
Views: 157

Re: different types of work

For work you should recognize expansion versus compression but that is as specific as you need to know. The other equations, such as those in the beginning of chapter 8 in the textbook, are not on the equation sheet.
by Sally Nason - 1K
Mon Mar 12, 2018 5:41 pm
Forum: *Organic Reaction Mechanisms in General
Topic: Functional Groups
Replies: 7
Views: 653

Re: Functional Groups

If you are able to draw the Lewis structure of molecules and know what to look for within the structure to determine the functional group, you should be set.
by Sally Nason - 1K
Mon Mar 12, 2018 5:40 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: slow step
Replies: 3
Views: 153

Re: slow step

There can only be one "slow" step because it is considered the slowest of all the steps. Slow is a relative term and describes the speed of one step of reaction in relation to the speed of the other steps.
by Sally Nason - 1K
Mon Mar 05, 2018 9:25 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: What is Molecularity?
Replies: 9
Views: 760

Re: What is Molecularity?

You can also think about molecularity in regards to the order of a reaction, as first order is unimolecular, second order is bimolecular, and third order is termolecular.
by Sally Nason - 1K
Mon Mar 05, 2018 9:24 pm
Forum: First Order Reactions
Topic: 15.23C
Replies: 6
Views: 329

Re: 15.23C

If the question tells you what order the reaction is in, always go with that because it is directly stated in the question.
by Sally Nason - 1K
Mon Mar 05, 2018 9:22 pm
Forum: General Rate Laws
Topic: rate of change of species
Replies: 2
Views: 98

Re: rate of change of species

I think this refers to questions like 15.1 on the homework where you are given the equation: N2 (g) + 3H2 (g) -> 2NH3 (g) and are asked what is the relation between the rate of consumption of H2 and the rate of consumption of N2. In this case, the rate of consumption of N2 is 1/3 times the rate of c...
by Sally Nason - 1K
Mon Mar 05, 2018 9:19 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: derivations
Replies: 2
Views: 119

Re: derivations

The derived equations are on the equation sheet. Derivations are helpful for sure so be comfortable with them but you will not need to know how to derive them.
by Sally Nason - 1K
Tue Feb 27, 2018 12:53 pm
Forum: General Rate Laws
Topic: Homework 15.49
Replies: 1
Views: 83

Homework 15.49

Could someone please explain how we know that HBr, NO2, and HOBr are first order? All we are given are two reaction equations with no numbers.
by Sally Nason - 1K
Tue Feb 27, 2018 12:51 pm
Forum: General Rate Laws
Topic: initial rate law
Replies: 5
Views: 169

Re: initial rate law

Also in a lot of the homework problems, you use the change in the concentration of the product to find the change in the concentration of the reactant.
by Sally Nason - 1K
Tue Feb 27, 2018 12:49 pm
Forum: General Rate Laws
Topic: Writing a rate equation
Replies: 3
Views: 142

Re: Writing a rate equation

I don't think you need to ever write it in terms of products because even if you are given the increase or decrease in concentration of products, you can convert it to reactants.
by Sally Nason - 1K
Sun Feb 25, 2018 9:46 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Homework 15.19
Replies: 3
Views: 189

Homework 15.19

/Is it always necessary to change units to mol from mmol?
by Sally Nason - 1K
Tue Feb 20, 2018 11:51 am
Forum: Balancing Redox Reactions
Topic: 14.5 part a
Replies: 2
Views: 153

Re: 14.5 part a

It will become more obvious that oxygen is being reduced once you balance the hydrogens and oxygens in the equation. The half reaction isn't balanced so it is difficult to tell what is happening. The balanced reaction should look like H2O (l) + O3 (g) + 2e- -> O2 (g) + 2OH- (aq). The oxygen is going...
by Sally Nason - 1K
Tue Feb 20, 2018 11:48 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15 given half reactions?
Replies: 3
Views: 172

Re: 14.15 given half reactions?

I also used the appendix. I think we will receive a sheet with half reactions that we will need for the test.
by Sally Nason - 1K
Tue Feb 20, 2018 11:46 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Example 14.3
Replies: 1
Views: 69

Re: Example 14.3

In this equation, n stands for the moles of electrons.
by Sally Nason - 1K
Tue Feb 13, 2018 9:49 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat Capacities of Gases
Replies: 3
Views: 288

Re: Heat Capacities of Gases

These are given on the equation sheet.
by Sally Nason - 1K
Mon Feb 12, 2018 7:44 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.101 (b) [ENDORSED]
Replies: 4
Views: 531

Re: 9.101 (b) [ENDORSED]

I had trouble with this one as well. You would think that since the outdoors is so much larger that any temperature change wouldn't affect the entropy as much but the question really just wants you to look at the equation.
by Sally Nason - 1K
Mon Feb 12, 2018 7:43 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Calorimetry
Replies: 2
Views: 124

Re: Calorimetry

Every time there is a temperature change or a phase change you need a separate equation. Usually when an ice cube has a phase change it first has to gain heat or lose heat to reach the right temperature for the phase change. That would be 2 equations. Then, if the ice cube changes temperature after ...
by Sally Nason - 1K
Mon Feb 12, 2018 7:41 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Delta U = 0
Replies: 2
Views: 156

Re: Delta U = 0

Usually when the question says the process is isothermal, delta U will be equal to zero. An isothermal system will also have a delta S(universe) of zero.
by Sally Nason - 1K
Thu Feb 08, 2018 10:55 am
Forum: Biological Examples (*DNA Structural Transitions, etc.)
Topic: problem 9.75
Replies: 2
Views: 393

Re: problem 9.75

For cis compounds there are 12 different orientations but for trans compounds there are only 3 different orientations. Therefore, the cis form will have higher residual energy (use Boltzmann entropy calculations).
by Sally Nason - 1K
Thu Feb 08, 2018 10:52 am
Forum: Phase Changes & Related Calculations
Topic: Phase changes with entropy?
Replies: 6
Views: 219

Re: Phase changes with entropy?

During a phase change, the temperature does not change but heat must still go into/leave the system in order to cause the phase change. Since heat is entering or leaving the system, the entropy will change and therefore needs to be accounted for.
by Sally Nason - 1K
Mon Feb 05, 2018 9:53 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Chapter 11 #17
Replies: 2
Views: 106

Re: Chapter 11 #17

I also got -2.7 kJ/mol! I think you are right and the answer key has an error in it!
by Sally Nason - 1K
Mon Feb 05, 2018 9:52 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.63
Replies: 4
Views: 159

Re: 9.63

If the value for delta G is negative then energy needs to be added in order to form the compound so therefore it would not be stable, as it doesn't occur spontaneously. However, a negative delta G for a compound means the compound can form without any stimulus or added energy and is therefore a stab...
by Sally Nason - 1K
Mon Feb 05, 2018 9:47 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 11.17
Replies: 10
Views: 372

Re: 11.17

I agree. I've put it in several different calculators and have gotten -2.7 kJ/mol every time.
by Sally Nason - 1K
Wed Jan 31, 2018 2:00 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: A spontaneous reaction
Replies: 7
Views: 216

Re: A spontaneous reaction

When you need to go up the hill (low to high), an input of energy is required to do so, so therefore this motion cannot happen on its own and it is not spontaneous. However, if you go down the hill (high to low), the reaction can happen on its own because no energy needs to be put into the system an...
by Sally Nason - 1K
Wed Jan 31, 2018 1:53 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Why should deltaG=0?
Replies: 4
Views: 382

Re: Why should deltaG=0?

You can also think of it like delta G goes from negative to positive with 0 as the "threshold" so to find at what temperature this threshold is at, you set deltaG equal to 0.
by Sally Nason - 1K
Wed Jan 31, 2018 1:52 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Dependence of free energy on pressure
Replies: 2
Views: 108

Re: Dependence of free energy on pressure

I was also confused by this in class but in the equation G = G^o + RTln(P/P^o), I believe that the P is the atmospheric pressure and the P^o is always 1.00 atm because it's standard. Therefore, when P is more than 1.00 atm, (P/P^o) is positive and when P is less than 1.00 atm, (P/P^o) is negative.
by Sally Nason - 1K
Wed Jan 31, 2018 1:19 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: Homework 9.25
Replies: 1
Views: 131

Homework 9.25

When I plug (1.38x10^-23)*ln(6^6.02x10^23) into my calculator, I get a result that says "overflow" but the answer is 14.9 J/K. How is this possible?
by Sally Nason - 1K
Thu Jan 25, 2018 12:04 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Confusion about entropy formula
Replies: 6
Views: 154

Re: Confusion about entropy formula

If there is a heat reservoir in the surroundings, the temperature will be constant and therefore heat will be reversible. If a system is doing work, it is losing energy but heat from the surroundings can be added to the system to replace the energy lost. For instance, during a phase change, work is ...
by Sally Nason - 1K
Thu Jan 25, 2018 12:57 am
Forum: Phase Changes & Related Calculations
Topic: Chapter 8 Homework #117
Replies: 1
Views: 94

Chapter 8 Homework #117

Why is the net production 2/3 mole of gas? There are 2 moles of reactants and 4 moles of products so how is the change 2/3?
by Sally Nason - 1K
Wed Jan 24, 2018 10:29 pm
Forum: Phase Changes & Related Calculations
Topic: Chapter 8 Homework #99
Replies: 2
Views: 192

Chapter 8 Homework #99

How do you figure out the enthalpies of formation? The enthalpy for ZnCL2 isn't in the appendix of the book.
by Sally Nason - 1K
Thu Jan 18, 2018 11:43 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.29 heat capacity
Replies: 3
Views: 131

Re: 8.29 heat capacity

From what I understand, there are three forms of movement that affect the internal energy of a molecule, and therefore the heat capacity. These are: translational, linear rotational, and nonlinear rotational. For more complex molecules, all three motions are occurring and therefore the molecule has ...
by Sally Nason - 1K
Thu Jan 18, 2018 11:40 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Q 8.13
Replies: 9
Views: 363

Re: Q 8.13

In this question, the part we are trying to figure out is the work that can be done by the system, or on the system. In general, if the system does work, such as expansion, the work value will be positive and if the system has work done on it, such as compression, the work value will be negative.
by Sally Nason - 1K
Thu Jan 18, 2018 11:31 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Hess's Law
Replies: 5
Views: 229

Re: Hess's Law

Since enthalpy is a state function, it doesn't matter what path is taken to the final point, the only values that matter are final and initial. That is why you can "skip" intermediate steps by adding equations.
by Sally Nason - 1K
Thu Jan 18, 2018 11:26 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: enthalpy
Replies: 3
Views: 175

Re: enthalpy

When you subtract the enthalpies of formation of the reactants from the enthalpies of formation of the products, the result is the enthalpy of the reaction.
by Sally Nason - 1K
Mon Jan 15, 2018 1:12 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Homework Problem 8.31
Replies: 2
Views: 136

Re: Homework Problem 8.31

For monatomic gases you can use the formula C[p][/M] = (5/2)*R as well as C[V][/M] = (3/2)*R
by Sally Nason - 1K
Thu Jan 11, 2018 11:53 am
Forum: Phase Changes & Related Calculations
Topic: Compression vs Expansion
Replies: 3
Views: 187

Compression vs Expansion

Is the work done during compression always positive and the work done during expansion always negative?
by Sally Nason - 1K
Thu Jan 11, 2018 11:51 am
Forum: Phase Changes & Related Calculations
Topic: Open, Closed, or Isolated Systems
Replies: 6
Views: 2884

Re: Open, Closed, or Isolated Systems

It would be a closed system because while the mercury cannot leave the thermometer, heat can be transferred to or from the thermometer, therefore not making it isolated.

Go to advanced search