Search found 36 matches
- Sat Mar 16, 2019 1:53 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate Law with Intermediates
- Replies: 1
- Views: 166
Rate Law with Intermediates
Is it okay to replace all of the constants in the final rate law (such as k1, k-1, k2, k-2) with one K to substitute the product of all of the constants?
- Sat Mar 16, 2019 1:52 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Using K to get rid of intermediate
- Replies: 1
- Views: 254
Re: Using K to get rid of intermediate
Yes! If trying to remove an intermediate in the rate law introduces another intermediate, use the rate law of the reaction that has the new intermediate as a product.
- Sat Mar 16, 2019 1:32 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Finding the Largest Standard Cell Potential
- Replies: 1
- Views: 228
Finding the Largest Standard Cell Potential
How would we know which two species, when we know their reducing power or oxidation power, would give the largest standard cell potential when assembled in an electrochemical cell? Would we pair the species with the strongest reducing power with the species with the strongest oxidation power?
- Sat Mar 16, 2019 12:58 pm
- Forum: Balancing Redox Reactions
- Topic: Acidic v basic conditions
- Replies: 3
- Views: 390
Re: Acidic v basic conditions
For acidic solutions, balance O using H2O, and then balance H using H+.
For basic solutions, balance O using H2O, and then balance H by adding H2O to other side of each half-reaction that needs H and adding OH- to the other side
For basic solutions, balance O using H2O, and then balance H by adding H2O to other side of each half-reaction that needs H and adding OH- to the other side
- Sat Mar 16, 2019 12:57 pm
- Forum: Balancing Redox Reactions
- Topic: Acidic v basic conditions
- Replies: 3
- Views: 390
Re: Acidic v basic conditions
It does! Knowing whether it is basic or acidic will determine which method you use to balance the redox reaction, whether you use H+ or H2Os to balance the equation.
- Mon Mar 11, 2019 8:08 pm
- Forum: First Order Reactions
- Topic: 15.31 6th Edition
- Replies: 3
- Views: 361
15.31 6th Edition
For question 31 in Chapter 15 of the 6th Edition of the textbook, we are looking for the half life of iodine-131. I know that we use the equation t(halflife) = 1/(k*initial concentration), but the answer key shows that t(halflife) is (1/k) * ln2. Where does the ln2 come from? Is that supposed to be ...
- Sun Mar 10, 2019 10:02 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Unique Rate of a Reaction
- Replies: 2
- Views: 302
Re: Unique Rate of a Reaction
Hey! For this question, remember that the formula for reaction rate is
1/(coefficient of reactant or product) * rate of reaction for specific reactant or product.
I think the answer key uses NO2 to solve for it, so the answer would be:
1/2 * 6.5*10^-3 mol/L/s
or 3.3x10^-3 mol/L/s
1/(coefficient of reactant or product) * rate of reaction for specific reactant or product.
I think the answer key uses NO2 to solve for it, so the answer would be:
1/2 * 6.5*10^-3 mol/L/s
or 3.3x10^-3 mol/L/s
- Sat Mar 09, 2019 2:19 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.17 6th Ed
- Replies: 2
- Views: 341
Re: 15.17 6th Ed
Hey! The reason we compare experiments 2 and 3 to find the initial rate of B is because when we compare experiments 1 and 4, we can see that changing the concentration of C does not affect the initial rate, so even though the concentration of C changes between experiments 2 and 3, it will not affect...
- Sat Mar 09, 2019 2:13 pm
- Forum: General Rate Laws
- Topic: Positive/negative sign for rates
- Replies: 1
- Views: 211
Re: Positive/negative sign for rates
If we're referring to the rate of consumption as such, then yes, it is positive, because it is the rate at which the reactant is being used. However, if we simply say that it is the rate of the reactant, we would put a negative sign to indicate that reactant is being used up. Product is usually prod...
- Sun Mar 03, 2019 7:05 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Gases and Liquids in Cell Diagrams
- Replies: 3
- Views: 544
Re: Gases and Liquids in Cell Diagrams
Hey! I think that we only include gases and liquids into cell diagrams when they are either the anode or cathode where the element gets oxidized or reduced, or if they are oxidized or reduced to form the ions in the solution. Since H2O isn't usually directly part of the reaction, I think it's safe t...
- Sun Mar 03, 2019 7:02 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.7 (6th Ed.): Figuring out reactant rate order
- Replies: 1
- Views: 243
Re: 15.7 (6th Ed.): Figuring out reactant rate order
Hey! I think you're actually looking at problem 17 in the 6th edition. The reason why we can use experiments 2 and 3 to find the order for the B is because: The concentration of C doesn't change the rate of the reaction, as seen between experiments 1 and 4 (The concentration of C changes, but the ra...
- Sun Mar 03, 2019 6:57 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Kinetics HW help
- Replies: 1
- Views: 242
Re: Kinetics HW help
Hey! For this question, you have to focus on two formulas: t1/2 =0.693/k First use this formula to find the value of k, and then ln [A] = -k t + ln [A]o use this formula to find the time that has passed. Also, you can use ratios to help you with this problem. Such as for part a, since it wants to kn...
- Tue Feb 26, 2019 10:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Solid Electrodes
- Replies: 1
- Views: 170
Solid Electrodes
When not specified, can we assume that the electrode of a galvanic cell is solid platinum?
If so, why is it that sometimes in the answers for the homework, platinum is shown on the anode side of the cell diagram and not the cathode?
If so, why is it that sometimes in the answers for the homework, platinum is shown on the anode side of the cell diagram and not the cathode?
- Sun Feb 24, 2019 3:32 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: cell potential vs emf
- Replies: 2
- Views: 248
Re: cell potential vs emf
There is!
Cell potential is a potential energy, while electromotive force is a force.
The electromotive force is the maximum difference between two cell potentials (like between anode and cathode).
Cell potential is a potential energy, while electromotive force is a force.
The electromotive force is the maximum difference between two cell potentials (like between anode and cathode).
- Sun Feb 24, 2019 3:24 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: salt bridge
- Replies: 2
- Views: 253
Re: salt bridge
The purpose of the salt bridge is to keep replenishing the ions in the solution (or cell), so that the salt bridge will keep replenishing either the positive or negative charge in the solution (or cell), based on if it's with the cathode or anode.
- Sun Feb 24, 2019 3:21 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.67 6th edition
- Replies: 1
- Views: 235
Re: 9.67 6th edition
Hey! To answer your question: In order to find the temperature ranges for each of the reactions, you'll want to find at what temperature that delta G is 0, and then see whether increasing temperature from there or decreasing it would make delta G negative. For part B and C, you'll have to gauge if d...
- Fri Feb 15, 2019 3:19 am
- 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: n in degeneracy
- Replies: 8
- Views: 2504
Re: n in degeneracy
When calculating degeneracy, n should equal the number of particles (or in this case, molecules) that can exist in a certain state.
So in your question, we would to know about the entire molecule of O2 gas that is in a certain energy state, so n would equal 1.
So in your question, we would to know about the entire molecule of O2 gas that is in a certain energy state, so n would equal 1.
- Fri Feb 15, 2019 3:14 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: 6th edition 8.25
- Replies: 4
- Views: 3219
Re: 6th edition 8.25
Hey! Just an elaboration, we know that q(calorimeter) = - q(reaction) because any heat released from the reaction is going to be absorbed by the calorimeter.
Because of this, heat from the reaction should just be negative of the heat absorbed by the calorimeter.
Because of this, heat from the reaction should just be negative of the heat absorbed by the calorimeter.
- Fri Feb 15, 2019 3:06 am
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Cv vs. Cp
- Replies: 6
- Views: 1011
Cv vs. Cp
When is it appropriate to use Cv = 3/2 *R versus Cp = 5/2*R?
I know you are supposed to use Cv when the problem is dealing with a monoatomic gas, but do they have any relation to pressure and volume?
I know you are supposed to use Cv when the problem is dealing with a monoatomic gas, but do they have any relation to pressure and volume?
- Fri Feb 08, 2019 2:19 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Reversible vs. Irreversible processes
- Replies: 1
- Views: 300
Reversible vs. Irreversible processes
Can someone explain why for reversible processes, as volume increases, pressure decreases, and temperature remains constant, but for irreversible processes, temperature changes along the pathway, and pressure remains constant as volume changes?
- Fri Feb 08, 2019 2:13 am
- Forum: Calculating Work of Expansion
- Topic: Hess's Law
- Replies: 2
- Views: 234
Re: Hess's Law
Yes! The bond enthalpies will typically be given to us when solving for reaction enthalpy using bond enthalpies - I would be quite surprised if they were not. You just have to make sure you correctly add the bond enthalpies of all bonds broken and all bonds formed, respectively. Also, Hess's Law is ...
- Fri Feb 08, 2019 2:08 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Closed Systems and q
- Replies: 1
- Views: 238
Re: Closed Systems and q
Hey! I have the sixth edition of the textbook, so I'm not too sure if this is right, but I think you're referring to question 8.15 in the 6th edition! This question says that an adiabatic process, no energy is transferred at heat, and asks you about an adiabatic process in a closed system. To answer...
- Fri Feb 01, 2019 1:48 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpy Method Accuracy
- Replies: 1
- Views: 271
Re: Bond Enthalpy Method Accuracy
Hey! The reason for this is because bond enthalpies are calculated as the average of many bonds. Different molecules have different bond enthalpies for a bond, such as C-H, which will have a different bond enthalpy depending on the molecule. Because these bond enthalpies are only averages, they are ...
- Fri Feb 01, 2019 1:43 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57 6th Edition Units
- Replies: 1
- Views: 202
Re: 8.57 6th Edition Units
Hey! In response to your question: In the question in the textbook, kJ *mol^-1 is referring to the energy of the equation when involving only one mole of either C2H2, C2H6, or H2. However, in the solutions manual, when they use the units kJ, it is referring to the energy change of the entire reactio...
- Fri Feb 01, 2019 1:39 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Question 8.3 6th edition
- Replies: 2
- Views: 235
Re: Question 8.3 6th edition
Hey! To answer your question: A bicycle pump is like a cylinder, so to find volume, we'd want to use the volume equation for a cylinder, being pi * radius^2 * height (or in this case, distance depressed). Since radius is 1/2*diameter, the radius is 1.5cm. From this, you'll get pi*(1.5cm)^2*(20cm), o...
- Thu Jan 31, 2019 4:51 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Enthalpy Symbol
- Replies: 5
- Views: 3612
Enthalpy Symbol
The symbol for enthalpy is represented by triangle(H). However, sometimes the symbol has H naught, with a o or a circle on the top left, such as when signifying the standard enthalpy of formation, which has naught f, or standard enthalpy of combustion, which has naught c. What does that naught circl...
- Fri Jan 25, 2019 12:33 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: K and temp changes
- Replies: 2
- Views: 197
Re: K and temp changes
Hey! Just adding on to the last reply, in endothermic reactions, I like to think of heat (energy) like a reactant of the chemical reaction, since heat is being taken in. Therefore, increasing temperature would cause the "concentrations" of the reactants to increase, and therefore the react...
- Thu Jan 24, 2019 9:12 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Change in Pressure caused by Volume
- Replies: 3
- Views: 286
Change in Pressure caused by Volume
If there is a change in pressure caused by the volume to be halved, then the reaction will shift to the side of the equation with the smaller sum of coefficients. However, what exactly happens during that shift? For example, if it shifts towards the products, does that mean the concentration of prod...
- Thu Jan 24, 2019 5:09 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Strong vs. Weak Acids
- Replies: 3
- Views: 390
Strong vs. Weak Acids
Just as a clarification, if the concentration of H3O+ is greater than 10^-3, or the pH is less than 3, then it is considered a strong acid?
Alternatively, if the concentration of OH- is greater than 10^-3, or the pOH is less than 3, then it is considered a strong base?
Alternatively, if the concentration of OH- is greater than 10^-3, or the pOH is less than 3, then it is considered a strong base?
- Fri Jan 18, 2019 2:53 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Weak Acids
- Replies: 1
- Views: 124
Weak Acids
Can someone explain again why weak acids can have a pH of greater than 7 when they are technically still acids? I don't quite understand.
- Fri Jan 18, 2019 1:24 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: pressure
- Replies: 2
- Views: 329
Re: pressure
Hey! To answer your question, increasing pressure by adding an inert gas doesn't affect concentration because the volume of the system does not change (probably due to there not being a large enough difference in change in volume by adding the gas), and the moles of both reactants and products are n...
- Wed Jan 16, 2019 1:15 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.89 Part B 6th Edition
- Replies: 3
- Views: 181
11.89 Part B 6th Edition
Part B of question 11.89 in the 6th edition of the textbook asks to calculate the equilibrium constant for the reaction, which I got as 2A -> B + 2C. First off, do we need to show a double arrow for this reaction equation, since it reaches equilibrium, because the answer key does not show that. Seco...
- Sat Jan 12, 2019 9:16 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.43 in 6th Edition
- Replies: 2
- Views: 148
11.43 in 6th Edition
In the 6th edition of the textbook, the question says: Consider the reaction 2 NO(g) Δ N2(g) + O2(g). If the initial partial pressure of NO(g) is 1.0 bar, and x is the equilibrium concentration of N2(g), what is the correct equilibrium relation? When solving this question, does it not matter if we u...
- Thu Jan 10, 2019 2:53 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Kc and relative energy
- Replies: 2
- Views: 173
Kc and relative energy
Can someone explain how Kc reveals the relative energy? I know we discussed that if Kc is large, then it has lower energy, but why is that?
- Wed Jan 09, 2019 10:54 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Units for Kc and Kp
- Replies: 1
- Views: 1445
Units for Kc and Kp
For the equilibrium constant for concentration, Kc, I know that we omit the units of moles per liter because it is implied by the brackets [], but why do we omit the units for the equilibrium constant of partial pressure? Is it because it often does not cancel out to K having no units?
- Wed Jan 09, 2019 1:16 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium constants Kp notation
- Replies: 2
- Views: 263
Re: Equilibrium constants Kp notation
As Melissa mentioned, brackets, [], indicate concentration, and therefore we do not use them when referring to partial pressures. I think the parentheses are just included for clarification, to identify clearly which exponents belong to which partial pressures of certain gasses.