Search found 27 matches
- Sun Mar 11, 2018 11:22 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: K and K'
- Replies: 4
- Views: 311
Re: K and K'
k' (rate constant prime) is simply the 'reverse' of k (the rate constant). This means that k' is the rate constant for the reaction of products going back to the reactants.
- Sun Mar 11, 2018 11:12 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate constant
- Replies: 6
- Views: 399
Re: Rate constant
The rate constant is deemed a constant for a specific reaction at a specific, constant temperature. This differs from reaction rate because the rate of reaction is the speed in which a reaction is occurring. A good way to think about it is that concentration increasing or decreasing may change the r...
- Sun Mar 11, 2018 11:09 pm
- Forum: Calculating Work of Expansion
- Topic: Midterm 4a
- Replies: 5
- Views: 536
Re: Midterm 4a
I think you can identify it as irreversible because for that question, it says the external pressure is constant. For reversible changes, however, external pressure is ever so slightly changing.
- Sun Mar 04, 2018 11:10 pm
- Forum: First Order Reactions
- Topic: First order
- Replies: 4
- Views: 371
Re: First order
Generally if the concentration of a reactant doubles along with the reation's initial rate, it should be first order reaction. However, always do the arithmetic just to be sure :)
- Sun Mar 04, 2018 11:07 pm
- Forum: General Rate Laws
- Topic: Units for Rate Laws 15.3
- Replies: 2
- Views: 198
Re: Units for Rate Laws 15.3
I'm not sure what you meant by "here". However, the units for zero order reactions are (mol/L^-1 x (time unit)), for first order reactions is (time unit)^-1, and for second order reactions is (L x mol^-1 x (time unit)^-1).
- Sun Mar 04, 2018 11:02 pm
- Forum: General Rate Laws
- Topic: general rate laws
- Replies: 3
- Views: 261
Re: general rate laws
That's a good model to write rate laws after. But remember that some rate laws may omit some reactants and of course, the exponents (in this case m,n, and l) will be different. :)
- Sun Feb 25, 2018 5:04 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Initial rates
- Replies: 5
- Views: 454
Re: Initial rates
Also, there may be scenarios where your products are gases and they escape such that the only thing you can accurately rely on to calculate the rate is the reactants themselves.
- Sun Feb 25, 2018 4:58 pm
- Forum: General Rate Laws
- Topic: Writing the Rate Law
- Replies: 3
- Views: 239
Re: Writing the Rate Law
I don't think we do, especially since rate laws simply deal with the concentrations of substances
- Sun Feb 25, 2018 4:55 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Instantaneous Rate
- Replies: 4
- Views: 408
Re: Instantaneous Rate
We aren't just looking at initial (or at least I don't think so). There are two different equations we use to find the initial rate and then we use the Integrated Rate Laws to find rates after the start of the reaction :)
- Sun Feb 18, 2018 9:11 pm
- Forum: Balancing Redox Reactions
- Topic: "Rules"
- Replies: 7
- Views: 441
Re: "Rules"
In order to balance half-reactions, you often want to balance the O atoms in each first, which requires adding H2O on the other side of the equation. For instance, the half-reaction MnO4- -----> MnO2 has 4 O atoms on the left and 2 O atoms on the right. To balance, you add 2 H2O molecules to the rig...
- Sun Feb 18, 2018 9:00 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Platinum electrodes
- Replies: 4
- Views: 295
Re: Platinum electrodes
Since Platinum is an inert metallic compound, it doesn't affect the cell reaction. Moreover, Platinum is a good electric conductor, so I guess it's often used in this redox reactions.
- Sun Feb 18, 2018 8:08 pm
- Forum: Balancing Redox Reactions
- Topic: Add H+ or OH-
- Replies: 2
- Views: 1995
Re: Add H+ or OH-
You add H+ ions to redox reactions in acidic solution and OH- ions to redox reactions in basic solutions.
- Sun Feb 11, 2018 11:14 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Second Law
- Replies: 2
- Views: 207
Re: Second Law
If you look at example 9.1, it tells you that you can use that same equation to calculate the change in entropy when a system is being heated (reversibly). So if the question says that at constant temperature, a system is gaining some amount of energy in joules, it should be safe to use this formula.
- Sun Feb 11, 2018 11:09 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Stability and deltaG
- Replies: 2
- Views: 250
Re: Stability and deltaG
I think that if delta G is negative, the reaction is indeed spontaneous leading to the products being at a more stable energy state than the reactants.
- Sun Feb 11, 2018 11:06 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Cp v. Cv!
- Replies: 5
- Views: 461
Re: Cp v. Cv!
I remember from going to the midterm review session that unless the question specifically says at constant pressure, to assume using constant volume heat capacity. I think inuitively, however, that Cv should be used in the first question since yes, the pressure is the one that is changing. The secon...
- Mon Feb 05, 2018 1:05 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Entropy question [ENDORSED]
- Replies: 2
- Views: 185
Entropy question [ENDORSED]
I remember from discussion last week and hearing about the fact that entropy is always increasing in the universe. How is this so?
- Mon Feb 05, 2018 12:46 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Constant K and reaction quotient
- Replies: 3
- Views: 250
Re: Constant K and reaction quotient
To add, K, the equilibirum constant determines the relative concentrations (and partial pressures) of the products and reactants of the reaction at equilibrium while the Q, or reaction quotient says the same thing but when the reaction is not at equilibrium. A large K indicates that the concentratio...
- Sun Feb 04, 2018 10:17 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Sig Figs
- Replies: 6
- Views: 308
Re: Sig Figs
I'm sure sig figs are important when doing the calculations in the textbook.
- Sun Jan 28, 2018 11:47 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Adiabatic system [ENDORSED]
- Replies: 1
- Views: 246
Re: Adiabatic system [ENDORSED]
An adiabatic system is one where heat and matter does not flow in or outside of the system itself. A great example of an adiabatic system is a thermal container that is closed (kind of like a really good thermos). This means that if the internal energy of the system were to change in some way, then ...
- Sun Jan 28, 2018 11:37 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.41
- Replies: 3
- Views: 227
Re: 8.41
The way I would approach this problem is to see it in steps. First, you have to calculate the heat (q) released when the ice is melted and then calculate heat from the water changing its temperature in response to the ice cube's presence. After identifying the information you're given, you know that...
- Sun Jan 28, 2018 10:24 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: entropy and spontaneity
- Replies: 1
- Views: 174
Re: entropy and spontaneity
I think that a good way to think about this concept is to remember the picture of the gas contained in a flask (blocked by a valve that prevents it from diffusing to the rest of the flask). When you open the valve, the gas diffuses out into the rest of the flask and therefore, entropy has increased ...
- Sun Jan 21, 2018 3:29 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.115
- Replies: 2
- Views: 209
Re: 8.115
You could do final minus initial, but that would give you a negative change in moles. The question is asking how much H2 was burned, which already implies that the system is losing X amount of moles. The answer is positive because you wouldn't say -X moles of H2 was burned, but rather that X moles o...
- Sun Jan 21, 2018 3:04 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: How to calculate bond enthalpy
- Replies: 1
- Views: 326
Re: How to calculate bond enthalpy
Bond enthalpy calculation is reactant minus products because if you think about it, bond enthalpy is defined as the energy it takes to break 1 mol of a bond. Using this, positive values would be given to broken bonds (in the reactants) and negative values for bonds formed. Technically, enthalpy of r...
- Sun Jan 21, 2018 2:48 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Textbook Example 8.13
- Replies: 1
- Views: 272
Re: Textbook Example 8.13
I couldn't find the bond enthalpy of a Br-Br bond, so I just followed the enthalpy given on the example. After identifying the bonds broken and formed in the reaction as shown, you basically calculate the regular enthalpy of reaction using bond enthalpies. I'm not sure if you had stopped there (whic...
- Sat Jan 13, 2018 11:13 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpy vs. Bond Dissociation Enthalpy
- Replies: 3
- Views: 474
Re: Bond Enthalpy vs. Bond Dissociation Enthalpy
I believe there is a difference between the two. Bond enthalpy refers to the amount of energy it takes to break one mole of a bond. On the other hand, Bond Dissociation Enthalpy is referring to the enthalpy change as two atoms that were previously bonded retain the electron they put into the bond in...
- Sat Jan 13, 2018 10:56 pm
- Forum: Phase Changes & Related Calculations
- Topic: Extensive/Intensive
- Replies: 5
- Views: 848
Re: Extensive/Intensive
All the replies above are pretty accurate. I also think it's important not to get mixed up between chemical properties and physical ones. Just remember that intensive and extensive properties are just like sub-categories of physical properties too (that is because properties like color, mass, and vo...
- Sat Jan 13, 2018 6:01 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Reversible vs irreversible
- Replies: 2
- Views: 152
Re: Reversible vs irreversible
I believe the reason why reversible processes can output the maximum amount of work is, if you recall, a reversible process must be done slowly and this causes the system to lose energy as heat much less than an irreversible process. Irreversible processes tend to be done more quickly and therefore ...