CHEMISTRY COMMUNITY

Lindsay Kester 2L

Increasing temperature increases the energy of the atoms or molecules that are participating in a reaction, so it makes them collide more frequently and with higher energy. Because they collide more often and with higher energy, they overcome the activation energy barrier more often, so the reaction...

Lindsay Kester 2L

Mechanisms need to support the orders shown in the rate law- so if the rate law says that something is second order, then there should be two molecules used up in the slow step of its elementary reaction. However, mechanisms can only provide possible answers to the rate law, not prove one, and there...

Lindsay Kester 2L

Catalysts don't show up in the rate law because they are usually 0th order. Because they are not used up in a reaction, their concentration doesn't change the rate of a reaction and so it doesn't show up in the rate law.

Lindsay Kester 2L

Molecularity affects the order of the reaction as well- bimolecular reactions are second order, unimolecular is first order, etc.

Lindsay Kester 2L

You can derive all of the half life reactions, but since third and fourth order reactions aren't likely, we don't need to know them really.

Lindsay Kester 2L

You could also look at the slow step of a reaction and figure out it's molecularity. If it's A->B, then first order, A+A->B, second order, and so on.

Lindsay Kester 2L

Yes, the catalyst increases the rate of both the forward and the reverse reaction.

Lindsay Kester 2L

I think that since heterogeneous catalysts are usually used in the solid phase, their appearance in the rate law would depend on the phases of the other reactants and products.

Lindsay Kester 2L

When we graph K, we make it negative, because the reaction is using up reactants (at least initially). K itself is usually positive.

Lindsay Kester 2L

Enzymes are examples of catalytic reactions, where one molecule that isn't used up helps to facilitate the reaction of another set of molecules.

Lindsay Kester 2L

Essentially, at different temperatures and activation energies, K changes. You calculate it the same way, using concentrations, but you're going to get a different value at different states. We don't have a way to calculate new K's based on old ones yet, but we might learn one.

Lindsay Kester 2L

They're almost the same thing! ∆G° is just defined in terms of the standard molar Gibbs energies of the reactants and products; it's the difference in energy between the pure products and the pure reactants that won't change as the reaction proceeds.

Lindsay Kester 2L

For anything besides entropy (where it doesn't matter if something is reversible or irreversible), you should be told whether or not it's reversible or irreversible. If the problem just says volume changes from 2L to 5L, you assume it's irreversible. If it says something along the lines of "thr...

Lindsay Kester 2L

This depends on which C you're using. When you have to multiply C by n, it means that C is the specific heat per mole of substance, and when you don't, it's just specific heat. You can also tell if you look at the units; specific heat per mole will be in J/mol, while specific heat will just be J.

Lindsay Kester 2L

When delta G is negative, it means that the products are at a lower energy level than the reactants, and thus the reaction moves toward them. When it's positive, however, the reactants are at a lower energy level than the products, and so the reaction moves toward them.

Lindsay Kester 2L

However, entropy is a state function, so regardless of whether it's irreversible or reversible expansion, it's going to have the same entropy at the end, and you can use the same formulas.

Lindsay Kester 2L

You use them based on the information you're given, and depending on what is being changed (Is value constant, pressure, etc.).

Lindsay Kester 2L

The more moles of gas there are in the system, the greater the entropy because W=number of possible states^number of molecules, and when you increase the number of moles in a system you also increase the number of molecules.

Lindsay Kester 2L

Whenever heat is leaving something, the amount of possible spaces (W) that the molecules can take up decreases, so entropy does as well.

Lindsay Kester 2L

The K here is also a constant for all equations, and doesn't change with each reaction/molecule being used.

Lindsay Kester 2L

The equation is the number of possible states (In this case, 2) raised to the number of particles. If there were three possible states, it would be 3^n, and so on.

Lindsay Kester 2L

Gas molecules will never go back to a smaller volume because it would decrease the entropy, or amount of states, of that system. In a smaller volume, there are less possible organizations for the molecules. Objects always want to go to a state of greater entropy, and moving to a smaller volume would...

Lindsay Kester 2L

Yes, entropy is a state function, because the it describes the equilibrium state of a system, regardless of how it got to that state.

Lindsay Kester 2L

In class, I think we started with 8.11, which is the enthalpy of physical change.

Lindsay Kester 2L

Because an open system is connected to its surroundings, any heat added would be dispersed throughout its surroundings, so the total E of the system would remain the same. Therefore, it wouldn't work as a way to add energy to a system.

Lindsay Kester 2L

Radiation doesn't require the use of a medium- think of the Sun, for example, which heats us despite there being nothing directly connecting us. Convection requires a medium- when you put something less hot in something hot, the heat moves from the hot thing to the less hot thing.

Lindsay Kester 2L

In the reading for Chapter 8, it says that we should include table 8.2, which is a list of some the heat enthalpies. Do we need to memorize these for the test, or will they be provided?

Lindsay Kester 2L

I think this depends on your TA- a friend of mine took 14A last quarter and her TA understood that since most people have or have access to the solutions manual, we usually have the right answers, and just graded the homework on completion. Regardless, though, you should probably just check your ans...

Lindsay Kester 2L

To figure out if something is a closed system, you should first try to figure out what the system is. In a bomb calorimeter, the system is what is inside, the inner chamber. Because this system can't exchange heat or energy with its surroundings, it's a closed system.

CHEMISTRY COMMUNITY

Search found 29 matches

Re: Increasing Tempt

Re: mechanisms

Re: Catalysts

Re: Molecularity

Re: 3rd order, 4th order, etc.

Re: Order?

Re: Catalyst in a Equilibria

Re: Homogenous vs. Heterogenous Catalyst and Rate Law

Re: K constant

Re: catalytic reaction

Re: Rate Constant K [ENDORSED]

Re: delta G vs. standard Gibbs free energy

Re: Reversible vs Irreverisble

Re: Entropy equation

Re: Spontaneity and the Direction of a Reaction

Re: Irreversible Vs. Reversible Expansion

Re: Calculating deltaG

Re: Definition of Entropy

Re: 9.5

Re: Relationship between ΔG⁰ AND K

Re: gas with 2x positions

Re: Gas atoms

Re: Entropy [ENDORSED]

Re: Thermochemistry Sections in Ch 8

Re: Heating/Cooling A System

Re: Radiation vs. Convection

Heat Enthalpies We Need to Know

Re: Homework grading

Re: What type of system is a Bomb Calorimeter?