CHEMISTRY COMMUNITY

Masih Tazhibi 2I

What's an Organic Chemist's greatest fear?

A BACKSIDE ATTACK!

Masih Tazhibi 2I

This quarter is moving so fast! I really need to get it in........Kinetic control

Masih Tazhibi 2I

Suppose you have two possible products, A and B. If A is kinetically favorable, it means that it has a lower activation energy, and a higher rate constant. If B is thermodynamically favorable, it means that as a product, it resides at a lower energy state, and thus is more stable. The problem is, th...

Masih Tazhibi 2I

The slow step always determines the rate of the reaction. The problem is that when the slow step is not the first step in the rxn, the experimentally observed rate law doesn't always seem to match the rate law obtained by the slow step. This is because the slow step rate might include the concentrat...

Masih Tazhibi 2I

The pre equilibrium approach is used when the slow or rate determining step in a mechanism is not the first step, and thus you can't simply write out a rate law and be done. This is because writing out the rate law for the second step will not consider the concentrations of reactants in step 1, but ...

Masih Tazhibi 2I

You can use either version of the equation. It was discussed in class that scientists just prefer not having negative signs in their equations if they can avoid it, so they sometimes reframe the equation to that end. However, both are perfectly fine as long as you keep your k and T values consistent.

Masih Tazhibi 2I

Adsorption occurs when you have a heterogenous catalyst being used, that is, a catalyst that is in a different phase than the reactants. Usually, adsorption occurs when your catalyst is a solid, and the reactants, rather than going inside or being absorbed by the catalyst, simply sit on top of it as...

Masih Tazhibi 2I

Reaction mechanisms are found experimentally, and are very similar to putting together a puzzle. You know through experimental data that a mechanism will create the final reaction, but you need to (through trial and error) figure out how the initial reaction gets to its final form. Creating a mechan...

Masih Tazhibi 2I

I have heard that when plugging in values for Ea, you want to convert the units to joules, or else your answer will be incorrect. Is this because R is in J/mol*k?

Masih Tazhibi 2I

The slope of the first order reaction can be found through assessing the integrated rate law. The first order integrated rate law is ln(A)=-kt+ln(Ao), which is in the linear y=mx+b form. As such, the slope is -k.

Masih Tazhibi 2I

Lucian did a fantastic job explaining the rate orders here. I would also like to add that experimentally, you determine rate orders by isolating the impact of one reactant at a time on the overall reaction rate. If you have two reactants, you would keep one constant and change the concentration of t...

Masih Tazhibi 2I

As has already been stated, knowing the derivations really adds another layer of depth to your understanding of the topic. It's great to have an equation sheet to give you the formulas, and for most of the problems you do it's nice to have them on demand, but knowing the way these equations came abo...

Masih Tazhibi 2I

Another key point is that thermodynamics really buckled down on the initial and final states of a system, dealing mostly with state functions and just looking at relative changes in work, energy, and entropy. What seems to be the most different about kinetics is that it focuses on the very path port...

Masih Tazhibi 2I

One of the big things that I've noticed is that thermodynamics was heavily concerned with state functions, that is, the initial and final relative states of a system without respect to the path the reaction took. I think Kinetics is concerned with the very path dependency that thermodynamics ignored...

Masih Tazhibi 2I

When we are calculating cell potential, it is usually a means of calculating the Gibbs Free Energy of the redox reaction taking place. This is the maximum work that the cell can do in the form of electrical energy, which is what we use these cells for in the first place.

Masih Tazhibi 2I

We also use the initial values because it makes our calculations a lot simpler. This is because you only have to consider the reactant concentrations, rather than taking into account both products and reactants, as well as the extent of both the forward and reverse reactions.

Masih Tazhibi 2I

..Also, as the concentrations of the two half reactions change, the voltage also changes

Masih Tazhibi 2I

In general I think for the purposes of this class you are looking for the transfer of electrons from one half reaction to the next. There would thus be two cells in system.

Masih Tazhibi 2I

The idea behind this law is that the universe is always heading towards some increase in entropy. The entropy of the surroundings plus the entropy of the system is greater than zero because in a spontaneous process, the system has not reached thermal equilibrium. Since Entropy is Q/T, having the sam...

Masih Tazhibi 2I

I think in general, there are a couple things to look at when comparing molecules. The first is the state each molecule occupies, as we know that Sgas>Sliquid>Ssolid. Additionally, it is important to compare the relative size of the molecule, as an increase in the number of bonds and electrons leads...

Masih Tazhibi 2I

It is actually possible to have a negative change in entropy, but we don't typically see those in what we are dealing with. However, you do see negative entropies quite frequently in aqueous ions. It's just good to know they are out there, but the main thing you should know about entropy is that it ...

Masih Tazhibi 2I

The 3/2 and 5/2 is for ideal gases that have different properties. For gases that are not ideal, you use the one without Cp.

Masih Tazhibi 2I

A big trend in thermodynamics is that it is hard to conceptualize and analyze these concepts in and of themselves, as some quantity. It is much more useful to analyze concepts like enthalpy, entropy, and Gibbs free energy as relative quantities, where you look at changes with respect to differing co...

Masih Tazhibi 2I

While the internal energy is zero, the gas is expanding against a vacuum which has no resistance.

Masih Tazhibi 2I

Changes in entropy are based on the equation q/T. It intuitively makes sense that if you held q constant in both cases, lets say 100J, the change in entropy would be less with a higher temperature. For example, 100J/200K is greater than 100J/500k. This is logical because adding a constant amount of ...

Masih Tazhibi 2I

In this problem, you have both the volume changing and the temperature changing. Thus, there are two different factors impacting the total change in entropy. The first is positional entropy, due to a change in the volume. This entropy change takes into account the increase or decrease in entropy whe...

Masih Tazhibi 2I

The sign for the change in entropy comes from the q value in the numerator. Thus, if your system is releasing heat, the change in entropy of the system is decreasing, and vice versa.

Masih Tazhibi 2I

The integral is to show the more scientifically accurate way of writing your PdeltaV equation. They essentially are doing the same thing in a nonreversible expansion, but when things change, it is good to know conceptually where the integral is coming from, and how it could further be applied. I wou...

Masih Tazhibi 2I

Yes, adding matter would increase the internal energy of a system, because there are more substances to react in the system. Taking matter out of the system would limit the extent of reaction you would have had if the matter was not removed.

Masih Tazhibi 2I

It's important to also know that the reason diatomic molecules have standard enthalpies of formation at zero is not because they are diatomic in and of itself, but it is a consequence of the fact that diatomic elements are pure substances. For example, molecules like C(s) or K(s) or O2 etc., are all...

Masih Tazhibi 2I

Another important thing to note is that the bond enthalpies have a role in determining the overall Enthalpy of the reaction. If you take the summation of the enthalpies of bonds broken, and subtract from that value the summation of the enthalpies of bonds formed, you get the total enthalpy of the re...

Masih Tazhibi 2I

For all intents and purposes, I think it is just important to know the concept behind their interchangeability. q is used to describe values of released or absorbed heat, and using the subscript p to create "qp" just means you are looking at the change in heat at a constant pressure. This ...

Masih Tazhibi 2I

It is also important to know that a state property is independent of how the substance was prepared. Good examples of state properties include things like altitude, pressure, volume, temperature. Note that when looking at these values, it isn't important to track the path they have taken; with tempe...

CHEMISTRY COMMUNITY

Search found 33 matches

Re: Post All Chemistry Jokes Here

Re: Post All Chemistry Jokes Here

Re: Kinetically controlled reactions

Re: Slow Step Determining Overall Rate

Re: Pre-Equilibirum Approach

Re: Confusion on different versions of the Van't Hoff equation?

Re: Adsorption

Re: Kinetic information [ENDORSED]

Re: Arrhenius Equation [ENDORSED]

Re: slope

Re: Difference between zero, first, second [ENDORSED]

Re: Test 3

Re: How to tell difference btwn thermo and kinetics [ENDORSED]

Re: kinetics

Re: Electromotive Force

Re: Initial rates

Re: Concentration Cells

Re: Galvanic cells

Re: 2nd law -- conceptual

Re: Number of Bonds and Entropy

Re: negative entropy [ENDORSED]

Re: deltaS question

Re: Question about Change in Gibb's Free Energy

Re: Delta U (Concept Clarification)

Re: Entropy based on initial temp

Re: Practice Example 9.5

Re: Sign of ΔS

Re: Integral from Today's Lecture

Re: Concept Clarification

Re: Standard enthalpies on test 1 [ENDORSED]

Re: Bond Enthalpies

Re: qp=deltaH

Re: Enthalpy is a state property