CHEMISTRY COMMUNITY

Ridhi Ravichandran 1E

A catalyst and intermediate are both species that are consumed in the reaction, but a catalyst is present in the beginning of the reaction and an intermediate is formed at the end of a step.

Ridhi Ravichandran 1E

Complex molecules have higher entropies due to the extra bonds and electrons, which increase the overall potential energy and disorder of the system.

Ridhi Ravichandran 1E

Additionally, a reversible expansion does more work than a irreversible expansion because it is the sum of many small expansion steps.

Ridhi Ravichandran 1E

Will we have to know how to work with substitution reactions on the final? I know we covered it in class, but there were no homework problems based on it and I'm pretty sure they weren't covered in chapter 15.

If so, I'm confused about what exactly they are?

Thanks!

Ridhi Ravichandran 1E

You would look at standard reduction potentials. the key here is the word reduction, however. you MUST look at reduction potentials, not oxidation potentials, because the signs matter. if you're looking at a reduction potential, the more positive, the more oxidizing power, and the more negative, the...

Ridhi Ravichandran 1E

A rate constant is simply a proportionality constant included in the rate law. A reaction rate is the actual rate of a reaction, so it tells us how long it takes for reactant to be used up and product to be formed. While a reaction rate always includes units of time, the units of a rate constant var...

Ridhi Ravichandran 1E

Because diamond is thermodynamically unstable with respect to graphite, this means that delta G is negative for this process and diamond will spontaneously turn into graphite. However, because this process is also kinetically stable, this means that the process will occur very slowly. The book is ju...

Ridhi Ravichandran 1E

The rate law for a zero-order reaction is: rate=k[A]^0. However, anything to the power of 0 is just one, so the rate is simply equal to k. k is a constant that doesn't change throughout the reaction, and concentration is not even part of the equation, so zero-order reactions do not depend on the con...

Ridhi Ravichandran 1E

You use integrated rate laws when you need to find the time or final/initial concentrations. Most problems involve using the integrated rate law; we deal with the differential rate law when we need to find reaction orders, etc.

Ridhi Ravichandran 1E

Yes, you can use any units as long as they match. In fact, for one of the HW problems, the solutions manual encourages you to not change g to mol/l because it is extra work and unnecessary, especially because the final answer asks for g.

Ridhi Ravichandran 1E

You usually swap the sign of the equation that needs to be oxidized. Since the back of the book only lists reduction potentials, and a redox reaction has both reduction and oxidation half reactions, we need to flip one of the equations to turn it into an oxidation. Turning a half reaction from a red...

Ridhi Ravichandran 1E

Yes, in order to make the overall cell potential positive (which is what we want for a galvanic cell), the anode needs to be on the left and cathode needs to be on the right.

Ridhi Ravichandran 1E

It cannot be used for ln as well. For ln, there is a different constant (it is listed in the book).

Ridhi Ravichandran 1E

Why do we need the electrolytes in a galvanic cell? For example, there will be a copper electrode immersed in a CuSO4 solution, and a zinc electrode immersed in a ZnSO4 solution. Why is it necessary to have these solutions?

Ridhi Ravichandran 1E

delta U equals 0 for isothermal reactions that are both reversible and irreversible. Additionally, it does not need to be an isothermal expansion, it can be a compression as well.

Ridhi Ravichandran 1E

Wmax actually represents the maximum non-expansion work a reaction can do. Basically, an exothermic reaction will release a certain amount of energy, and this energy can be harnessed to do other kinds of work, such as power a non-spontaneous reaction.

Ridhi Ravichandran 1E

Yes, it can be -nRT. This formula is especially helpful if you don't know the external pressure and only know the net change in the moles of gas.

Ridhi Ravichandran 1E

Delta U is zero for an isothermal (temperature constant) expansion/compression. It can either be a reversible, irreversible, or free expansion. The key idea here is that it is isothermal, so the internal energy doesn't change.

Ridhi Ravichandran 1E

The reaction they are referring to has the compound as the product. For example, the reaction H2 + (1/2)O2 ----> H2O would have a certain Delta G, which would also equal the delta G of formation for H2O. If delta G were negative, the reaction would move in the forward direction and H2O would spontan...

Ridhi Ravichandran 1E

Yes, bond enthalpies are only valid for compounds in the gas state. If you want to use this method, you must convert all substances to the gas phase first, then solve for the delta H of the reaction.

Ridhi Ravichandran 1E

For this problem, they assumed the volume was constant during the temperature increase and thus used the Cv value to find the change in entropy. However, why did they assume volume was constant? Why couldn't you have assumed pressure was constant and used the Cp value?

Ridhi Ravichandran 1E

There is no such thing as negative entropy, but a negative change in entropy exists. For example, a reaction that condenses from a gas to liquid would have a negative delta S because the liquid would occupy less possible states than the gas due to the decrease in temperature and volume.

Ridhi Ravichandran 1E

1 bar is usually standard pressure, but 1 atm is sufficient because it is close enough to one bar. 25 degrees celsius is not technically standard temperature, but it is the temperature at which most reactions take place. Another requirement for "standard state" is that the elements must be...

Ridhi Ravichandran 1E

The reason q=-w comes down to the fact that in isothermal expansions, delta U is equal to zero. (The temperature is constant, so the molecules have the same kinetic energy and same potential energy. Thus, there is no change in internal energy). We know that delta U equals q+w, so if delta U is zero,...

Ridhi Ravichandran 1E

If it takes more energy to break the bond, that means the bond is very stronger, and the attraction between the nuclei of the atoms is also very strong. According to Coulomb's law, stronger attractions mean a lower energy arrangement, which means more stability. Thus strong, high energy bonds have a...

Ridhi Ravichandran 1E

Work is defined as motion against an opposing force. In Chapter 8, we focus on expansion work, such as a gas expanding against a piston. The piston provides an opposing force. If it weren't there (if the gas was expanding into a vacuum, for example), no work would be done because there would be no o...

Ridhi Ravichandran 1E

Basically, a reversible pathway is one that is in equilibrium. The external and internal pressures are equal, so work is only done if the external pressure decreases slightly, causing the gas to expand to decrease the internal pressure, allowing the system to come back to equilibrium. That is why fo...

Ridhi Ravichandran 1E

For part b on 8.67, why do you have to include the heat released when methanol changes from a gas to a liquid? Isn't it already in the liquid phase? According to the solutions manual, the equation is C(gr) + 2H2 + (1/2)O2 ----> CH3OH(l). Doesn't this indicate that methanol is already a liquid? Also,...

Ridhi Ravichandran 1E

Yes, the lower the energy a molecule has, the more stable it is. That is why you see the dip in the potential energy diagram for atoms in a bond- atoms in a bond have lower energy, which account for the dip.

Ridhi Ravichandran 1E

I believe we do not have to know these equations. But it is basically because the different types of molecules possess different types of energy (such as rotational or vibrational), which is why the equations for each type of molecule are different.

Ridhi Ravichandran 1E

Heat is not a state property because it is defined as the energy transferred to/from an object due to a temperature change. This can be accomplished in different ways (such as mixing vigorously or actually heating the object through an external source), so the path matters. Thus, it is not a state f...

Ridhi Ravichandran 1E

Why does a reversible process do the most work?

Ridhi Ravichandran 1E

also, we know that bonds (versus independent atoms) are more stable arrangements. Since stability corresponds with lower energy, it makes sense that forming bonds releases energy and thus lowers the energy of the system.

Ridhi Ravichandran 1E

To add, while temperature should usually be in kelvin, you can still use celsius with this formula because it is only the change in temperature, not the actual temperature. So, not having to convert to kelvin will save you time.

Ridhi Ravichandran 1E

Additionally, you can manipulate equations and add them together to get the final desired equation. When you reverse an equation, the delta H changes signs, and when you multiply an equation by a constant, the delta H is multiplied by that constant. Then, you can add all the new enthalpy values toge...

CHEMISTRY COMMUNITY

Search found 35 matches

Re: Catalyst and Intermediate

Re: complex molecules

Re: Final Question W18: Reversible vs. Irreversible External Pressure

SN2 Reactions

Re: Ranking elements

Re: Rate constant

Re: Diamond/Graphite

Re: Dependence on concentration

Re: Differential and Integrated Rate Laws

Re: Integrated Rate Law Units

Re: swapping signs of E values

Re: Galvanic cell set up

Re: NERNST

Electrolytes in Galvanic Cells

Re: Isothermal, irreversible?

Re: Delta G= Wmax

Re: Work when pressure is constant

Re: delta U

Re: 9.13 Gibbs free energy of formation

Re: Bond Enthalpies

Example 9.5 (page 325)

Re: negative entropy [ENDORSED]

Re: standard states

Re: q=-w [ENDORSED]

Re: Bond Enthalpies

Re: work definition

Re: Reversible and Irreversible Pathways

8.67

Re: Kekule structures, Resonance, and stability

Re: Heat Capacity for atoms and molecules

Re: Heat not being a state property..

Reversible process

Re: Bonds Formed Exothermic

Re: q=mCdeltaT

Re: Hess' Law