CHEMISTRY COMMUNITY

Ramya Lakkaraju 1B

We can make out which compounds are intermediates since they are the ones that are produced in one step and then used up in another step; they are not present to begin with or left over after the reaction is over.

Ramya Lakkaraju 1B

I don't think we have to worry about what components make up the total energy.

Ramya Lakkaraju 1B

You can only have 1 slow step and that is the step that matches the overall reaction's rate law.

Ramya Lakkaraju 1B

Yes, reaction mechanisms and pseudo reactions will not be on test 3.

Ramya Lakkaraju 1B

The rate constant is a constant and therefore must stay the same throughout.

Ramya Lakkaraju 1B

I think you should just know that thermodynamics controls whether or not a reaction will happen (whether or not it is favorable). Kinetics controls how fast the reaction proceeds.

Ramya Lakkaraju 1B

It is important to choose 2 reactions in which only the concentration of one reactant is changing. This is so that you can say for sure that the change in rate is due to the change in the concentration of that one reactant.

Ramya Lakkaraju 1B

The more accurate method would be to analyze changes in both the reactant and product concentrations.

Ramya Lakkaraju 1B

There are no products at the beginning of a reaction since none of them have been formed yet.

Ramya Lakkaraju 1B

The solutions manual is just trying to distinguish between the different reactants and products so that the reader doesn't get confused as to why there are multiple same answers for different reactants/products.

Ramya Lakkaraju 1B

I think its main use is to integrate this and use it as the integrated rate law.

Ramya Lakkaraju 1B

All the unique rates of reactions for all the reactants and products in a reaction are the same.

Ramya Lakkaraju 1B

Yes, you should not ignore the units of anything when using equations because the units must make sense in the final answer.

Ramya Lakkaraju 1B

I think it might have to do with the fact that delta G is a state function so you can add each step's delta G to get the total.

Ramya Lakkaraju 1B

I looked in the appendix for that one since I don't think there is a straight forward way to come up with them.

Ramya Lakkaraju 1B

I think since the bromine is going from -1 to 5 (meaning it's oxidation), the oxygen would have to be reduced for this to be a redox reaction.

Ramya Lakkaraju 1B

I think it is because the change in oxidation state of H2 is associated with the hydrogen part of hydrochloric acid. And the oxidation state of Cl2 is associated with the chlorine part of hydrochloric acid. Due to this, you have to keep them separate.

Ramya Lakkaraju 1B

Once we determine that delta U = 0, we can manipulate q and w and set q=-w.

Ramya Lakkaraju 1B

Make sure to distinguish between delta G and delta G knot since I think that is what is confusing.

Ramya Lakkaraju 1B

This fact that delta U = 0 allows us to set q = -w and do other calculations involving heat and work.

Ramya Lakkaraju 1B

This is also true for enthalpy and entropy, also state functions. If the initial state and final states are the same, then the change in that quantity is zero.

Ramya Lakkaraju 1B

Do we have to know this theorem for the midterm? I do not think Dr. Lavelle discussed it in very much detail in lecture.

Ramya Lakkaraju 1B

Increasing volume also results in an increase in entropy since the gas can now have more microstates.

Ramya Lakkaraju 1B

Is there ever a situation in which the entire beaker would be the system and for example, the Earth would be the surroundings?

Ramya Lakkaraju 1B

Sometimes, when you are trying to calculate values for a state function, it does not matter whether you use data from reversible or irreversible reactions since all you need is final minus initial; the path does not matter.

Ramya Lakkaraju 1B

In the equation S=qrev/T, why do we need to use the q for a reversible reaction?

Ramya Lakkaraju 1B

Since q and w are not state functions, you cannot do calculations with just final minus initial.

Ramya Lakkaraju 1B

The system's q sign and the surrounding's q sign will always be opposites since whatever the system is losing is what the surroundings gain.

Ramya Lakkaraju 1B

Gibbs free energy is basically the amount of energy that is available to do useful work in the system. The equation for Gibbs free energy takes the total enthalpy of the system and subtracts whatever energy is lost through the t delta s term. The remaining energy is the amount of energy left to do u...

Ramya Lakkaraju 1B

I just wrote the equation, delta U = q + w.

Ramya Lakkaraju 1B

Whenever the sign of something is not immediately obvious think about whether the system or the surroundings are gaining or losing heat; losing heat is always signified by a negative sign.

Ramya Lakkaraju 1B

I think the difference is that when you want the standard entropy, you are forming 1 mole of the product with everything in its standard states.

Ramya Lakkaraju 1B

Yes, if no other temperature is given in the problem, it is safe to assume that the reaction occurred at room temperature which is 298 K.

Ramya Lakkaraju 1B

Yes and Dr. Lavelle also said we will be expected to know concepts related to work and internal energy even though we do not need to perform calculations related to them.

Ramya Lakkaraju 1B

In problems like these (with the subscript c), you have to set up a combustion equation with the reactant and oxygen on the left and carbon dioxide and water on the right.

Ramya Lakkaraju 1B

I find it easy to find the desired reactants and products and arrange them with reactants on the left and products on the right. From there, I just multiply the various equations by whatever is necessary to cancel other substances out with each other.

Ramya Lakkaraju 1B

In most problems, if a specific temperature is not given, you should use 298K as it is considered standard.

Ramya Lakkaraju 1B

It is useful to use the specific heat capacity so that the mass becomes irrelevant. Specific heat capacity is an intensive property.

Ramya Lakkaraju 1B

In the table of enthalpies, there is a specific number for the bonds found in a benzene ring.

Ramya Lakkaraju 1B

In state properties, the path taken does not matter; only the final and initial values matter when doing calculations related to state properties.

Ramya Lakkaraju 1B

The most stable form of a compound is when the electrons are as far away from each other as possible, which is what occurs in this example.

Ramya Lakkaraju 1B

So far, we can also do problems 37 and 39 before number 55.

Ramya Lakkaraju 1B

Yes, whenever the mass of a substance is given in grams, use specific heat capacity. If it is given in moles, use molar heat capacity.

Ramya Lakkaraju 1B

Another mistake I ran into while solving this question was forgetting to convert the enthalpy of fusion into Joules because in the tables, it is given in kJ.

Ramya Lakkaraju 1B

From what I understand, the only reason that nitrogen is not included is because its enthalpy is 0 if you look at the table. I don't think it has to do with the fact that it is in its gas phase.

Ramya Lakkaraju 1B

Look at formal charges, you know that phosphorus likes to make 5 bonds since it has a valence number of 5, however in the case of PF6 it is making 6 bonds, resulting in a -1 charge.

Ramya Lakkaraju 1B

How strict is the 5% rule? If the approximation is slightly more than 5% off, would we have to recalculate everything with more accuracy?

Ramya Lakkaraju 1B

The transition metals act as Lewis acids, accepting the donated electron pair.

Ramya Lakkaraju 1B

The usual prefixes are bi for 2, tri for 3, tetra for 4, etc. However, if the ligand is polydentate, you use bis for 2, tris for 3, tetrakis for 4 and so on.

Ramya Lakkaraju 1B

When an endothermic reaction is heated, product formation is favored.
When an endothermic reaction is cooled, reactant formation is favored.
When an exothermic reaction is cooled, product formation is favored.
When an exothermic reaction is heated, reactant formation is favored.

Ramya Lakkaraju 1B

In lecture, Dr. Lavelle mentioned that it does not matter in which order you write them.

Ramya Lakkaraju 1B

Although working with the hybridization of carbon is easy since we know for sure that carbon makes 4 bonds, the hybridization of other elements works in a similar way.

Ramya Lakkaraju 1B

Can anyone explain the reasoning behind number 43? The question is:
Do you expect the bond angle between 2 hybrid orbitals to increase or decrease as the s-character of the hybrids is increased?

Ramya Lakkaraju 1B

We are only expected to know the general bond angles (such as less than 120, etc) since the exact bond angles can only be determined from experimental observation.

Ramya Lakkaraju 1B

From what I understand, there will only be one radical since if there were two, they would combine to form an electron pair.

Ramya Lakkaraju 1B

The exceptions to the bond angles are caused by repulsion by the lone pairs of electrons. In lecture, Dr. Lavelle mentioned that we only had to know what the bond angles should be for most shapes and that we did not have to memorize the exceptions.

Ramya Lakkaraju 1B

Resonance shouldn't have an effect on the shape since technically, a resonant structure is the same structure put together in a different way. However, you should use formal charges to decide on the best lewis structure before you decide on the shape.

Ramya Lakkaraju 1B

The resonance structure in which you can get the most number of FC to equal 0 is the most stable and should be used.

Ramya Lakkaraju 1B

What is meant by a sea of electrons?

Ramya Lakkaraju 1B

All the transition metals that are in the same group as chromium and copper will want to make both their s and p subshells half full so instead of s2 p4, it will be s1 p5

Ramya Lakkaraju 1B

Generally atoms that have half or full orbitals are the most stable so they will not want to add electrons that easily.

Ramya Lakkaraju 1B

The frequency should always be positive, so subtract the bigger number from the smaller number to get a negative number and multiply it by the Rydberg constant times -1.

Ramya Lakkaraju 1B

Yes, if the energy of the photon exactly matches the work function of the metal, the electron that is emitted will have zero kinetic energy.

Ramya Lakkaraju 1B

Another common problem concerning the De Broglie Equation is one in which they give you the wavelength and mass and ask you to calculate the speed. In all these problems, they give you 2 of the 3 variables (Planck's constant is known) and you are asked to calculate the third one.

Ramya Lakkaraju 1B

The term black body came about since these objects appear black in color at room temperature. They absorb all light and do not reflect any light.

Ramya Lakkaraju 1B

Assuming you use temperature in your calculations, it does count for sig figs because it is a measurement. In general, anything you measure that does not have 100% accuracy counts for sig figs and any exact values, like conversion factors, do not count for sig figs.

Ramya Lakkaraju 1B

Hi, the reason you would start "fresh" is that when you balance an equation, you are simply trying to make sure that for that particular equation, the two sides are in fact equal to each other. When you start with another equation, you do not include any of the stoichiometric coefficients ...

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