CHEMISTRY COMMUNITY

Harjas Sabharwal 1G

Why is the entropy change at constant pressure greater than the entropy change at constant volume? I understand mathematically by looking at the formula, but I was curious why conceptually?

Harjas Sabharwal 1G

Haocheng Zhang 2A wrote:I asked my high school chemistry teacher this question, he said we don't have to convert them to the same unit becasue equilibrium constant is unitless, so I am wrong. Sorry about that.

Do we still use molarity and bars then or do we convert it to atm or some other pressure unit?

Harjas Sabharwal 1G

You need to convert bar to M. We need to keep unit the same. Think of the idea gas equation PV=nRT, then we have P/RT=n/V. The number of mole divided by volume is molarity, so P/RT=Molarity. That's how we convert bar to M. Hope it helps! You had to assume standard conditions, so temperature = 298 K...

Harjas Sabharwal 1G

In this class do we us potential energy, enthalpy or gibb's free energy for the y-axis?
Do we use reaction progress or reaction coordinate for the x-axis?

Harjas Sabharwal 1G

Do we use J/mol or J for activation energy?

Harjas Sabharwal 1G

From the Wednesday discussion electro-chemistry test question 8, how do you figure out which units of pressure to use in the mixed equilibrium expression? We can't use PV=nRT to find molarity because we don't know temperature.

Harjas Sabharwal 1G

How do you know temperature? It isn't stated in the problem.

Harjas Sabharwal 1G

I'm pretty sure that standard refers to the standard state of the molecule (Ex: gas, liquid or solid). It is also at 273.15K and 1 atm pressure.

Harjas Sabharwal 1G

The half life for first order is constant whereas the half life for second order reactions are double the previous half life. Hence, you can not apply the method to second order reactions.

Harjas Sabharwal 1G

By manipulating the equation to the y = mx + b format, you can compare your equation to the integrated rate laws for each order, and then determine the order of your expression.

Harjas Sabharwal 1G

The first equation you pointed out and the second equation are equal. It is just converting between ln and log.

Harjas Sabharwal 1G

the rate constant is constant for a specific reaction at a specific temperature. If the reaction is different then the activation energy would be different, and if the temperature is different then that would influence K as well.

Harjas Sabharwal 1G

The answer to your question depends on the specific rate law for the reaction you are discussing. Generally if you plug in the change and then compare the resulting rate laws, you should be able to figure out what the change in rate is.

Harjas Sabharwal 1G

It really doesn't matter whether you convert it to mol/L or not. Its pretty much up to personal preference and whatever helps you solve the problem.

Harjas Sabharwal 1G

The rest of the reaction proceeds much faster than the slow step, hence the slow step is the only thing holding every thing back. For example, if you have a bunch of funnels that pour water into each other. The smallest funnel hole will determine how fast the water flows even if the funnels before a...

Harjas Sabharwal 1G

the values themselves did undergo non-linear growth or decay, but we can modify the values to make the graph linear.

Harjas Sabharwal 1G

If product is present then it can influence the reaction rate. Hence, we study the initial rates (when little to no product has formed) so that we can ignore the effects of products on the reaction rate.

Harjas Sabharwal 1G

If you think about radioactive decay it almost always involves a large atom decaying into smaller ones or giving off gamma rays (energy), so it is probably safe to say that radioactive decay only has one reactant and thus is first order.

Harjas Sabharwal 1G

The poster above did a good job explaining the concept, so here is an example using an enzyme. Enzymes bring their substrates together and then alter their formation to make them more likely to react and thus start the reaction. However, it would be horribly energy inefficient if a cell had to creat...

Harjas Sabharwal 1G

A catalyst just alters the environment or makes it more likely that a reaction will occur. Effectively, it reduces the activation energy "hill" so that the reactants can react with each other. This can be shown by enzymes. Enzymes bring their substrates together and then alter their format...

Harjas Sabharwal 1G

When writing rate laws do we include solids and liquids in the rate law expression?

Harjas Sabharwal 1G

When do we include H+ in the cell diagram? Additionally, how do we know whether it is there to balance the charge or to partake in the redox reaction?

Harjas Sabharwal 1G

Ah sorry. I think I meant only reversible expansion.

Harjas Sabharwal 1G

Why is the ΔS_tot = 0 for a reversible free expansion?

Harjas Sabharwal 1G

I understand that ΔH⁰_f is 0 for an element in its standard state.
But then by using the equation ΔG⁰_f = ΔH⁰_f - TΔS⁰_f, shouldn't ΔG⁰_f have some other value than 0? because elements in there standard states do have some ΔS⁰_f value.
Ex: ΔG⁰_f = 0 - (298K)ΔS⁰_f ≠ 0

Harjas Sabharwal 1G

I previously understood that a reaction was spontaneous when ΔG < 0?
But I was reading through the textbook and it stated that a reaction was spontaneous if ΔS_tot > 0. I though that a reaction did not have to be spontaneous if ΔS > 0?

Harjas Sabharwal 1G

W represents degeneracy and is the various micro states that the molecule can have at a specific energy.

Harjas Sabharwal 1G

Some real life examples of spontaneous processes are:

water freezing below 0°C

ice melting about 0°C

a boulder rolling down a hill

Harjas Sabharwal 1G

When the volume of the system changes then the system has to do or have work done upon the system. For example if the system expands, it is doing work to more the surroundings out of the way. Hence, work is being done on the system. If a system is compressed, the surroundings are making the system s...

Harjas Sabharwal 1G

The textbook uses 1 bar for standard states but if I remember correctly we used 1 atm in class. Which one should we use on the exam?

Harjas Sabharwal 1G

Why is internal energy independent of volume for an ideal gas?

Harjas Sabharwal 1G

In section 8.10, it states "Note that for a monatomic ideal gas, both CP and CV are independent of temperature and pressure"(281). Why is this so?

Harjas Sabharwal 1G

What is the difference between internal energy and molar internal energy conceptually and mathematically?

Harjas Sabharwal 1G

When do we use the following equations/what do they mean?
Ideal gas, Cp = (5/2) R
Ideal gas, Cv = (3/2) R
and what does R refer to?

Harjas Sabharwal 1G

When do we use U = 3/2RT and when do we use U = 5/2RT and what does each equation represent? Additionally when do we take into account translational, rotational linear and rotational non-linear energies?

Harjas Sabharwal 1G

Why is reversible expansion is called reversible expansion because it is usually in one direction? Also why does the pressure on the outside have to be the same as that on the inside?

Harjas Sabharwal 1G

I don't understand the conceptual difference between heat(Q), internal energy(U), and enthalpy(H). I understand that from a mathematical standpoint U contains Q but I am confused as to what exactly the difference is.

Harjas Sabharwal 1G

Will we be given the values to convert between the different pressure units, converting from L*atm, and the various R values on the exams?

Harjas Sabharwal 1G

Intensive properties are properties that are not related to the size or mass of the system. (Think intrinsic). Extensive properties change with the mass or the size of the system. For example, for water will always boil at 100°C. Hence boiling point is an intensive property. However, the amount of e...

Harjas Sabharwal 1G

State properties are pathway independent. Meaning regardless of how the final product is achieved it will always have that particular value (in this case enthalpy). For example, if you wanted to make a NO 2 . The enthalpy of NO 2 will always be the same, regardless of how it was formed. Hence, we ca...

Harjas Sabharwal 1G

I was curious why fusion and melting mean the same thing with regard to ΔH_fus? Fusion and melting seem almost the opposite of each other seeing as how when something melts, some of the bonds in the solid break apart.

Harjas Sabharwal 1G

In lecture today we learned that bond enthalpy is equal to the amount of energy required to break the bond. I just wanted to clarify if bond enthalpy the same as dissociation energy?

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