CHEMISTRY COMMUNITY

Samuel G Rivera - Discussion 4I

To calculate total change in entropy (S) you add change in S of the surroundings and change in S of the system.

Samuel G Rivera - Discussion 4I

A catalyst does not affect equilibrium constant. Only temperature, in think, affects K.

Samuel G Rivera - Discussion 4I

I'm not sure if early or later steps are more common but I've seen more problems with the later step being the slower reaction rate which would create a "bottle neck" or a back up in intermediate reactants

Samuel G Rivera - Discussion 4I

The second order reaction has a positive slope on a graph with 1/[A] on the y axis and 1/time on the x axis

Samuel G Rivera - Discussion 4I

The anode has the reduction potential that is more negative I believe

Samuel G Rivera - Discussion 4I

The 0th, 1st, and 2nd order equations are provided on the equation sheet but the equation sheet doesn't specify which one is which. So if you are doing your test with the equation sheet I recommend labeling each one so you don't mix them up.

Samuel G Rivera - Discussion 4I

A way to think about this is that if the reaction in the battery/cell is non spontaneous and needs energy added to it, it defeats the purpose of a battery which is supposed to release energy spontaneously.

Samuel G Rivera - Discussion 4I

I think that we use it to figure out the rates of reactions with different temperatures.

Samuel G Rivera - Discussion 4I

How would you determine the cathode or anode of a cell by only knowing the E value of the half reactions (assuming the you don't know which electrode is being oxidized or which is being reduced)?

Samuel G Rivera - Discussion 4I

When you take the equation E = E(standard) - (RT/nF)lnQ and you change the concentration of products or reactants then you change the value of Q which is equal [P]/[R]. Therefore if you change a concentration, whether it be of products or reactants, your will change the cell potential (E).

Samuel G Rivera - Discussion 4I

That section in the notes reads "Step 1 must be the rate determining (slowest) step, since overall rate is proportional to the concentration of [NO2]^2.
Slowest step determines rate of reaction: rate = k1[NO2]^2"

Samuel G Rivera - Discussion 4I

I was under the impression that standard cell potentials were supposed to be positive but in 6N.1 the standard cell potentials of the cathode and anode cause it to be negative. Did I confuse two different types of problems? Is there a type of problem where the answer (in volts) needs to be positive?

Samuel G Rivera - Discussion 4I

I don't think that we will be tested on deriving since in the last midterm we were given the already derived equations we would need.

Samuel G Rivera - Discussion 4I

I remember hearing in lecture, regarding galvanic cells, that a difference in concentration between two solutions causes an electric flow. I don't really understand what this means. Could someone elaborate?

Samuel G Rivera - Discussion 4I

I'm confused about how you would know which metal in the galvanic cell acts as the cathode that receives electrons and which acts as the anode that gives electrons. I'm assuming that the more electronegative element becomes the cathode, but in a cell with zinc and copper, copper is more electronegat...

Samuel G Rivera - Discussion 4I

Are redox reactions balanced differently when they're taking place in an acidic solution?

Samuel G Rivera - Discussion 4I

Diamond is kinetically stable in relation with graphite because there is an activation barrier for it to become graphite and energy would need to be added. It is thermodynamically unstable in relation to graphite because it is at a higher energy level then graphite.

Samuel G Rivera - Discussion 4I

I'm in the 3pm lecture, but if it's the same song he played for us it's called Hazey by Glass Animals

Samuel G Rivera - Discussion 4I

The Van't Hoff equation helps us find how K (the equilibrium constant) changes when temperature changes.

(P.S. I was confused about what equation you were referring to in your post. Even though it was in the Van't Hoff equation section, I recommend being specific in the posts just to avoid confusion)

Samuel G Rivera - Discussion 4I

I don't think that work is necessarily 0 in a reversible, isothermal process. The values that are equal to 0 are delta U (change in internal energy) and delta H (change in ethalpy).

Samuel G Rivera - Discussion 4I

I understand that the salt bridge keeps both sides neutral, but why do both sides need to be neutral? What would happen if the salt bridge wasn't there?

Samuel G Rivera - Discussion 4I

A voltage (a potential difference in charge) causes a current. A current is the transfer of electrons from the anode (oxidized) to the cathode (reduced).

Samuel G Rivera - Discussion 4I

If E is positive, then the reaction goes in the forward direction and is spontaneous.

Samuel G Rivera - Discussion 4I

E(standard) is under standard conditions (1 atm pressure and 0 degrees C or 273 K) while E is more general and under any conditions depending on the problem.

Samuel G Rivera - Discussion 4I

For reversible you would use delta S = nRln(V2/V1). For isothermal irreversible free expansion you would use the same equation. The difference is that the change in entropy of the surroundings is - delta S for reversible and 0 for irreversible.

Samuel G Rivera - Discussion 4I

The state functions so far in class are the ones that appear with the Delta sign in front. Just remember that once in a while they don't have the delta sign in front. By the way these are G, H, U, and S so far. You could also use the word G.U.S.H. to remember.

Samuel G Rivera - Discussion 4I

Why does work equal 0 when, for example, a system pushes a piston out within a vacuum? If the piston is moving wasn't work done to move it?

Samuel G Rivera - Discussion 4I

I would have to see an example of a problem but if it is a reversible, isothermal process and temperature is constant, then Delta H would be 0 in all cases because there would be no heat transfer in and out of the system.

Samuel G Rivera - Discussion 4I

My TA sent us a list of equations and Cv wasn't changed to Cp when pressure is constant so Delta U = n*Cv,m*deltaT under constant pressure and constant volume. I was just wondering why since I thought you would need to use Cp under constant pressure.

Samuel G Rivera - Discussion 4I

Why does Delta U = n*Cv,m*deltaT apply for both constant pressure and constant volume?

Samuel G Rivera - Discussion 4I

When calculating Enthalpy of a reaction, its the other way around. You add up the bond enthalpies of the products and then subtract the sum of bond enthalpies of reactants. When you are given enthalpies of formation, the products keep the same sign because they are being formed while you subtract th...

Samuel G Rivera - Discussion 4I

Entropy really mean how many different ways stuff in a system can be reconfigured. So hypothetically if you have a box with a lot of space and objects inside of it, the system could have a lot of entropy because there are many ways you can rearrange the objects. On the other hand, something that's l...

Samuel G Rivera - Discussion 4I

You accounted for the for Um(rotation, nonlinear) but also need to take Um(translation) into account. They are both the same but you need to add them together in order take all motion into account. Adding them would give you 3RT which would then give you the answer in the book.

Samuel G Rivera - Discussion 4I

So if you take the heating curve, you have Q on the x axis and temperature of a substance on the y axis. If you add a little bit of heat energy (Q) and the temperature increases a lot, then the slope will be steep. This means that the substance has a low heat capacity because it did not take much he...

Samuel G Rivera - Discussion 4I

I believe that when you are balancing a combustion equation you usually want to have a 1 for the coefficient of the hydrocarbon, or the thing you're burning, even if it means that the O2 has a coefficient that's not a whole number, such as 1/2.

Samuel G Rivera - Discussion 4I

When you get a specific value, that is supposed to be an exact heat capacity so your answer would be more accurate. When you're not given that value, you can estimate using 3R/2, 5R/2, etc. The way you know which one to use is my checking if the molecules are monoatomic (1 atom), diatomic (2 atoms),...

Samuel G Rivera - Discussion 4I

What is the difference between reversible and irreversible processes in terms of thermodynamics. Does it have to do with volume changing back to what it was before? I'm confused on the text book's explanation. And also why do reversible processes do more work than irreversible ones?

Samuel G Rivera - Discussion 4I

I believe that is correct because DELTA H depends on the amount of the substance that reacted.

Samuel G Rivera - Discussion 4I

I'm not sure what "max" heat capacity refers to but a system's heat capacity is the amount of pressure needed to increase the temperature of 1 gram or 1 mole of the substance by 1 degree Celsius. If dealing with mass, its how much energy is needed to increase 1 gram of a substance. If deal...

Samuel G Rivera - Discussion 4I

I don't understand state functions other than that its just final minus initial, but examples of pathway functions are heat and work.

Samuel G Rivera - Discussion 4I

You are right that you subtract the energy from heat (492 J) from the total energy (982 J) to get the measure of work. The answer should be +490 J. You probably made a small mistake when you calculated for this problem.

But you're right: delta U = w + q

Samuel G Rivera - Discussion 4I

I know that for the 2nd method (Bond enthalpies) only the the bond enthalpies in diatomic molecules are accurate (i.e. O2, N2, F2). All the other ones are averages for various molecules making it the least accurate.

Samuel G Rivera - Discussion 4I

I'm a little confused about enthalpy. From what I understand, enthalpy is the capacity to transfer heat. Is this the same as energy being the the capacity to do work? The lines between these concepts are a little blurry for me.

Samuel G Rivera - Discussion 4I

In thermodynamics, heat is the energy transferred as the result of energy difference. Heat transfers from high temperature regions to low temperature regions. Work is the process of achieving motion against an opposing force, such as gravity. Energy is the capacity of a system to do work. You can ad...

Samuel G Rivera - Discussion 4I

I believe that there is a correlation. It would make sense since if there is a higher bond enthalpy, it would be expected for the atoms to be held closer together and therefore have a shorter bond length. I'm not 100% sure though.

Samuel G Rivera - Discussion 4I

Q is the state of the relationship between the concentration of product and reactant at a given time. K is the equilibrium constant which is where the reaction is approaching. Usually K is given. You can also find it by dividing [P]/[R] when the reaction is at equilibrium (and the concentrations of ...

Samuel G Rivera - Discussion 4I

I'm confused on how to find [OH-] from only knowing [H30+]. In problem 6B.9, when [H3O+] is 1.5M, the answer for [OH-] is supposed to be 1.5 x 10^-14 but I keep getting 6.67 x 10^-15. What I'm doing is taking Kw = [H3O+][OH-] and then dividing [H3O+] on each side but it gets me the wrong answer.

Samuel G Rivera - Discussion 4I

For this problem, I understand that the expected pH is 1.6 because you plug in the concentration of HCl(aq) into -log(x). However, I'm confused why the actual pH is 1.7. Is the solution 200mL or is it 250mL?

Samuel G Rivera - Discussion 4I

I keep hearing the 5% rule, but I'm still confused about what it is and when to use it.

Samuel G Rivera - Discussion 4I

Q and K have the exact same formula. Q can be any value depending on the state of the reaction. K is Q when the reaction reached equilibrium.

Samuel G Rivera - Discussion 4I

A higher pH means a lower concentration of Hydrogen ions (protons). To go from pH to [H3O+], you calculate 10^(-pH). Therefore the higher the pH value, the lower the concentration of [H3O+] and the lower the acidity.

I'm not sure if this answers what you were asking, but I hope it helps.

Samuel G Rivera - Discussion 4I

If you dilute a solution that is, for example, 1 mL to 10 mL, will [H3O+] and [OH-] both be divided by 10? My question is based off 6B.11.

Samuel G Rivera - Discussion 4I

Can you have a negative pH? If so, what does that mean conceptually? Does a concentration of 1 mean that 100% of the solution is H30+? And if that is true then how does that make sense?

Samuel G Rivera - Discussion 4I

For problems that don't say what Kw is, do we just assume that it is 1.0*10^-14 and that the pH scale is 1 to 14? Or is there some way to calculate Kw when just given, for example, [H30+]?

Samuel G Rivera - Discussion 4I

When going from an acid to a conjugate base or a base to a conjugate acid, is the difference in H ion always 1 or can multiple H ions be added or removed when becoming the conjugate base/acid?

Samuel G Rivera - Discussion 4I

I believe the question you are working on will usually tell you whether you are finding Q or K. From the problems I've done, you are often given a K value at the beginning and are told to find Q in order to compare it to K.

Samuel G Rivera - Discussion 4I

When calculating Q , does it matter whether it's written as R<->P or R->P? Is it different because R->P doesn't have a reverse reaction rate(or does it)?

Samuel G Rivera - Discussion 4I

This is Daniela Shatzki 2E:

There is a button that says "New Topic" that looks similar to the button that says "Post Reply." It should be near the top and bottom of the page and shows up on pages like "Equilibrium Constants & Calculating Concentrations."

Samuel G Rivera - Discussion 4I

I'm not really sure why either, but I think it makes some sense when you think about adding two equations that both have K=1. If you multiply like you're supposed to, then K is still 1 and you still have an equal number of reactants and products. If you add the K's, then you would get 2 which means ...

Samuel G Rivera - Discussion 4I

Also, if K is less than 10^-3, then the equilibrium lies towards the left and favors the reactants more. For this class specifically, I believe that a K value between 10^-3 and 10^3 isn't considered to significantly favor either reactants or products.

Samuel G Rivera - Discussion 4I

I understand that if the equilibrium constant, K, is 10 and you multiply 2 through the equation, then K becomes 10^2. For example if you have Br2 + 3F2 <--> 2BrF3 and multiply it so its 2Br + 6F2 <--> 4BrF3, then you would make the initial K to the power of 2. However, does this work with every numb...

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