CHEMISTRY COMMUNITY

Uisa_Manumaleuna_3E

Another important equation for first order that my TA said we should know too, is the derivation of the 1st order equation (noted above). It makes it easier to work out problems in the HW too. When you derive the above equation for 1st order you get:
[A] = [A] initial x e^(-kt)

Uisa_Manumaleuna_3E

i think this is a typo in the book, because I saw that me and some others got 2.85 with the same units you did, just not with the 10^12 part. I think your answer is fine

Uisa_Manumaleuna_3E

What i did for this one was set up an ICE table for the equation, 2A ---> B + C, with the "end" category being after 115secs, rather than at equilibirium like we normally do. Once you put all the information in, you see that the end concentrations of A and B are .153 - 2x and x respectivel...

Uisa_Manumaleuna_3E

the above comment is absolutely correct. Always remember that the rates are gonna be different depending on the changing concentrations, so if you look at two concentrations changing at once, you can't really calculate the impact the concentration has on the overall rate change. So if you want to se...

Uisa_Manumaleuna_3E

When you right the overall reaction, all you do is look at the two steps and add up all the reactants on one side and all the products on the other. Any intermediates (anything that was made as a product of one step that was later used up as a reactant in the following step) should be cancelled out ...

Uisa_Manumaleuna_3E

try to remember the order in which we balance everything: 1. balance all the elements (that aren't Oxygen or Hydrogen) 2. add H20 to balance oxygen 3. add H+ to balance the Hydrogens 4. ADD ELECTRONS TO BALANCE THE CHARGES 5. scale the reactions so electrons cancel out like it was said before you ha...

Uisa_Manumaleuna_3E

the test specified for us if it will be basic or acidic, and I imagine that they'll advise us also on the final, because of the extra steps we have to calculate in basic solutions.

Uisa_Manumaleuna_3E

Your Q should be .005/.15, giving you the correct E(cell) to plug into the work equation.

Uisa_Manumaleuna_3E

In regards to any rule, just remember that this particular half reaction represents a sort of ground level for the other half reduction reactions. In other words, the electronegativities were calculated by comparing their interaction with a hydrogen electrode.

Uisa_Manumaleuna_3E

Also on the same topic of strength of reducing agents, sometimes you see metals that have two different E values (eg. Ag+/Ag2+ E=.80 or E=1.98),and sometimes this can make things a little confusing. But when comparing these kinds of elements I try to just look at these different electronegativities ...

Uisa_Manumaleuna_3E

if you are ever worried about what material will on a test or exam, just check the announcements on the class website page, that helps keep you updated.

Uisa_Manumaleuna_3E

the way I understand it, both the cold water and the hot water have the exact same specific heat capacity, so when we set them to one another (q of cold = - q of hot) it just makes it simpler knowing that we can divide both sides without having to do any extra calculations I imagine that if you do e...

Uisa_Manumaleuna_3E

table 4J2 in the textbook shows examples of the most stable forms of some elements. It shows us examples for H2, O2, Cl2, and a few others. Just really helpful in case you wanted to know!

Uisa_Manumaleuna_3E

well, technically speaking, the reaction wasn't at equilibrium at all. when we calculate the reaction quotient, Q, with the given pressures, we realize that Q is a different number than K so its not at equilibrium. the previous response is correct; the problem is just a matter of plugging in the var...

Uisa_Manumaleuna_3E

So it took me a while to work this one out but you basically do it in 3 steps 1. use the partial pressures to calculate the reaction quotient, Q 2. use the equation delta G naught = - RTlnK and substitute it for delta g naught in the equation delta g = delta g naught + RTlnQ, so you get: delta G = (...

Uisa_Manumaleuna_3E

You only reverse one of the reactions and change the sign of one of the standard potentials with the first method. If you try reversing and changing the signs with the second method, you get a different value entirely.

Uisa_Manumaleuna_3E

So I ended up getting to lecture a little late today and I missed Professor Lavelle's explanation on cell diagrams. His example in class was the equation: 2 Fe (3+)(aq) + Cu(s) -> Cu(2+)(aq) + 2 Fe(2+)(aq) with an inert conductor of Platinum. I understand that you need to write down the Anode half r...

Uisa_Manumaleuna_3E

I would kind of like to know if there is some sort of cut off professor lavelle could give us. It just helped me learning the Reaction Quotient (K) knowing what was "large" and "small", and making assumptions and all. If the professor ever specifies anything further about "v...

Uisa_Manumaleuna_3E

I might be wrong about this but I think have a larger reaction quotient actually makes the E smaller, considering the Nernst equation. We know that E is equal to the E standard minus (RT/nF)*(lnQ). So if the lnQ was to grow in number, the E standard would keep subtracting larger and larger values. I...

Uisa_Manumaleuna_3E

the equation itself is explaining to us how the nature of a battery is, essentially. if you want to know how much work (how much energy aka its potential power), then you have to consider three things - the max potential energy possible (E), faraday's constant (F, aka the charge of 1 mole of electro...

Uisa_Manumaleuna_3E

professor lavelle said that every time we're asked whether or not something is spontaneous we know that delta g has to be negative. that's it really.

Uisa_Manumaleuna_3E

in a discussion of mine for another class we also studied the Gibbs Free Energy equation and what conditions make it more negative and more positive. We made a cross diagram of when delta h was + or - crossed with when delta s was + or - to see which ones were more likely to create a negative delta ...

Uisa_Manumaleuna_3E

Enthalpy of formation of O2 is zero because it's in its most stable form of oxygen How would you know that it's the most stable form? Would you have to memorize it? Yeah, it's kind of a memorization thing. It's a gas, bonded between two identical atoms. And O2 as a gas is literally one of the simpl...

Uisa_Manumaleuna_3E

just remember that delta h / enthalpy is a state function. Therefore calculations will always be the Final State minus the Initial State.

Uisa_Manumaleuna_3E

Also, it also kinda helps to look at the answer you have and see which one (J/KJ) just makes more sense. Like, for some answers I would get something long like 1000000 J and that would be appropriate for the question. But like it just made more sense to shorten it to 1000 kj. But yeah, they're both ...

Uisa_Manumaleuna_3E

for calorimeters we're usually just worried about the heat capacity, q / delta t, so you don't have to worry about any other formulas like that. Calorimeters are nice because usually we have a ton of info to plug in with other systems and whatnot, but calorimeters are literally just simple specific ...

Uisa_Manumaleuna_3E

Expansion and compression both denote work, but they exert that work in opposite directions. It helps me keep in mind the visual of a balloon expanding and deflating. Both require a change of energy or movement, but one obviously has a positive growth and the other a negative. You just have to remem...

Uisa_Manumaleuna_3E

Friday's (today's) lecture IS included on the midterm, but it cuts off there, at Gibbs free energy. He explained in class that the Midterm will include everything up to today. In the outline of thermodynamics, there are two sheets of learning objectives; the midterm will have everything on the first...

Uisa_Manumaleuna_3E

try to think of the - and + signs as how we express the energy change from one perspective. Like it was said before, the reaction is exothermic so it'll leave to it's surroundings, lessening the amount of heat available. From this perspective, heat is being taken away and moved somewhere else.

Uisa_Manumaleuna_3E

Delta H is the change in enthalpy and it can be used to calculate the delta u (aka the change in internal energy of the system). The equation is delta u = delta h - (temperature)(delta S or entropy).

Uisa_Manumaleuna_3E

Understanding that you calculate the enthalpy of sublimation with the enthalpy of fusion and of vaporization lets you understand that enthalpy is additive, a state property. So for our purposes, it is helpful in that sense. However, you'll probably see that we don't actually calculate the enthalpy o...

Uisa_Manumaleuna_3E

Yup, internal energy is the delta u = q + w. just remember, if the reaction has a constant volume, then no expansion work is done, and delta u is equal to q subscript v.

Uisa_Manumaleuna_3E

You can also think of it on a physical level - open: literally open to pass energy in between system and environment (open-flask experiment etc) closed: physically closed (with a lid or piston) but clear to see that elements such as heat can still pass in and out of reaction isolated: physically clo...

Uisa_Manumaleuna_3E

It is a little confusing because we know Kelvin and Celsius to be different, but just remember that a majority of the calculations of temperature we do in this section are just asking for the CHANGE OF TEMPERATURE. Temp change the same way in Celsius and in Kelvin, by degrees. So it doesn't really h...

Uisa_Manumaleuna_3E

My TA explained that when we think of state reactions, it's like asking someone what floor of a building they're on. You say, "What floor are you on?" and the person responds with "first floor" or "second floor" or something like that. In order to know the answer to you...

Uisa_Manumaleuna_3E

State properties, as said above, are not dependent on the "in between"; rather, we only focus on the initial and final measurements to make calculations and assumption. State properties include energy, pressure, volume, temperature, density and heat capacity Non state properties (or proper...

Uisa_Manumaleuna_3E

It helps for me to understand why steam burns more than liquid water if I remember the graph he showed us in class. I think everyone else did a fantastic job already of explaining it, but when dealing with the phase changes, at least for me, it helped seeing the comparison of energy change water und...

Uisa_Manumaleuna_3E

Q and K require aqueous solutions because their concentrations do change in the system depending on the rate of reaction

Uisa_Manumaleuna_3E

You can usually tell if you have to make an ICE box because the question will specify an initial concentration of some sort ("x amount of moles was place into x amount of liters and allowed to reach equilibrium....") and then itll ask you for the equilibrium concentration. If you're ever a...

Uisa_Manumaleuna_3E

Reaction quotient is just you looking at the same system, at a certain point in time, and comparing it to what it should be at equilibrium. If they match up, great! your system's chilling at equilibrium. If not, you can see the direction in which its progressing to equilibrium.

Uisa_Manumaleuna_3E

Pure liquids and solids are always left out. And to find the change column, its just -x/+x with the molar ratios considered

Uisa_Manumaleuna_3E

Do yourself a favor and just practice the quadratic formula as much as you can. it just helps make it that much easier, solving everything

Uisa_Manumaleuna_3E

Definitely use molarity which is moles/L. This is important because for the K and Q expressions work with concentrations and partial pressures, and for the most part the hypothetical experiments use varying volumes for their processes.

Uisa_Manumaleuna_3E

I understand it as being less than 10^-3. It's a little tricky with 10^-4, but i think we're safe just saying that its less than 10^-3

Uisa_Manumaleuna_3E

In class Dr Lavelle told us to use approximation for sure when K is less than 10^-3. But its a little tricky if we have 10^-4. Definitely use approximation with 10^-5, but its a little uncertain for 10^-4

Uisa_Manumaleuna_3E

In terms of ideal gases, I just know we need it to make assumptions and judgments about what's happening to the gases during equilibrium. Someone else asked which gases were considered to be more "ideal" but I don't think we need to focus so much on the gases being ideal, rather we should ...

Uisa_Manumaleuna_3E

I know that in the Video Modules and pre/post assessments, the questions make us convert the PV=nRT equation to show how we can get Concentrations from given pressure; and then in another question we convert the equation, but this time to get Pressure from given concentration. So I imagine that it i...

Uisa_Manumaleuna_3E

You don't have to worry too much though about which one of the R constant values to use, because (like everyone else said) the units will match up to the one we're supposed to use. So if you end up plugging in one of the other R values to the equation, not all of the units will cancel out and your a...

Uisa_Manumaleuna_3E

Just think of the Reaction Quotient like you proofing the equation. You do just calculate it the same way you find K (with the concentrations raised to the power of their coefficients), but you're trying to see if the value of Q, with the given concentrations, will give you K. Because if it gives yo...

Uisa_Manumaleuna_3E

From what I understand, yeah, pretty much we learn homogeneous and heterogeneous to know what to include in the K or Q calculations. Because, really, we already recognize the phases of the molecules anyway, it just helps specifying where we include all the concentrations/partial pressures (as we do ...

Uisa_Manumaleuna_3E

Uisa_Manumaleuna_3E

The R constant has a lot of different values and whatnot, but we mostly only use the first one, 8.314 J*K^(-1)*mol^(-1). I think we mostly use this one because we focus on primarily on energy change while the other R values pertain to volume in Liters. Joules and kilojoules are usually the units we ...

Uisa_Manumaleuna_3E

Another way you can look at the differences between transition states and intermediates is in their role when it comes to writing out reaction mechanisms. Transitions states, when drawn out in mechanisms, usually show whatever compounds and elements in the reaction are reacting with one another and ...

Uisa_Manumaleuna_3E

I missed class the day we went over Ethane and Butane and how to calculate energy as a function of a dihedral angle (page 106 of the course reader). Can someone explain to me where we got the values plotted on the two graphs? Were they values we calculated in class? Or were they simply given to us?

Uisa_Manumaleuna_3E

small question on finding out which step of a reaction is the rate determining step.

The rate determining step IS the slowest step. Does that mean that this step takes more energy to be carried out in comparison to the faster step?

Uisa_Manumaleuna_3E

Uisa_Manumaleuna_3E

From what I heard in lecture intermediates have to be produced in the reaction and then used up. Whereas, catalysts are needed for the reaction to get started and then are somehow regenerated later in the reaction? Can someone verify that this is accurate for me? I don't want to go onto the next top...

Uisa_Manumaleuna_3E

Maybe I'm not reading the Course Reader correctly but I'm having a hard time figuring out the orders for each reactants just by looking at the initial concentrations and Initial rates given to us in the tables??? How do we know which two experiments to use when comparing the concentrations and tryin...

Uisa_Manumaleuna_3E

Uisa_Manumaleuna_3E

I was just wondering if we would have to memorize some standard reaction enthalpies for quizzes or the final? Maybe not for some of the intense or complex formations, but for things like the formation of water molecules maybe???

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