CHEMISTRY COMMUNITY

Yashaswi Dis 1K

It can because rate law is experimentally determined so if the experiment yields a result that allows for a negative reaction order then that is possible. Typically I don't think we will be dealing much with negative reaction order because it is not as common but it is good to know that it can exist.

Yashaswi Dis 1K

delta G is equal to w_max with a process at constant temperature and pressure I believe. I am not entirely sure.

Yashaswi Dis 1K

I don't think you can use the stoichiometric coefficients from the slow determining step but you can use the reactants and see if what type of molecular reaction it is (i.e. bimolevular, trimolecular, etc.). We use it from the slow-rate determining step because that is the reaction that if it's slow...

Yashaswi Dis 1K

Well, rate is usually considered in terms of the consumption of reactants which is easier to measure. So if you had A + B <----> C + D, a hypothetical forward reaction's rate law would like rate = k[A][B], for example. For the reverse reaction, where the reaction mechanism does not change, now it wi...

Yashaswi Dis 1K

Sure. So here's how it goes: Rate = -d[A]/dt = k[A] 0 for zero order reactions where [A] 0 = 1, so rate = -d[A]/dt = k. Then you have to do a method called separation of variables which is used in calculus where you multiply the dt to the other side to get all the d(stuff) on each side (kind of like...

Yashaswi Dis 1K

It's because it is specified in the question that the two following equations will be second-order so you just have to use the second-order integrated rate law because it explicitly says that: 1/[A]_t = 1/[A]₀ + kt.

Manipulate to solve for t and you should get 3.3 x 10³ min.

Hope that helps!

Yashaswi Dis 1K

Yeah for similar problems like that, I suggest looking at the given experimentally determined rate law and then look at all the reactants in the equation and if any reactant is not shown in the rate law shown, then it's most likely zero order because [Reactant]^0 = 1. Hope that helps!

Yashaswi Dis 1K

Yeah, I don't think it's always the case that the number of steps equals the number of reactants present. I am note sure however.

Yashaswi Dis 1K

I am not sure if I will be able to exactly answer your question, but I think each order has it's own integrated rate law. To find the order, I suggest using the given concentrations and if time is also given, then plot each type of integrated rate law and see what the order is based off of that.

Yashaswi Dis 1K

I am not sure if I will be able to exactly answer your question, but I think each order has it's own integrated rate law. To find the order, I suggest using the given concentrations and if time is also given, then plot each type of integrated rate law and see what the order is based off of that.

Yashaswi Dis 1K

I am not sure but based on what we have seen in class, then yes.

Yashaswi Dis 1K

I don't have the textbook in front of me but if you could post the question, it might be easier. Thanks!

Yashaswi Dis 1K

Yeah, I am confused on that too. I don't remember the exact question, but I think if it's in a basic solution you might choose something like KOH to act as a salt bridge? Hopefully someone else can explain better than me.

Yashaswi Dis 1K

I heard O-chem is difficult. Those who study it have alkynes of trouble.

Yashaswi Dis 1K

Disclaimer: A Corny Acid-Base joke
What did H₂PO4^- say to H₂O?

"Why you gotta be so BiPoLaAr!"

Water: "As an amphoteric molecule, I like to have multiple affairs!"

H₂PO4^-: "Can't believe I am dating a weirdo."

Yashaswi Dis 1K

I am not entirely sure but whichever one has the correct balanced equation (I think it's the one in solutions manual?). It makes sense for 3/2 to be a coefficient for O₂ to get 3 moles of oxygen in the product. Hope that helps a little!

Yashaswi Dis 1K

Yes. Technically speaking, the bigger the molecule is implies that the molar entropy will be higher because there are more atoms that can have different positional arrangements (microstates - W), more atoms means more bond vibration energy, and a higher rotational energy etc. all contributing to the...

Yashaswi Dis 1K

Voltage is the unit! Remember to help with conversions, 1 J/C (C=coulumb) = 1V.

Yashaswi Dis 1K

Yes, but you use -delta H_fus for delta H because it's freezing, not melting and keep the temperature the same. Hope that helps!

Yashaswi Dis 1K

Hi, So delta S according to the Second Law of Thermodynamics will always be increasing in the universe. Delta S of the system can decrease, in other words, if you freeze liquid water to ice, you are reducing the possible number of arrangements (microstates) that the H 2 0 molecule can be in. So delt...

Yashaswi Dis 1K

I am not sure exactly on the lowest value of G, but I do know that delta G can have a positive value, negative value, and a value of 0 because it is a state function so you have to think about final minus initial.

Yashaswi Dis 1K

Hi,

Yeah so to find Q, you can just plug in the partial pressures which is directly proportional to concentrations, so Q = (.98)^2/(.13). I hope that helps! That should then give you the right answer hopefully.

Yashaswi Dis 1K

I think they are essentially the same when we do the computation. The delta G of formation is the energy it takes to make the molecule you want to find from its base elements, which is different from molar Gibbs' free energy which is for molecules in their standard states at 1 mole (it's not necessa...

Yashaswi Dis 1K

Heat results because of a temperature difference. We know temperature is a state function because you can do T (final) - T (initial), but it doesn't really make sense to find delta q = q(final) - q(initial) because q is dependent on the pathway. Think of it like this: Every molecule motion due to th...

Yashaswi Dis 1K

Work done by system: negative sign for work Work done on system: positive sign for work. Another way to help differentiate the two: Think of the equation: w= -P * (delta V). ---->If volume expansion occurs, then final volume minus initial volume is going to be positive so with the negative sign in t...

Yashaswi Dis 1K

-q cal = q (of the main reaction). You can find delta U from there since w=0 because there's constant volume (delta V is 0) so delta U is equal to q. There's a thread on this same question too which I answered in a little bit more detail and can check out here: https://lavelle.chem.ucla.edu/forum/vi...

Yashaswi Dis 1K

In part a, the process is irreversible meaning there is an external, constant pressure and the amount of work is usually less than in part b, in which it's a reversible process, where more work is done. Equation for part a to use: w = -P*deltaV and for part b it's: w = -nRT ln (V f /V i ). In a reve...

Yashaswi Dis 1K

The way I understood was that there can be heat present in the system already but no further heat transfer is allowed; only work energy is transferable. Since delta U = q + w, then U = q if w = 0.

Hope that helps!

Sincerely,
Yashwi

Yashaswi Dis 1K

Here's Question 8.21: A piece of copper of mass 20.0 g at 100.0 degrees Celsius is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 degrees Celsius. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings. I understand how...

Yashaswi Dis 1K

I think the value that we really need to find here is change in internal energy (delta U). We know that delta U is equal to q + w. Since the question involves a bomb-calorimeter (constant volume), we know that w should equal zero since w=-P*deltaV and delta V is zero. So we only need to really find ...

Yashaswi Dis 1K

Hi, I think you brought up a good point. I am a bit confused by your question but I think what they are saying essentially is during a reversible process, a system is in thermodynamic equilibrium with its surroundings. So for every reduction in external pressure, the volume usually changes infinites...

Yashaswi Dis 1K

I think an easier way to think of it is like this: If a molecule has electron lone pairs on it, according to the Lewis definition, it will be a Lewis base because it is donating an electron pair, while the Lewis Acid is the molecule that accepts the electron lone pair. So for instance, if I have: NH...

Yashaswi Dis 1K

Hi, So generally weak acids or weak bases when they are mixed with water, they don't dissociate all their protons and that's why we write them as molecules instead of separating them like we do for Ca(OH)2 for example. HF is a weak acid so we leave it, we don't need to separate it. Hope that helps! ...

Yashaswi Dis 1K

Using shift is fine, but it is better to describe the reaction as either sitting to the left, if there are more reactants, or sitting to the right, if there are more products. Maybe, I am not sure considering we didn't do LeChatelier's Principle yet, if you have to describe which way the reaction wi...

Yashaswi Dis 1K

Hi, I don't have the textbook with me right now. Maybe if you could please post the question up? For your general question, say I have 3H 2 + N 2 <---> 2NH 3 . To set up the ice table, let's assume that we have 0 NH 3 initially and want to form that product. In the change section of the ICE table, w...

Yashaswi Dis 1K

@Michael So a chelate is, by definition from the textbook: "A complex containing one or more ligands that form a ring of atoms that includes the central metal atom. " Basically a chelate is like a name for a type of structure where a ring of ligands are attached to the central metal atom. ...

Yashaswi Dis 1K

Hi yes, So these are the steps that I like to follow: 1) Identify the cation and the anion and make sure you write cation first then the anion. In this case Sodium is the cation and the anion is bisoxalato(diaqua)ferrate (III) anion. 2) Focus on writing the anion first and make sure you open bracket...

Yashaswi Dis 1K

Yes, If you are taking out an electron from a trigonal bipyramidal shape, then take it from the equatorial position so that the lone-pair electron can have room for itself and push the other equatorial bonds closer to each other. If it is however, an octahedral shape, then remember that there are 6 ...

Yashaswi Dis 1K

I believe they are a similar question. If it asks to state/identify, you can provide sp^3 for CH4 because it has 4 electron regions as a tetrahedral figure. Maybe the solutions manual could help if you are still stuck on it. Hope that helps!

-Yashwi

Yashaswi Dis 1K

It is not required to write out the coefficient I believe, but it would be better practice to do so. As long as the hybridization matches up with the number of electron regions. For instance sp^3 = for 4 electron region densities.

Yashaswi Dis 1K

An easy way to help figure out where the lone electron goes is to see which element will have a formal charge close to zero depending on the placement of the lone electron. For instance, in CH3, the lone electron would go on the C since the formal charge would be 0 as it is more favorable too. H can...

Yashaswi Dis 1K

Hi, I know in lecture we mentioned that the arrow for the dipole moment points toward the negative partial charge with the plus sign on the positive partial charge. In the textbook, it mentions that the way I described above is the conventional method and the book will use the modern convention with...

Yashaswi Dis 1K

Hi, so I know that chromium's e-config is: [Ar](3d^5)(4s^1) as it is one of the exceptions to the e-configuration. When that element is ionized and becomes an ion (Cr^+1), which is the correct e-config?: (1) [Ar](3d^5)...because the electron is removed from the s-orbital? or (2) [Ar](3d^4)(4s^1)...b...

Yashaswi Dis 1K

Basically, atoms love to either have completely filled orbitals or half-filled orbitals. Nitrogen loves its half-shells, so it will be very hard to remove an electron, thus it will take a high amount of energy to remove one of the electrons from a half-orbital, making Nitrogen have a high ionization...

Yashaswi Dis 1K

Yes, electron affinity is the energy that's released when an electron is added. Ionization energy is energy needed to remove an electron. Really important to know (maybe not for now but helpful for later too) is the following: Electron Affinity (Mostly Exothermic)...For the first electron affinity e...

Yashaswi Dis 1K

Yes, electron affinity is the energy that's released when an electron is added. Ionization energy is energy needed to remove an electron. Really important to know (maybe not for now but helpful for later too) is the following: Electron Affinity (Mostly Exothermic)...For the first electron affinity e...

Yashaswi Dis 1K

The f-orbitals start with the lanthanides and the actinides, in the n=6 row and the f-orbitals have an n=4 as the principle quantum energy number. Specifically, the f-orbitals start with the element 58. n=4 is the energy level, b/c I believe it's in a lower energy state than the d-orbitals, which is...

Yashaswi Dis 1K

I am not exactly sure but I think for an electron to penetrate a nucleus means the electron can be attracted to the positive force of the nucleus and in effect shields the outermost electrons. Hope this helps!

Yashaswi Dis 1K

Hi, I know the ground state configurations for elements but does the excited state mean any deviation from the ground-state configuration? My interpretation of an excited state was that some electrons will jump to higher subshells or orbitals. In the textbook in one of the problems, they showed the ...

Yashaswi Dis 1K

In class today, we discussed how the spin of parallel electrons can either be up or down. Whenever I draw electron configurations for ex. Nitrogen's, for the p subshell, I draw the electrons as if their m(s) is +1/2. For ex: for nitrogen I would do: [He] (in 2s-orbital: up arrow and down arrow) (in ...

Yashaswi Dis 1K

When looking at a lewis structure diagram, there will be lone pair electrons (dots around the element) which you will count each dot as one and then there will also be the single bond lines, which instead of counting as two electrons, count them as one. So add them up and then subtract it from the v...

Yashaswi Dis 1K

I am not exactly sure but as far as I know, the speed of light is the fastest speed on earth, unless research shows otherwise. Thus, if you get a speed less than 3.00 * 10^8 m/s, I am pretty sure it's reasonable b/c it's less than speed of light, which is the fastest speed known so far. That's my th...

Yashaswi Dis 1K

Hi,

A Joule is the definition as mentioned above. However, to help with your calculations and unit conversions, A Joules is also equal to kg((m/s)^2). Hope that helps!

Yashaswi Dis 1K

I think your answer is like correct, meaning approx. close to the answers given b/c when you round the answer you got and change it to scientific notation, you should get 1.00 * 10^5 m/s and since it's way less than speed of light, which if the fastest speed known so far, answer should be reasonable...

Yashaswi Dis 1K

Hi, Just a quick question...when writing chemical reactions and balancing them, if there is a catalyst involved or heat being utilized for the reaction to occur, do we have to also place the catalyst's name above the arrow and place the triangle (representing heat being involved) above the arrow too...

Yashaswi Dis 1K

Sure. So Combustion Reactions generally involve organic compounds or hydrocarbons (they can also involved other molecules too) that burn in the presence of oxygen. So whenever you hear the words burning, combustion, etc. always think of the compound given reacting with oxygen gas (O2). Now, the prod...

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