CHEMISTRY COMMUNITY

Alyssa Pelak 1J

During lecture, Dr. Lavelle explained that a higher frequency factor, A, results in a higher rate constant, K. I understand this when I look at the Arrhenius Equation. However, I am a bit confused on why this is so.

Alyssa Pelak 1J

The reaction can only go as fast as its slowest step. Think of this as a funnel. Imagine a 2 step reaction where Step 1 is faster than Step 2. Step 1 can produce reactants as fast as it wants, but if Step 2 is slow then ultimately the products of Step 1 will be turned into the final products only as...

Alyssa Pelak 1J

Yes! A larger molecule has a higher molar entropy. This is because it is more complex and has more electrons.

Alyssa Pelak 1J

Yes! So if you are given rate=k[A]^2[B]^-1 the overall order would just be 1!

Alyssa Pelak 1J

you would use the integrated rate law for this question.

1/[A]=kt+1[A]0
1/[A]-1/[A]0=kt

You know that 1/[A] is 1/16[A]0 so you sub that into the equation:
1/(1[A]0/16)-1/[A]0=kt

In order to remove the fraction in the denominator you get
16/[A]0-1/[A]0=kt

Hope this helps!

Alyssa Pelak 1J

Typically with a spectrometer you would be able to measure the absorbance through different times in the reaction. You could put your solution in the spectrometer and determine the absorbance at a certain time. From that, you could use Beer's Law (A=elc) to determine the concentration at that point.

Alyssa Pelak 1J

The simplest way is to write out the rate law and divide. 1st order: rate=k[A] and divide rate by [A]. rate is written in mol.L-1.s-1. and [A] is mol.L-1. If you divide the two you are left with s-1. 2nd order: rate=k[A]^2. as with first order divide rate (mol.L-1.s-1.) by [A]^2 to get L.mol-1.s-1. ...

Alyssa Pelak 1J

If they want a specific unit they will likely state it in the problem. If not, you can use the easiest unit.

Alyssa Pelak 1J

The question asks you to express the units for rate constants when the concentrations are in moles per liter and time is in seconds. You are using the equation rate= k[A] since it is first order. Rate is stated in mol L(-1)s(-1) and concentration is in mol L(-1). Therefore, in order to find the unit...

Alyssa Pelak 1J

I do not believe that you would be able to because you are not supposed to round to sig figs until the final answer. But, I would bring it in to your TAs office hours to check anyway!

Alyssa Pelak 1J

It indicates first order because the concentration change of the reactant directly correlates to the change in the rate. So, if the concentration was doubled the rate would double too.

Alyssa Pelak 1J

In the anode, oxidation is occurring. In oxidation electrons are leaving the cell, making the charge on the cell less negative and more positive. For the cathode, reduction is taking place meaning that the cell is gaining electrons and gaining negative charge.

Alyssa Pelak 1J

In order to determine if a metal would dissolve in a solution you would calculate the standard cell potential using the balanced half reactions. Then you would use the calculated standard cell potential and plug it in to the equation delta G (standard)=-nFE. If the calculated standard Gibbs free ene...

Alyssa Pelak 1J

The anode is the one being oxidized. This can be remembered with "an ox" "red cat"

Alyssa Pelak 1J

Yes, all diatomic molecules have 0 Gibb's Free energy

Alyssa Pelak 1J

You use Cv when calculating the temperature change because you are ignoring the change in volume for that part. So, in this question you are calculating the total change in entropy by calculating temperature change and volume change independent of one another. Therefore, when you change volume you c...

Alyssa Pelak 1J

Can someone please explain how we calculate work and heat for question 4 on the practice midterm? The question reads, "You have a system consisting of 0.60 moles of an ideal gas contained in a 50.0L container at 1.0 atm. You just love chemistry to a fault, so you perform a series of steps to th...

Alyssa Pelak 1J

The question reads.. Assuming that the heat capacity of an ideal gas is independent of temperature, calculate the entropy change associated with raising the temperature of 1.00 mol of ideal gas atoms reversibly from 37.6 C to 157.9 C at (a) constant pressure and (b) constant volume. Can someone plea...

Alyssa Pelak 1J

No you do not need to use the integral to solve the actual problems. The point was to show the derivation and why irreversible and reversible use different formulas.

Alyssa Pelak 1J

Iodine is larger and has more electrons than Bromine. That means than iodine has more possibly states for its electrons to be in, giving it a higher standard molar entropy.

Alyssa Pelak 1J

Additionally, entropy is positive when the reaction spontaneous. It means that the system has more possible states. Entropy is negative when the system has less possible states.

Alyssa Pelak 1J

Reversible reactions are at equilibrium meaning that they are slow. They occur when P is not constant, so the equation, w=-nRTln(V^2/V^1), is used. More work is done because these systems have to work on equilibrium. Irreversibly reactions are fast and take less work. These reactions occur a constan...

Alyssa Pelak 1J

Gibb's Free Energy is a state function because it is defined by other state functions (G=H-TS). H and S are both state functions. Additionally, Gibbs Free Energy is a state property because it is defined by final- initial and is not dependent on the pathway.

Alyssa Pelak 1J

I don't have the answer, but Dr. Lavelle mentioned in class that there was a very complicated way to prove this that was way too advanced for the class. So we definitely do not need to know why.

Alyssa Pelak 1J

How do we know when work is done by the system or the surrounds or when work is done on the system or surroundings?

Alyssa Pelak 1J

Can someone please explain how we know we need to add two things together to get q versus when they are equal(but opposite signs) to each other? For example in 8.19 the kettle and the water are added together and 8.21 they are equal (but opposite signs.

Thank you!!

Alyssa Pelak 1J

(A) true because the equation U=q+w. Since they give you that q=0 and no work is done then U=0 (B) true because they tell you that there is no heat transfer (C) false because it says there is no heat transfer so it is not losing heat (D) true because when work is 0, U=q+w and w=0, U=q (E) true becau...

Alyssa Pelak 1J

In question 8.17, there are two pictures. One of melting and one of condensation. In each case, indicate whether heat is absorbed or given off by the system, whether expansion work is done on or by the system, and predict the signs of q and w for the process. For melting why is work negative and for...

Alyssa Pelak 1J

I don't believe one is used more than the other. It just depends on what the given units are in the problem.

Alyssa Pelak 1J

This simply means that enthalpy is dependent only on the initial and final values, or the product and reactants. The path taken to obtain that is not important. That explains why we can use the equation delta H= H products-H reactants.

Alyssa Pelak 1J

Can someone help me with question 8.3 please. Air in a bicycle pump is compressed by pushing in the handle. If the inner diameter of the pump is 3.0 cm and the pump is depressed 20. cm with a pressure of 2.00 atm, (a) how much work is done in the compression? (b) Is the work positive or negative wit...

Alyssa Pelak 1J

Yes! The energy goes into breaking the bonds so the temperature does not increase. For example when a solid melts into a liquid the heat is going into breaking the bonds of the solid!
Hope this helps!

Alyssa Pelak 1J

When do we need to add the word ion to the end of the coordination complex name? Is it just when it has a + or - charge, or is it all the time?

Thank you!

Alyssa Pelak 1J

Why do PO43-, PCl5, and SF6 all not have dipole moments? And, for the future, how can you determine by looking at a lewis structure which molecules will have dipole moments?

Alyssa Pelak 1J

These are the same ideas. When there is resonance the bond is delocalized.

Alyssa Pelak 1J

sorry the alignment for the ICE box got messed up! Hopefully you can still understand it :-)

Alyssa Pelak 1J

Hello! For part a) i) The amount of NO2 will increase because if you increase the amount of NO the reaction will proceed toward the products in order to minimize the effects of the change ii) Removing the SO2 will also cause the reaction to proceed toward the products, therefore producing more NO2 i...

Alyssa Pelak 1J

Hello!
I was curious if someone could explain why AX2E3 is linear?

Thank you!

Alyssa Pelak 1J

For a coordination complex such as [Cr(NH3)3(H2O)3]Cl3 is the anion at the end named with a prefix or not?

So for this one, would the name be triamminetriaquachromium(III) trichloride or triamminetriaquachromium (III) chloride.

Thank you!

Alyssa Pelak 1J

Hello!

You know to use partial pressures because the given expression is all gases.

Hope this helps!

Alyssa Pelak 1J

Do bent and angular mean the same thing? In some online VSEPR charts I have found they are used as synonyms and on some they have been shown as completely different.

Alyssa Pelak 1J

I am still a little confused on how we determine whether a ligand is polydentate or not. I understand what polydentate is, I am just unsure of how I would know whether it can bond more than one time?
Can someone please explain how you can determine this? For example in textbook problem 17.33.

Alyssa Pelak 1J

Can someone please explain step by step how 17.31d is done?

write the formula for each of the following coordination compounds: sodium bisoxalato(diaqua)ferrate(III)

Thank you!

Alyssa Pelak 1J

No the radical is not supposed to defy the octet rule. The element with the radical will likely not have a complete octet. Radicals are unstable so this is expected. An example of this would be HOCO from textbook problem 3.123. The C has an incomplete octet.

Alyssa Pelak 1J

Yes, it does. The highest polarizing power comes from the smallest, most highly charged cations. And, the highest polarizability comes from the largest anions.

Alyssa Pelak 1J

The expanded octet is possible due to the available d orbital. Therefore, I think that maybe the maximum would be 18electrons so 9 bonds, but I doubt that would ever exist.

Alyssa Pelak 1J

Se has two unpaired electrons because its electron configuration is [Ar] 3d^10.4s^2.4p^4. The p subshell has 3 orbitals. The first three electrons in the 4p would each be placed in their own orbital because of Hund's Rule. Then, the fourth electron in the p subshell would be placed in an orbital alr...

Alyssa Pelak 1J

I do not understand question 5d on Test 3. 5. Answer the following questions related to the ground state electron configuration of Cadmium. d.) What is the highest value of l possible? I put the answer as 4 because l has the possible values of 0 to (n-1). Cadmium's ground state electron configuratio...

Alyssa Pelak 1J

A radical is when a compound has an unpaired electron. So, in a Lewis Structure, you would have just one dot representing an unpaired electron instead of the usual two representing paired electrons. But with an incomplete octet a element can exist in a stable compound without the usual 8 electrons. ...

Alyssa Pelak 1J

Write the Lewis structure of each of the following reactive species, all of which are found to contribute to the destruction of the ozone layer, and indicate which are radicals: (a) chlorine monoxide, ClO

Why is the unpaired electron on the Cl instead of the O?

Alyssa Pelak 1J

According to the solutions manual Ge actually have 2 unpaired electrons. You are correct it is 4p^2 and each electron would be in its own orbital meaning that 2 are unpaired.

Alyssa Pelak 1J

Hello! I am slightly confused as to why we are drawing the Lewis Structures differently in 3.39. Write the complete Lewis structure for each of the following compounds: (a) ammonium chloride; (b) potassium phosphide; (c) sodium hypochlorite. Why are we drawing each part separate? Is it because they ...

Alyssa Pelak 1J

When you draw aufbaw diagrams for transition metals in the first row do you draw 3d above the 4s or the other way around? I know that when you write out the electron configuration 3d is written before 4s, but I am unsure of how this works for Aufbaw/ Building up diagrams. For example, if I were to w...

Alyssa Pelak 1J

It is true that values of l range from 0 to n-1. Also, each orbital (s, p, d,f) has a corresponding l value. So l=0 is s, l=1 is p, l=2 is d, and l=3 is f. That would explain how l=0 is s as you stated in your question. In example, if you are given a problem to state n and l for 6d. You know that n=...

Alyssa Pelak 1J

So coulomb potential energy is proportional to (q1)(q2)/r The two q values are each charges and r is the distance between the two charges. The general thing you need to understand when it comes to this equation is that larger charges have more potential energy and longer distances have weaker attrac...

Alyssa Pelak 1J

The easiest way I have found to do these is to go through each quantum number step by step. Given 6d you know that n=6 and because it is d l=2. The values of ml allowed range from -l to l, therefore a 6d sub shell would have values -2,-1,0,1, and 2 (5 values). Then, for 7s following the same process...

Alyssa Pelak 1J

Hello, I am aware that we do not have to know Silver's electron configuration, but out of general curiosity can someone please explain how we determine this.
It appears in 2.43a...What is the ground-state electron con guration expected for each of the following elements: (a) silver?

Alyssa Pelak 1J

I am curious as to whether Wein's law will be on the exam. I remember the Lavelle said in lecture that we didn't need to go over it, but it falls into the 3-41 homework problems he said we are responsible with knowing. Does anyone know for sure?

Alyssa Pelak 1J

wavelength=h/mv
h is J.s but the SI based units for J are kg.m^2.s^-2
Therefore, it would be (kg.m^2.s^-2)(s) / (kg)(m.s^-1)
kg cancels, then one m^2 on the top gets canceled to m, and s^-1 on the bottom can be brought to the top as s^1 making the s^-2 and now s^2 cancel. This leaves just m left.

Alyssa Pelak 1J

Lavelle said that we will be given anything we need on the test so I wouldn't worry about it too much. I'm assuming that probably means well be given
c=(v)(wavelength)
E=hv
E=-hR/n^2
v=r(1/n^2-1/n^2)
E(k)=1/2mv^2
E(K)=hv- work function
wavelength=h/mv

Then the values for c, R, and h

Alyssa Pelak 1J

Balmer series indicates that the electron has moved from another shell to n=2 or from n=2 to another shell. The wavelength that is absorbed or emitted falls into the visible light range. For Lyman series the electron has moved from another shell to n=1 or from n=1 to another shell. The wavelength is...

Alyssa Pelak 1J

Yes, essentially Balmer series indicates that the electron has moved from another shell to n=2 or from n=2 to another shell. The wavelength that is absorbed or emitted falls into the visible light range. For Lyman series the electron has done the same but for n=1 and therefore the wavelength is shor...

Alyssa Pelak 1J

I believe we just need to know Balmer and Lyman because that is what we discussed in class. I think just knowing that Balmer series are in the visible light range and involve n=2 and Lyman series are in the UV range and involve n=1 is more than enough. He could put a problem that similar to that of ...

Alyssa Pelak 1J

So 140.511 keV must be converted to J in order for us to use it. 1 keV is equal to 1.602x10^-16 J. Once you convert it you plug in that energy into E= hc/wavelength to find the wavelength.
Hope this helps!

Alyssa Pelak 1J

Lavelle uses 3.00x10^8 when solving his problems, so I think it would be fine to use this one. Either will work though!

Alyssa Pelak 1J

Hello, I have read the solution guide and cannot understand the answers for question 1.3. Which of the following happens when the frequency of electromagnetic radiation decreases? Explain your reasoning. A. The speed of radiation decreases. B. The wavelength of the radiation decreases. C. The extent...

Alyssa Pelak 1J

Also, just to help with future questions, this can be assumed with all of the diatomic molecules. Br, I, N, Cl, H, O, and F. You can remember them as BrINClHOF (brinkl-hoff) or Have(H) No(N) Fear(F) Of(O) Ice(I) Cold(Cl) Beer(Br). Hope this helps

Alyssa Pelak 1J

I came across a similar issue and I believe we are supposed to round up if the value is .05 or higher... so 1.04 would become 1, but anything like 1.05 or 1.06 become 1.1!

Alyssa Pelak 1J

I used a stoichiometric approach because the mmol given are Na+ whereas the g and mL given are sodium carbonate. Therefore, you are supposed to find L of Na+ I notice you already found the molarity of the sodium carbonate (Na2CO3) to be .079M. I then took the 2.15 times 10^-3 mol Na+ and multiplied ...

Alyssa Pelak 1J

I can't seem to figure out this question... Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is this claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active substan...

Alyssa Pelak 1J

If you start with the Fe there is 2 on the reactant side and 3 on the product side. Therefore, multiply the Fe2O3 by 3 and the Fe3O4 by 2 to make a balanced 6.
3Fe2O3 + CO ---> 2Fe3O4 + CO2

And that just so happens to balance everything else out too.

Alyssa Pelak 1J

In this problem, you first want to start with balancing N because it is only present once on the reactant side and once on the product side. Notice that N is balanced with one on each side, so you move on to the O. Because there is an NO2 present on the product side you know that there must be an ev...

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