CHEMISTRY COMMUNITY

Andrea Grigsby 1I

I'm looking at the answer for this question and I understand how to do everything, except for the value '-0.59'
I'm assuming that's the lnA from the Arrhenius Equation, however, how do we calculate that with the values given?

Andrea Grigsby 1I

How is the equation listed in the answers derived?
I understand it's from lnk=-Ea/RT with changes in T, but my question is where did lnA go?
Thanks!

Andrea Grigsby 1I

you would use the equation dG = - RT lnK for this question rearrange and substitute in the values you have lnK=- -200000/(8.314*298) = 80.7 solve for K K=1.14*10^35 the question says the equilibrium constant of the first step is 10 times greater than that of the second step so you simply divide the ...

Andrea Grigsby 1I

Ti < Mn < Ag
Al < Pb < Cu

Andrea Grigsby 1I

Im not sure but maybe it has to do with the states each compound is in. AgBr is a solid while Br- is aqueous. And just to keep things neat the solids (Ag and AgBr) are placed next to each other.
I've seen this done in other examples

Andrea Grigsby 1I

yes, it is a constant

Andrea Grigsby 1I

Why is the formula used for this question ln[A]0 / [A]t , instead of ln [A]t/[A]0?

Andrea Grigsby 1I

use the equation deltaG = -RTlnK
and rearrange
lnK= - 200000/(8.314*298)
= 80.7
K = 1.14*10^35
divide by 10 so K2 = 1.14*10^34
and then use it in the same equation as above
deltaG2 = -8.314*298*ln1.14*10^34
= -194 kJ

Andrea Grigsby 1I

I don't know what the question is exactly?
But maybe if the question also suggests/says the change in internal energy is 0, you could use q=-w
and just substitute the given values into w=-PdeltaV ?
Just a thought

Andrea Grigsby 1I

nope. because none of the questions pertaining to trouton's rule are on the syllabus/homework problem list

Andrea Grigsby 1I

so use q=mCdelta T to calculate the heat produced by the water when the temperature is raised from 22.6 to 34.6 so it would look something like this: 115.43*4.184*(34.6-22.6) = 5795.51 J and then for the metal because Qsys=-Qsurr 245.7*C*(34.6-75.2)= -5795.51 J C=-9975.42/-5795.51 =0.580979 =0.581 J...

Andrea Grigsby 1I

they are the same thing you just have to inverse the sign

Andrea Grigsby 1I

Will we have to know what the entropy for aqueous ions are mostly negative
(like in question 9.89)?

Andrea Grigsby 1I

I understand how we need to use deltaG = deltaH - TdeltaS for this question, and how we need to use the tables in the textbook to find the values for deltaH and deltaS, however, why is the value of delta S listed in the solutions "97.6J/K/mol" ?
Where does that value come from?

Andrea Grigsby 1I

I think they would give nearly the same result as the average bond enthalpy values are averages of that said bond in a bunch of different molecular structures.
But i'm not 100% sure

Andrea Grigsby 1I

what does the state (gr) mean?

Andrea Grigsby 1I

yep it's decimeter
where 1 decimeter^3 = 1 Liter

Andrea Grigsby 1I

because we need the amount of mol of product, which would depend on the limiting reagent of the reaction

Andrea Grigsby 1I

if it's calculating temperature change, then the difference will be the same whether it's in kelvin or celcius

Andrea Grigsby 1I

Will there be questions involving average kinetic energy (like question 8.103) in the upcoming test?
I don't recall Dr Lavelle going over it in lecture, nor do I see it on his outline 1...

Andrea Grigsby 1I

torr is another way to express pressure, but thermodynamics answers are usually always in atm so the conversion from torr to atm is on the formula sheet where 1 atm=760 torr!
and for the 1000ml, it's because work is calculated with liters

Andrea Grigsby 1I

reaction enthalpies may not just be indicating 1 mol of substance so it's only kJ. If it were enthalpy of formation or enthalpy of combustion, then it would be kJ/mol as it's the reaction enthalpy of the formation/combustion of 1 mole of a substance

Andrea Grigsby 1I

specifically part b, why is the final step in the answer book shown as the mole (from after calculating it with PV=nRT) being multiplied by 180.6kJ? I would've thought it would be multiplied by 90.3kJ which is reaction enthalpy that produces 1 mole of NO.

Andrea Grigsby 1I

oh yea, I completely forgot about that.
Thanks so much!

Andrea Grigsby 1I

if you have deltaH/Q, you can look at the sign it has -deltaH means that energy is released by the system (exothermic; the system is losing energy, thus the minus sign) and so energy is being transfered from the system to the surroundings. eg combustion reactions +deltaH means that energy is gained ...

Andrea Grigsby 1I

for the first equation, usually the question says "isothermic" (as in the assigned text book questions)
but it's not going to be on this coming test! :)

Andrea Grigsby 1I

the example Dr lavelle gave in class was just to think of 3 test tubes - Open system: both matter and energy can travel out, which would be an open test tube with no insulation - Closed system: only energy can travel out, not matter, which would be a closed test tube - Isolated system: neither matte...

Andrea Grigsby 1I

Im pretty sure this is the only type of work because a) Dr Lavelle didn't talk about the rest in lecture and b) the practice questions he set aren't about any other types of work (to the extent of my knowledge)

Andrea Grigsby 1I

for question 8.25, I understand everything up until the point where the answer key says Qreaction + Qcalorimeter =0, which then leads to the final answer. Can someone explain why the heat from the reaction + the heat of the calorimeter equals to 0?

Thanks :)

Andrea Grigsby 1I

the two are different but that's how you would do a calculation of the enthalpy of reaction through bond enthalpies

Andrea Grigsby 1I

the most common example would be diethylenetriammine where if it was (dien)2 then it would be didiethylenetriammine, so instead of having "didi-" you switch the first 'di' out and replace it with 'bis' so it's clearer
--> bisdiethylenetriammine

Andrea Grigsby 1I

because a representative equation for acid reactions is HA --> H3O+ + A- where Kc=[H3O+][A-]/[HA] if you were to use an ice table to calculate molarities, Kc would become: Kc= x^2 / [HA]-x as both have the same amount of moles and because the concentration of [H3O+] and [A-] are the same, it can be ...

Andrea Grigsby 1I

I think as far as what was taught in class, yes. molecules might have angles of <109.5 if there are lone pairs present (eg in bent/angular shapes) as they have greater repulsion than normal bonds. molecules with angles of >109.5.. I would assume there's only molecules with 120 degrees (as far as wha...

Andrea Grigsby 1I

you add the change in N2 to the original value so now PN2=4.68 and the rest of the initial values are the values they give you.
and from here you just solve for x as you would any other ICE table.

Andrea Grigsby 1I

the reaction will favor the reactants and thus shift left as K must remain constant
and K=[P]/[R] so if P increases, R will increase to keep K remain the same

Andrea Grigsby 1I

so you would use di, tri, tetra on ligands that don't have said prefixes in their names already, eg dichlorido while if you were to write (dien)2, instead of it being didiethlynetriamine which would be confusing, it's bisdiethylenetriamine, same with (en)2 for example, it would be bisethylenediamine...

Andrea Grigsby 1I

I think either one is ok, the (OH2) just makes the fact that it is O that's bonded to the metal clearer

Andrea Grigsby 1I

use the equation PV=nRT and substitute the values in

Andrea Grigsby 1I

https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=1415

Andrea Grigsby 1I

adding inert gas doesn't change the concentration of the reactants/products as the moles and the volume of R and P remain constant while increasing the pressure, and thus decreasing the volume does - C=n/V so because the number of moles of the reactants/products remain the same but the volume decrea...

Andrea Grigsby 1I

change in concentration doesn't affect K, and the only way this is possible is that if the concentration of reactants are increased, the concentration of products will increase as well to keep the P/R ratio the same

Andrea Grigsby 1I

If you're talking about (en)2, it would be called bisethylenediamine and yes; while diethylenetriamine (dien) would have 3 bonding sites each

Andrea Grigsby 1I

K can either indicate Kc (equilibrium constant with concentrations) or Kp (equilibrium constant using partial pressure)

Andrea Grigsby 1I

use the method of how you would find the equilibrium constant
K=[X]^2 / [X2]
=[(12/17)*0.1]^2 / [(5/17)*0.1]
=0.17

the 12/17 and 5/17 are the number of molecules that have/ haven't dissociated, as given in the diagrams of the 4 containers

Andrea Grigsby 1I

Andrea Grigsby 1I

So first, draw out the Lewis Structure - use valence electrons of each atom to determine the bonding (eg every carbon makes 4 bonds), then once you have the structure drawn out it'll be more straightforward to determine the hybridization - basically the number of electron densities of each carbon, a...

Andrea Grigsby 1I

when the formal charge = 0

Andrea Grigsby 1I

[Ar] 3d10

basically Ga Z=31
but it's Ga3+ so 31-3=28

Andrea Grigsby 1I

Maybe the ones where we're able to work out from the periodic table? Like the electron configuration of nitrogen is 1s2 2s2 2p3 (stable, half filled) while Oxygen has an electron configuration of 1s2 2s2 2p4 (one extra electron from being stably half filled), so taking an electron from the oxygen is...

Andrea Grigsby 1I

Ionisation Energy for atoms would be different when they're in the gaseous state than in other states - which is the easiest way to compare the energy required between different atoms as there is very little attraction with other atoms. In liquids/solids, the intermolecular forces will distort the v...

Andrea Grigsby 1I

electrons in the Lyman series emit light in the UV region, while electrons in the Balmer series emit light in the visible light region.
So you would look at the wavelength of the light emitted and deduce it from that.

(Im not sure if I answered what you were asking, but yeah)

Andrea Grigsby 1I

I think in the case of ~Atomic Spectra~, it's not so much the attraction of electrons to protons, but instead the absorption and emission of energy.

n=1 is the ground state/ energy level for electrons, which is the furthest "way down" an electron can go after emitting energy.

Andrea Grigsby 1I

there is also a "Brackett series" which is n=4 but i'm 99.9% sure we don't need to know this

Andrea Grigsby 1I

An experiment showed that if a light beam had high enough frequency it could make a surface eject an electron. The emphasis is on frequency here as the intensity of the light had no effect on the electron, which showed that the light was not acting like a classic wave. --> this boils down to the ide...

Andrea Grigsby 1I

whenever you need to use an equation with the constant "h" in it.
eg E=hv

good thing is, the value is on the formula sheet so we don't have to memorize it :)

Andrea Grigsby 1I

1. convert the wavelength given to m 102.6x10^-9m 2. figure out the frequency by rearranging c=wavelength*frequency v=c/wavelength = 2.998*10^8 / 102.6*10^-9 = 2.922x10^15 3. as the question says "in the ultraviolet spectrum", one can concur that the final energy level is n=1 4. put known ...

Andrea Grigsby 1I

So it's posted on the website that there's going to be a test on Chapter 1 up to 1.5 in the textbook this week during the discussion section you're enrolled in. I'm in discussion 1B on Tuesday mornings so I was just wondering if the test was going to be on this week's discussion section (October 10)...

Andrea Grigsby 1I

my bad i only did half the question, but yes what @Ivy Lu 1C said

Andrea Grigsby 1I

@Hyein Cha 1K

I do it the same way

Andrea Grigsby 1I

I'm assuming this is the white phosphorus to phosphorus oxide question. P4 + 3O2 --> P4O6 masses for both the reactants = 5.77g get the molar mass of the elements from a periodic table and divide the mass with it to get the number of moles of each particle n (P4) = 5.77/(30.97*4) = 0.0466 n (O2) = 5...

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