Photon336
  • Photon336
Chem-question
Chemistry
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Photon336
  • Photon336
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Photon336
  • Photon336
318
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sweetburger
  • sweetburger
Ill start with 321 because its easy. If you added NaCN more AgCN would precipitate. B
Photon336
  • Photon336
@sweetburger because it's at equilibrium already can we assume that it's saturated solution if so then adding more NaCN would precipitate in my opinion
sweetburger
  • sweetburger
Agreed
Photon336
  • Photon336
yes it's B
sweetburger
  • sweetburger
For 320 [.2]^2/([.1][.2])= 2=Keq
sweetburger
  • sweetburger
So 320 is B. I think.
Photon336
  • Photon336
319. we have \[N _{2} + O _{2} --> 2NO \]
Photon336
  • Photon336
319. A. equilibrium constant is temperatuer dependent so if we changed the temperature we should also change the equilibrium concentrations because we would have a new Keq value. C adding more NO will shift to the reactants if it's already at equilibrium D N2 we don't know if we added or removed it but that would cause a shift. B. is the answer because if we have the same number of moles on either side of the equation increasing the pressure does not cause any change in the equlibrium concentrations. remember increase in pressure favors the side with fewer number of moles , while decreasing pressure it's the opposite.
sweetburger
  • sweetburger
Completely agreed.
sweetburger
  • sweetburger
318. Adding a Catalyst would lower the activation energy of the reaction which in turn would increase the rate of the forward reaction, but would also increase the rate of the reverse reaction as the activation energy of the reverse reaction is also lowered showed in this diagram below.|dw:1438548218094:dw|
sweetburger
  • sweetburger
So adding a catalyst would have minimal affect on the equilibrium concentrations.
sweetburger
  • sweetburger
effect*
Photon336
  • Photon336
yeah that's a great way to explain that.
Photon336
  • Photon336
\[K _{eq} = \frac{ [0.20]^{2} }{ [0.20][0.10] } = 2 \] for the other one like they gave us equilibrium concentrations so we didn't have to do that ICE table stuff.
sweetburger
  • sweetburger
Haha, the ICE Table is very annoying to write out...
Photon336
  • Photon336
I hate. that..
Photon336
  • Photon336
287. A. magnitude of the activation energy, the higher the Ea the harder it will be to get the reaction going. B.(nope) this is our answer C. lowering activation energy is what a catalyst does and that increases the rate of the reaction. D. absolute temperature yes. because increasing temperature increase the rate of our reaction \[k = Ae ^{Ea/RT}\] you increase the temperature and lower Ea the k value goes up. so B for 287
Photon336
  • Photon336
286/288 left
sweetburger
  • sweetburger
Ahh i see it now. I was somewhat confused on question 287, but you clarified it quite nicely.
Photon336
  • Photon336
thanks haha
sweetburger
  • sweetburger
For 286. I'm not sure, but I am leaning towards answer choice D. I think answer choices A and C are effectively the same in which the same process is occurring in which the activation energy is being lowered. I know that increasing the temperature does indeed effect the rate constant. TBH I thought that only temperature truly could change the rate constant.
Photon336
  • Photon336
288. A. \[-\frac{ 1 }{ 2 } \frac{ \Delta [KMnO _{4}] }{ \Delta T}\]
Photon336
  • Photon336
288. Notice that once Mn forms the reaction is now fast. to write our rate law our reaction is going to depend on our slowest step. so that's how I eliminated C and D. now for A and B. we know that in our rate law, our concentration of [KMnO4] is going to decrease and when you write the integrated rate law the coefficients let' say two becomes (1/2) so that eliminates B.
Photon336
  • Photon336
that's why we consider KMnO4 and not MnSO4
Photon336
  • Photon336
for 286 i agree with you.. that's funny because whenever two answers are the same they can't be the answer they are looking for
sweetburger
  • sweetburger
Yea I agree with your thought process on question 288.
Photon336
  • Photon336
did they teach you this equation \[k = Ae ^{Ea/RT}\]
sweetburger
  • sweetburger
That is the Arrenhius? If i remember.
Photon336
  • Photon336
yeah
sweetburger
  • sweetburger
I have to go, but I will gladly help anytime.
Photon336
  • Photon336
also \[r = k[A]^{x}[B]^{y}\]
Photon336
  • Photon336
it's d doubling the concentration of one of the reactatns
Photon336
  • Photon336
that affects the rate of the reaction but not the rate constant k
sweetburger
  • sweetburger
Ahh so for 286 the answer was D?
Photon336
  • Photon336
yep
Photon336
  • Photon336
Anyway I have to go cya around
sweetburger
  • sweetburger
Now looking at the Arrenhius I see how obvious it is XD *facepalm*
Photon336
  • Photon336
=D
sweetburger
  • sweetburger
Alrighty seeya :)

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