Chem-question

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Ill start with 321 because its easy. If you added NaCN more AgCN would precipitate. B
@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
Agreed
yes it's B
For 320 [.2]^2/([.1][.2])= 2=Keq
So 320 is B. I think.
319. we have \[N _{2} + O _{2} --> 2NO \]
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.
Completely agreed.
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|
So adding a catalyst would have minimal affect on the equilibrium concentrations.
effect*
yeah that's a great way to explain that.
\[K _{eq} = \frac{ [0.20]^{2} }{ [0.20][0.10] } = 2 \] for the other one like they gave us equilibrium concentrations so we didn't have to do that ICE table stuff.
Haha, the ICE Table is very annoying to write out...
I hate. that..
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
286/288 left
Ahh i see it now. I was somewhat confused on question 287, but you clarified it quite nicely.
thanks haha
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.
288. A. \[-\frac{ 1 }{ 2 } \frac{ \Delta [KMnO _{4}] }{ \Delta T}\]
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.
that's why we consider KMnO4 and not MnSO4
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
Yea I agree with your thought process on question 288.
did they teach you this equation \[k = Ae ^{Ea/RT}\]
That is the Arrenhius? If i remember.
yeah
I have to go, but I will gladly help anytime.
also \[r = k[A]^{x}[B]^{y}\]
it's d doubling the concentration of one of the reactatns
that affects the rate of the reaction but not the rate constant k
Ahh so for 286 the answer was D?
yep
Anyway I have to go cya around
Now looking at the Arrenhius I see how obvious it is XD *facepalm*
=D
Alrighty seeya :)

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