Photon336
  • Photon336
Questions (rate(s) and rate laws) Enjoy
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Photon336
  • Photon336
#1
1 Attachment
Photon336
  • Photon336
#2
1 Attachment
Photon336
  • Photon336
#3
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Photon336
  • Photon336
@cuanchi @rushwr
Rushwr
  • Rushwr
#1 I haven't still done the environmental chemistry at school. But I'll go with D none of the above!
Photon336
  • Photon336
#1 Graph I exothermic Graph II endothermic Agree/disagree?
Rushwr
  • Rushwr
oh sorry my answer goes to the second one !
Rushwr
  • Rushwr
that Ozone and NO reaction
Rushwr
  • Rushwr
for the exothermic endothermic question i'll go with C In an exothermic reaction the products are having less energy cuz in an exothermic reaction heat is lost to the surroundings !
Rushwr
  • Rushwr
the otherway round for the enothermic reaction.
Photon336
  • Photon336
One thing you notice is that NO remains unchanged
Photon336
  • Photon336
it's used in the first step of the reaction and then remains at the end of the reaction in the last step unchanged. so I would say that it's acting as a catalyst.
Photon336
  • Photon336
#2 A is definitely wrong because it does change the concentration of O3
Rushwr
  • Rushwr
so its gonna e B for the #2 question ?
Rushwr
  • Rushwr
259. I'll go with enthalpy change of the reaction
Photon336
  • Photon336
I think the issue is that NO is not present in the slow step so i thought it wouldn't cause any change. with that.
Photon336
  • Photon336
I think i'll go with C. catalyst only lowers activation energy but doesn't do anything else in terms of overall energy. that stay the same.
Photon336
  • Photon336
|dw:1438191437577:dw| the energy difference between products & reactants is the same
Rushwr
  • Rushwr
yeah I agree! that's what I was saying . and for the 260. is it the 3rd answer!
Photon336
  • Photon336
For #1 we are correct.
Photon336
  • Photon336
#2 i was between C and D and it was D. apparently NO does affect the overall energy change.
Photon336
  • Photon336
@Ciarán95 thoughts on #2?
Rushwr
  • Rushwr
Then that was the 1st answer I gave is correct? Okai now this is how I argued to take that answer. Rate determining step is determined by the slowest reaction out of all right?
Photon336
  • Photon336
right
Rushwr
  • Rushwr
There u can see that NO has no effect on it! Hence it says that even if we use NO as a catalyst its useless cuz it does not effect the rate determining step right?
Rushwr
  • Rushwr
Hence it won't lower the activation energy neither does it lower the overall energy !
Photon336
  • Photon336
overall rate of reaction is dependent on the rate determining step..
Rushwr
  • Rushwr
yes ! So I think that is the way to approach it ?
Photon336
  • Photon336
Thing is it does affect the overall energy choice C the one that says it doesn't is false.
Rushwr
  • Rushwr
The rate determining step is the slowest step . So if NO doesn't affect the rate of the slowest step that means it's useless as a catalyst cuz the rate of the reaction is not increased. if u want to increase the rate of the reaction u need to increase the rate of the slowest step. Since NO doesn't do that it doesn;t act as a catalyst.
Photon336
  • Photon336
so.. it technically doesn't act as a catalyst at all.. so it does affect the overall energy change making C false..
Photon336
  • Photon336
we still haven't done #3 (there's two questions for that)
Rushwr
  • Rushwr
for three 259. I have slight confusion with A and B But I would go for B
Rushwr
  • Rushwr
260. I would go with C
Photon336
  • Photon336
#3 our rate constant is \[r = k[A]^{x}[B]^{y}\] \[Lnk =Ae ^{Ea/RT}\] if we look at the bottom equation if we change the temperature we also change the value of K , b/c RT is in the denominator. a catalyst lowers the activation energy and thus increases the rate of the reaction. so that would affect K as well. Ea would be smaller and RT would remain unchanged. so for this one I would say it's B. Rate constant is independent of the enthalpy change.
Photon336
  • Photon336
any thoughts on #260 #3?
Rushwr
  • Rushwr
yey ! so it is the 2nd answer!
Rushwr
  • Rushwr
I though it would be that
Photon336
  • Photon336
yeah
Rushwr
  • Rushwr
for carbonates to release CO2 at a higher rate the concentrations of the acids should be high so we can eliminate A and B I choose C over D cuz they say the concentration of H is is directly connected to the rate.
Rushwr
  • Rushwr
the no. of H in C is greater than in D so I choose C.
Rushwr
  • Rushwr
@Ciarán95 are we arguing right?
Photon336
  • Photon336
\[H _{2}SO _{4}\] That's polyprotic? maybe more H+ So it could react faster.
Photon336
  • Photon336
oh *** wait that's H\[H _{2}SO _{3}\] There are weak acids with the exception of HCL
Photon336
  • Photon336
@Rushwr said the answer was HCl
Rushwr
  • Rushwr
NO I choose C
Photon336
  • Photon336
Because it's a strong acid.
Photon336
  • Photon336
I chose C too but i thought it was sulfuric acid
Rushwr
  • Rushwr
So they have said it as HCl?
Photon336
  • Photon336
yeah because they said it' a strong acid so it will produce more H+
Photon336
  • Photon336
\[Ka >>>> 1\]
Rushwr
  • Rushwr
okai then. But the concentration is small nah !

Looking for something else?

Not the answer you are looking for? Search for more explanations.