destroyer chem 166

[arpitpatel86]

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Consider the rxn A + B---->C Which of the following would not affect the rxn rate of the IRREVERSIBLE RXN.....

A. decreasing A
b. increasing B
c. Adding catalyst.
d. REMOVING C
E.increasing temp

ok normally if u decrease A the rxn shifts to the left but since its irreversible this should have no effect on the rxn right?

they said the answer was D......this doesnt make much sense because whether the rxn is reversible or not removing C will shift more towards the right regaurd less....what am i doing wrong. to be B AND D seem the same because if u increase B it'll shift to the right and if u decrease C it will still shift to the right.....and i know all this "shifting' occurs in equilibrium but even in equilibrium if u decrease A the it would shift to the left however since it cant because it is IRREVERSIBLE shouldnt A be the answer?

 

[DDSguyLA]

Hey... I just finished the DAT destroyer Gen Chem section tonight...lol.
Wow i must say. I remember every question and this one too. I got it wrong as well.
Believe me. I know what you're talking about and I had to think about it for 10 mins to figure it out.
I guess you have to look at this question a bit differently.
That's the thing with Destroyer, its questions make you think about everything and that's really good I think.

Think of it this way.

There is ONLY 5 choices.

Decreasing A
Decreasing B
Catalyst
Removing C
Increase in Temp

Ok.
Rule Number 1: Catalyst ALWAYS changes the rate. So "C" is out.
Rule Number 2: Temperature ALWAYS Changes the rate. So "E" is out.
If you decrease A, there would be less stuff to make suff, so the rate will come down even if its a little or what-not.
Same goes with if you decrease B.

The ONLY thing that is left is Choice D.

Think about it......... IF you remove C, what does that have to do with A and B. They will still make C at the same rate they did before (Since NOT a reversible reaction)

Hope this helps.

 

[tncekm]

....
Think about it......... IF you remove C, what does that have to do with A and B. They will still make C at the same rate they did before (Since NOT a reversible reaction)...

Yeah, that last sentence is the gist of it IMHO.

That is a pretty tricky question. But, I believe that it comes down to the fact that property D has to do with thermodynamics, not kinetics (rate). The two are not connected as far as the DAT is concerned. So, by adjusting the concentration of C you will adjust the thermodynamics of the situation and increase yield, but not the kinetics (rate).

 

[DRHOYA]

Think of it this way. Since its not reversible, Le Chatlier's principals of equilibriums goes out the window. Now, since the stoichiometry of A to B is 1:1, decreasing A WOULD effect the rxn rate, because you would not have enough A to even react. Choice D says that removing C has no effect on the rate, which is correct. This is because like I said above, Le Chatlier's principals do not apply, and A to B is 1:1, so removing C isn't going to speed up anything. The rxn proceeds from left to right ALWAYS, and as long as there is 1:1 ratio ot A to B to react together, it will keep forming C no matter what.