Chemical kinetics, changing reaction rate

[ashtonjam]

TBR General Chemistry Section 9 Kinetics, Passage VII, #46.

There appears to be conflict between what the passage says and the answer key says.

46. Addition of water to the reaction would have what effect?
A. It would increase the reaction rate.
B. It would have no effect on the reaction rate.
C. It would decrease the reaction rate.
D. It would make the reaction third-order.

Answer: C

What I highlighted in the passage says that if the species is not in the rate determining step, altering its concentration shouldn't affect the reaction rate. The answer explanation says that adding water DOES affect reaction rate, even though water isn't in the rate determining step. Anyone want to clarify why adding water affects the rate? I'm not totally understanding their reasoning. 🙁 #49 is a similar case, adding OH- (increasing pH) somehow makes reaction rate go up, so it's not just this problem that is weirding me out.

Choice C is correct. Adding water to the reaction mixture would push the Step I reaction backwards, generating more reactants ([(NH3)5CoCl]2+ and OH") and consuming the intermediate product ([(NH3)4Co(NH2)Cl]+). The rate-determining step, Step II, depends on the concentration of the product from Step I [(NH3)4Co(NH2)Cl]+. Thus, a decrease in products from Step I slows down the rate-determining step, thereby decreasing the overall reaction rate. The best answer is choice C. Choose C for sensations of correctness and satisfaction.

EDIT: May have reasoning as to why the bolded stuff is true at the bottom of the thread.

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[smiley27]

Think of Le Chatelier's principle, how does adding water affect the concentration of what IS in the rate determining step? Additionally, think of Le Chatelier's principle, how does adding OH- affect the concentration of what is in the rate determining step?

 

[ashtonjam]

Ok that makes sense. I just wasn't sure if equilibrium was an issue and was relying on the passage info. Is the passage info wrong, then? And is water a zero order reactant? It just seems weird that adding a greater amount of a zero order reactant would change the reaction rate because that's what shouldn't happen.

 

[FeinMS]

Water does not directly affect the kinetics, but does so indirectly by changing the [(NH3)4Co(NH2)Cl+]. Adding water increases the amount of product in Step 1, which pulls the Step 1 toward the reactants. This decreases the [(NH3)4Co(NH2)Cl+], therefore slowing Step 2. The reaction is still zero order relative to water, first order relative to (NH3)4Co(NH2)Cl+

 

[ashtonjam]

Thanks, that makes sense. I'll just assume that the passage is wrong and that equilibrium does play a role in reaction rate.

Does this only work because the first reaction is a reversible one with k1 and k-1? If it was only a forward reaction, adding water wouldn't do anything?

 

[sps27]

Thanks, that makes sense. I'll just assume that the passage is wrong and that equilibrium does play a role in reaction rate.

Does this only work because the first reaction is a reversible one with k1 and k-1? If it was only a forward reaction, adding water wouldn't do anything?

yup I remember this question now. I did it long time ago. Dang!!!!! I still got it wrong. Anyways, as said above, LeeChatliers principle plays a part here.

 

[ashtonjam]

Do you know if using Le Chatelier's principle is only possible because there is both a forward and backwards reaction (k1 and k-1)?

 

[sps27]

Do you know if using Le Chatelier's principle is only possible because there is both a forward and backwards reaction (k1 and k-1)?

I really don't think so. I mean if I look at the guts of what LeeChaterlier is saying i.e., if you stress a system, a rxn (reaction), then the rxn proceeds exactly in the direction to minimize the stress......so really it does not have forward and backward in it at all. And if it did, I am sure the books would have stated that in bold letters. So my understanding is, it is immaterial of forward or backward. The forward and backward really has to do with equilibrium, I think.

The concept of LeeChatterlies in my mind is similar to the concept of entropy. The concept of entropy is to spread out energy rather than bring it together or hold it, and LeeChatteliers is kind of the same, that it tries to oppose anything that stresses the system i.e,, if we considered holding energy to be distressing the system, which would mean all of us biological entities are anti-entropy and the universe will eventually end us some day. ha ha ha ha......

Anyways I am getting off topic here.

 

[ashtonjam]

Small bump, still trying to figure out why adding OH- or water alters the reaction rate in a mathematical sense. I may have an answer now. Thanks for input from previous posters.

This link from Chemguide had a bit of info about fast equilibria in the first step of a mechanism near the bottom of the page. I have a feeling that if the first reaction is faster than the second, there will be an equilibrium no matter what. So, the overall reaction rate may have a lot to do with the first.

Overall Reaction, and solving for the rate expression:

[(NH3)5CoCl]2+ = X, [(NH3)4Co(NH2)Cl]+ = Y, [(NH3)4CoNH2]2+ = Z

• Rate=k[Y]

But Y is in equilibrium in Step I, so writing out the equilibrium is necessary to solve for what [Y] really is.

• Keq=[Y]/[OH][X]
• [Y]=Keq*[OH][X]

Plugging this expression in for [Y] will give a different rate law than expected.

• Rate=k[Y]
• Rate=k*Keq*[OH][X]

Combine both k and Keq into a new rate constant, k'.

• Rate=k'[OH][X]

This implies that the rate is not dependent on compound Y at all? It would explain why adding more OH- would drive the reaction forward, but since water doesn't appear in equilibrium expressions, that's tripping me up as to why water affects the rate.

Maybe water actually does appear in the Keq expression. If it did, the rate expression would be divided by [H2O] at the end, which would actually make sense because adding water in the denominator would decrease the rate. Possible answer?

• Rate=k'[OH][X]/[H2O]

 

[BerkReviewTeach]

Thanks ashtonjam for providing the full reaction. Pretty much everything of intereste has been well discussed here already except for the sovent influence on concentration.

There are two factors at play here, one which fits well with the sentence from the passage in question and the second that walks the line of violating that sentence.

The first factor is that water is the solvent, which many people overlook. When the solvent is involved as a reactant or product, we generally ignore it because it is in such high concentration that it has little to no impact on the reaction rate (appears to be zero-order) and no impact on the equilibrium (its concentration is essentially constant). But the role it can play here is to dilute all species in the reaction mechanism, including [(NH₃)₄Co(NH₂)Cl]⁺. That effect will be minimal in all likelihood, but adding water will dilute the concentration of [(NH₃)₄Co(NH₂)Cl]⁺ and thus reduce the reaction rate.

The second factor (mentioned already) is that water has an indirect effect on the rate-determining step by pushing the initial equilibrium to the left and reducing the amount of [(NH₃)₄Co(NH₂)Cl]⁺ in solution. That reality makes the last sentence in the yellow highlighted portion of the OP's post invalid. It's generally true, but poorly worded.

Given that, I'm very impressed at how well this very tough passage was handled.

 

[ashtonjam]

Thanks so much for the response! I'll try to remember this for any future passages. Hope they aren't as tricky, though.