Chemistry Question

[RenaissanceBiker]

I'm probably overcomplicating this but I'm hung up on a chemistry problem. Say the hydrolysis of a chemical proceeds in two steps as follows:

P₃O₁₀^5- + H₂O > P₂O₇^4- + PO₄^3- + 2H⁺
P₂O₇^4- + H₂O > 2PO₄^3- + 2H⁺
The respective reaction rates are k₁ = 2.3E-6 s^-1 and k₂ = 1.0E-6 s^-1.

These reactions can be combined and written as:

P₃O₁₀^5- + 2H₂O > 3PO₄^3- + 4H⁺
Is the combined reaction rate simply k = k₁ * k₂ = 2.3E-12 ?

That would mean the half life of P₃O₁₀^5- in the first reaction is 3.5 days, but in the overall reaction the half life is over 9,000 years.

ETA: On second thought, that would not be dimensionally correct. The units of k would be s^-2. Also, I meant to put this in Math and Science. Can Mod help me out there?

 

[Madalch]

I'm probably overcomplicating this but I'm hung up on a chemistry problem. Say the hydrolysis of a chemical proceeds in two steps as follows:

P₃O₁₀^5- + H₂O > P₂O₇^4- + PO₄^3- + 2H⁺
P₂O₇^4- + H₂O > 2PO₄^3- + 2H⁺
The respective reaction rates are k₁ = 2.3E-6 s^-1 and k₂ = 1.0E-6 s^-1.

These reactions can be combined and written as:

P₃O₁₀^5- + 2H₂O > 3PO₄^3- + 4H⁺
Is the combined reaction rate simply k = k₁ * k₂ = 2.3E-12 ?

No. The rate of the overall reaction, to a first approximation, is the rate of the slowest step.

 

[Madalch]

I'm probably overcomplicating this but I'm hung up on a chemistry problem. Say the hydrolysis of a chemical proceeds in two steps as follows:

P₃O₁₀^5- + H₂O > P₂O₇^4- + PO₄^3- + 2H⁺
P₂O₇^4- + H₂O > 2PO₄^3- + 2H⁺
The respective reaction rates are k₁ = 2.3E-6 s^-1 and k₂ = 1.0E-6 s^-1.

These reactions can be combined and written as:

P₃O₁₀^5- + 2H₂O > 3PO₄^3- + 4H⁺
Is the combined reaction rate simply k = k₁ * k₂ = 2.3E-12 ?

That would mean the half life of P₃O₁₀^5- in the first reaction is 3.5 days, but in the overall reaction the half life is over 9,000 years.

In more detail, let's see. I'm a little rusty at this, but...
Each of the steps is first order in the polyphosphate ion. To make it easier to type, let's call P₃O₁₀^5- "A" and P₂O₇^4- "B". So your reaction sequence is:

A + H2O --> B + 2 H+
B + H2O --> products.

The rate of appearance of products is = k2[.B.] (extra dots added to prevent bolding). However, that's not suitable because we can't have an intermediate in our rate law.

We can throw in the "steady-state assumption", which is to simply assume that the concentration of B is small but constant. If it's constant, then the rate of formation must be equal to the rate of consumption. If the first reaction is irreversible, that means that the second reaction must go at the same rate as the first, and the overall rate law is rate = k1[A].

If the first reaction is in equilibrium, then the rate of consumption of B would be the sum of the rates of the second reaction (k2[.B.]) and the rate of the reverse of the first reaction (k-1[.B.]).

k1[A] = k(-1)[.B.] + k2[.B.], so [.B.] = k1[A]/(k2+k(-1)).

The rate of the formation of products, therefore, is rate = k2*k1*[A]/(k2 + k(-1)). If k(-1) is very small, this simplifies to rate = k1[A].

 

[Madalch]

For irreversible reactions, my phys-chem text gives the rate of the overall reaction (no simplifying assumptions such as steady-state or fast equilibrium) as:

rate = k1[A]o(k1e^(-k1t) - k2e^(-k2t))/(k2-k1), where [A]o is the initial concentration of A and t is time.

Thank Atkins for simplifying assumptions.....

 

[RenaissanceBiker]

It's actually:

A + H₂O = B + C + 2H⁺B + H₂O = 2C + 2H⁺
Simplified to:

A + 2H₂O = 3C + 4H⁺
The literature says if one rate is much slower than the other then it will govern. Fine, but k₁ and k₂ are in the same order of magnitude. I think I will go with the slower one, but I really want to know what it is. I hate it when a graduate level textbook says "it's complicated" then moves on to the next subject.

 

[Madalch]

It's actually:

A + H₂O = B + C + 2H⁺B + H₂O = 2C + 2H⁺
Simplified to:

A + 2H₂O = 3C + 4H⁺

Yes, but that's really irrelevant to the kinetics, unless the rate is pH-dependent. If it's first order or pseudo-first order, then the rate law is rate = k[A], and the fact that it's reacting with water or giving other products makes no difference at all.

The fact that you get a C in each step does make a difference to the rate of appearance of C (which I should have paid attention to). The rate of appearance of C will be the rate of the first reaction plus twice the rate of the second reaction.

rate = k1[A] + 2k2[B.]

Now, the first step probably isn't very irreversible, and the steady-state assumption probably isn't very useful in this case (it's best when the first step is a slow one), and I'm unsure how to proceed without either of these simplifications.

The literature says if one rate is much slower than the other then it will govern. Fine, but k₁ and k₂ are in the same order of magnitude. I think I will go with the slower one, but I really want to know what it is. I hate it when a graduate level textbook says "it's complicated" then moves on to the next subject.

Which text are you using?

That would mean the half life of P3O105- in the first reaction is 3.5 days, but in the overall reaction the half life is over 9,000 years.

The half-life of A, if we assume that the first step is irreversible, will be determined solely by the rate of the first reaction. It's the rates of change for the other two species of interest that are complex.

 

[RenaissanceBiker]

I'm using Water Chemistry by Benjamin.

Now that I look back at the homework question, He wants the rate expression for the disappearance of A and the half life of A. That makes the second reaction irrelevant, right?

I'm hesitant to think it's this simple. My instructor hasn't made anything simple yet.

Out of curiosity, let's assume I can calculate a combined rate constant and thus a half life for the combined reaction. Am I right to assume that it would be the time it takes half of A to become 3C + 4H₂O, not the half life of A?

 

[Madalch]

I'm using Water Chemistry by Benjamin.

Now that I look back at the homework question, He wants the rate expression for the disappearance of A and the half life of A. That makes the second reaction irrelevant, right?

That's correct.

Out of curiosity, let's assume I can calculate a combined rate constant and thus a half life for the combined reaction. Am I right to assume that it would be the time it takes half of A to become 3C + 4H₂O, not the half life of A?

That would be correct.

 

[pgwenthold]

I'm using Water Chemistry by Benjamin.

Now that I look back at the homework question, He wants the rate expression for the disappearance of A and the half life of A. That makes the second reaction irrelevant, right?

I'm hesitant to think it's this simple. My instructor hasn't made anything simple yet.

Given all your confusion over it, it sounds to me like he/she did make it pretty complicated for you.

If that is the question, it sounds like you are sorting it out. The rate of disappearance of A in the first expression (assuming first order kinetics, which could be pseudo first order) is given by -dA/dt = k[A]

Notice that the rate of that reaction does not depend on the concentration of anything else, such as [C]. As such, the fate of [C] has no bearing on the rate of reaction of A, in this case.

Now, if C were to react in a way that ultimately led to the formation of A, it could have an effect, but that is not the case here.

OTOH, the second reaction WILL have an effect on the rate of the reverse reaction, and ultimately the makeup of the system at equilibrium (I love le Chatelier's principle)

 

[Madalch]

OTOH, the second reaction WILL have an effect on the rate of the reverse reaction, and ultimately the makeup of the system at equilibrium (I love le Chatelier's principle)

This doesn't strike me as a reaction that's particularly reversible- I wouldn't expect phosphate ions to form diphosphate or triphosphate under ordinary, aqueous conditions. The CRC doesn't even give thermodynamic data for the diphosphates, and they can generally be prepared only by cooking the solid hydrogen phosphates.

 

[pgwenthold]

This doesn't strike me as a reaction that's particularly reversible-

Every reaction is reversible. The only question is what are the relative rates of the forward and reverse processes.

 

[Madalch]

Every reaction is reversible. The only question is what are the relative rates of the forward and reverse processes.

There are some reactions with an effective rate of zero.

 

[pgwenthold]

There are some reactions with an effective rate of zero.

"Effective rate" is a relative term, and depends on the circumstances. If the rates are so low that half lives are on the scale of the age of the universe or something like that, then sure, that is "effectively" zero, but is that the case here? What's the equilibrium constant?

 

[Madalch]

"Effective rate" is a relative term, and depends on the circumstances. If the rates are so low that half lives are on the scale of the age of the universe or something like that, then sure, that is "effectively" zero, but is that the case here?

I don't know, but I suspect that the rate of the reverse reaction is negligible.

What's the equilibrium constant?

I don't know- the CRC doesn't have thermodynamic data for the diphosphate or triphosphate ion, so I haven't tried calculating K (nor have I yet resorted to Google). But the equilibrium constant won't tell us anything about the rates of the reaction.

 

[RenaissanceBiker]

The professor canceled class tomorrow and gave us another week to do the homework. That means it will be two weeks before I will get it back graded. Thanks for your help. I'll let you know how it turns out.