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I found this problem in my book and I'm not sure how to assess this question.

It is as follows:

Regarding the relationship between the rates of the chemical species in the following reaction. Which of the following alternatives is the correct one?

$\ce{2H2 + O2 -> 2 H2O}$

$\begin{array}{ll} 1.&v_{H_2}=2v_{H_{2}O}\,, v_{O_2}=2v_{H_{2}}\\ 2.&v_{O_2}=2v_{H_{2}O}\,, v_{H_2}=2v_{O_{2}}\\ 3.&v_{H_2}=2v_{O_{2}}\,, v_{O_2}=v_{H_{2}O}\\ 4.&v_{H_2}=v_{H_{2}O}\,, v_{O_2}=\frac{1}{2}v_{H_{2}}\\ 5.&v_{H_2}=v_{H_{2}O}\,, v_{O_2}=v_{H_{2}}\\ \end{array}$

In the original source (which doesn't list a specific authorship) doesn't indicate the state of matter for each species. So I'm assuming perhaps that the reagents are gases and the products is liquid?.

Anyways what I've attempted to do is to establish the rate law as follows using the law of mass action:

$\ce{2H2 + O2 -> 2 H2O}$

For Hydrogen gas:

$v_{H_{2}}=-k[H_{2}]^2[O_{2}]$

For Oxygen gas:

$v_{O_{2}}=-k[H_{2}]^2[O_{2}]$

From this it can be established that:

$v_{H_{2}}=v_{O_{2}}$

For water (since it is on the products it must be established as if it were a reagent):

$v_{H_{2}O}=k[H_{2}O]^2$

And that's where I'm stuck.

Supposedly the answer is

$v_{H_2}=2v_{O_{2}}\,, v_{O_2}=v_{H_{2}O}$

But I don't see any way of how to get there. Can somebody point me exactly what did I missed?.

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    $\begingroup$ Rate of consuming of reagents ( or production of products) must be in ratio of their stoichiometric coefficients ( for case of a single step, or a pseudo single step reaction) $\endgroup$ – Poutnik Dec 26 '19 at 8:36
  • $\begingroup$ @Poutnik I recall that if water is a solvent then it can be left out of the rate law. But in this case it is a product. By comparing to what I attempted to do I placed the stoichiometric coefficients in the rate law for each species, were you referring to this?. $\endgroup$ – Chris Steinbeck Bell Dec 26 '19 at 8:51
  • $\begingroup$ Hm, unless it is in context of some fuel cell, water will be hardly in liquid state.But for fuel cells, different reactions would occur. $\endgroup$ – Poutnik Dec 26 '19 at 8:54
  • $\begingroup$ @Poutnik You're right about fuel cells. But in a general context as (I think it was intended in this question) does it exist a way to relate those three rate laws as presented in the alternatives?. $\endgroup$ – Chris Steinbeck Bell Dec 26 '19 at 9:02
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Rate law vs. definition of rate

For this problem, you don't need the rate law. And, just for the record, you can't get the rate law from the stoichiometry - it depends on the mechanism and has to be determined empirically.

Definition of rate of reaction

The rate of reaction is based on the change of concentrations. For a homogeneous reaction (i.e. all species in the same gas phase, or all species part of the same solution), the definition is the following:

$$\mathrm{rate} = \frac{d[\ce{X}]}{\nu_X \cdot dt}$$

where $\nu_X$ is the stoichiometric coefficient of the species $\ce{X}$. This does not apply to heterogeneous reactions. For one, if a species is a pure solid or liquid, its concentration will not change. Also, the stoichiometry of a reaction refers to amount of substance - only if all species are in the same solution (or the same gas phase) does this translate into easy relationships of concentration changes (or partial pressure changes).

Definition of "rates of the chemical species"

This is not a technical term I am familiar with. In order to be able to answer the question with answer 3., you would have to define it as:

$$v_\ce{X} = \frac{\mid dn_\ce{X} \mid}{dt}$$

Once you define it such, it does not matter what the physical state of reactants and products is, it does not matter whether a species is a reactant or a product, the only thing you need to related the different rates are the stoichiometric coefficients.

So for example, the ratio of the amount of hydrogen and oxygen used is equal to the ratio of stoichiometric coefficients:

$$\frac{\Delta n_\ce{H2}}{\Delta n_\ce{O2}} = \frac{\nu_\ce{H2}}{\nu_\ce{O2}} = \frac{2}{1}$$

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  • $\begingroup$ The answer according to the answers sheet in my book says it is the answer $3$ which would be the same as answer $4$?. From your answer the idea is that the rate of reaction by definition is the inverse of the stoichiometric coefficient?. Then why in literature is it always presented with the law of mass action instead?. Did I missed something. Can you explain this part or the relationship between these two concepts?. Btw I was aware that the rate law depends on the mechanism. $\endgroup$ – Chris Steinbeck Bell Dec 27 '19 at 4:35
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    $\begingroup$ @ChrisSteinbeckBell My mistake, answer 3 is the one that matches the stoichiometry. I edited my answer - thanks for paying attention. $\endgroup$ – Karsten Theis Dec 27 '19 at 22:45
  • $\begingroup$ You're welcome but there's still one of my doubts unnatended which is regarding on why in the books rate law is defined by the law of mass action. I mean $aA+bB\rightarrow cC$ being for $A$ rate law $\dfrac{dA}{dt}=-k[A]^a[B]^b$ and not by $\textrm{rate (A)}=\dfrac{d[A]}{adt}$?. You used the word rate but isn't rate the same as $\frac{d[X]}{dt}$ why is the stoichiometric coefficient in the denominator?. why is it the inverse?. Is it due a definition? $\endgroup$ – Chris Steinbeck Bell Dec 28 '19 at 19:40
  • $\begingroup$ @ChrisSteinbeckBell You want to have a single rate for a reaction even if not all species have 1:1 stoichiometry. So the reaction rate is defined with the stoichiometric factor in the denominator. No matter which species concentration you monitor, you will find the same rate. $\endgroup$ – Karsten Theis Dec 28 '19 at 23:32
  • $\begingroup$ @ChrisSteinbeckBell "In the books, rate law is defined by the law of mass action"? The rate law usually starts with rate =, and uses coefficients m, n, q etc. that are distinct from the stoichiometric coefficients, e.g. openstax.org/books/chemistry-2e/pages/12-3-rate-laws. If your textbook says otherwise, ask a specific question showing the portion of the textbook. $\endgroup$ – Karsten Theis Dec 28 '19 at 23:36
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The fourth possibility is the only correct answer, because the numbers it contains are proportional to the stoichiometric coefficients, whatever the physical state. The fourth possibility is the only one where O2 is consumed at a lower rate than H2. And this is exactly what the equation says : when 1 H2 is consumed, only 1/2 mole of O2 is destroyed.

And by chance the other relationship in 4th answer is also correct. The number of H2O produced is always legal to the number of H2 consumed, whatever the physical state.

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    $\begingroup$ Im still confused about what you mentioned about velocities proportional to their stoichiometric coefficientes, can you put this as a relationship for a generic situation so i could see it better? What ive attemped to do is follow the definition for rate law based on law of mass action but as you can see i could not get to an answer. There's still my question about water, should it account for rate law? How should it be wrote that velocity in terms of the other species? Is what i did correct? $\endgroup$ – Chris Steinbeck Bell Dec 26 '19 at 12:08

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