Why is the stoichiometric coefficient in the denominator in the rate expression?

For the reaction: $$\ce{2A + B → C}$$

The rate expression is given as:

$$\ce{-1/2 * Δ [A]/Δt}$$ = $$\ce{-1 * Δ [B]/Δt}$$ = $$\ce{Δ [C]/Δt}$$

I can't get my head around why it is $$\ce{-1/2 * Δ [A]/Δt}$$ and not $$\ce{-2 * Δ [A]/Δt}$$

If A is consumed at twice the rate that B is consumed, why is the rate for A half (and not double) the rate for B?

The only answer I could find on the internet is "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." Source: http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Kinetics/Rate_Laws/The_Rate_Law

But this doesn't make much sense to me. Why does the rate with respect to all substances have to be equal when one substance (A) is used up twice as fast as another substance (B)?

• It is a rate expression, which means how fast the reaction occurs. Think about it that way: if the rate of reaction corresponds to the rate of formation of C, then it corresponds to half of the rate of consumption of A. A is consumed faster, so to get the rate of reaction you divide by its stochiometric coefficient. – cfcief Jul 10 '16 at 8:23
• Hi cfcief. Thanks for the response. Why does the rate of reaction correspond to the rate of formation of C (as opposed to the rate of consumption of A)? – K-Feldspar Jul 10 '16 at 8:29
• You wrote the rate of reaction with respect to A, B and C, which is correct. You have two molecules of A, one molecule of B which will react to give one molecule of C. In this case, how would you measure the rate of reaction? One reaction = one molecule of C formed, one molecule of B consumed, two molecules of A consumed. So the rate of reaction correspond to the rate of formation of C, to the rate of consumption of B, to half of the rate of consumption of A. Remember that A is consumed twice as fast as B. Hope that clears things up. – cfcief Jul 10 '16 at 8:40
• Ah I see now! Thanks for explaining that cfcief. – K-Feldspar Jul 10 '16 at 9:03
• @CurtF. I don't think this is a duplicate. This question here is not concerned with the elementary reactions, reaction order and rate constants but with the general relationship between the reaction rates of the different compounds in the reaction. – Philipp Jul 10 '16 at 14:28

From the stoichiometric coefficients $\nu_{i}$ in your reaction equation

$$\ce{2A + B → C}$$

you can see (as you already correctly stated) that for each molecule of $\ce{B}$ consumed, $\nu_{B} = -1$, two molecules of $\ce{A}$ must be consumed as well, $\nu_{A} = -2$. Similarly for each molecule of $\ce{B}$ consumed one molecule of $\ce{C}$ must be formed, $\nu_{C} = +1$ (and analoguously for $\ce{A}$ and $\ce{C}$). This will also be reflected in the changes of the amounts of those species, $\Delta n_{i}$, during the reaction, e.g. if the amount of $\ce{B}$ changes by $2 \, \mathrm{mol}$ you can expect a change in the amount of $\ce{A}$ by $4 \, \mathrm{mol}$. So, the ratios of the $\{\Delta n_{i}\}$ and the $\{\nu_{i}\}$ will be equal, i.e.

\begin{align} \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{B}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{B}}} \ ; && \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{C}}} \ ; && \frac{\Delta n_{\ce{B}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{B}}}{\nu_{\ce{C}}} \end{align}

Rearranging these equations a bit leads you to:

\begin{align} \frac{\Delta n_{\ce{A}}}{\nu_{\ce{A}}} = \frac{\Delta n_{\ce{B}}}{\nu_{\ce{B}}} = \frac{\Delta n_{\ce{C}}}{\nu_{\ce{C}}} \ . \end{align}

Now, divide these equations by the reaction volume $V$ and use $\frac{\Delta n_{i}}{V} = c_{i} \equiv [i]$ to introduce concentration changes

\begin{align} \frac{\Delta [\ce{A}]}{\nu_{\ce{A}}} = \frac{\Delta [\ce{B}]}{\nu_{\ce{B}}} = \frac{\Delta [\ce{C}]}{\nu_{\ce{C}}} \ . \end{align}

Finally, if you divide now by the reaction time $\Delta t$ and use the values for the stoichiometric coefficients, you get the equations from your question:

\begin{align} \frac{1}{\nu_{\ce{A}}} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{\nu_{\ce{B}}} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{\nu_{\ce{C}}} \frac{\Delta [\ce{C}]}{\Delta t} \\ \frac{1}{-2} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{-1} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{1} \frac{\Delta [\ce{C}]}{\Delta t} \\ -\frac{1}{2} \frac{\Delta [\ce{A}]}{\Delta t} &= -1 \frac{\Delta [\ce{B}]}{\Delta t} = 1 \frac{\Delta [\ce{C}]}{\Delta t} \ . \end{align}

• Thanks for the response Phillip. That helps my understanding. I can follow all of that except I am a little confused as to the difference between ΔnA and vA. Va is the amount of moles of A, with the units moles (given by the coefficient in the reaction)? How did you come up with ΔnA and how is it different to vA (or a scaled up version of vA)? – K-Feldspar Jul 10 '16 at 21:04
• @Gems9 $\Delta n_{i}$ is the change in the amount of substance $i$ during the reaction time $\Delta t$. It has the unit of $\mathrm{mol}$. The stoichometric coefficient $\nu_{i}$ is unit-less. It's just a number. It is the coefficient in the reaction equation (except that it is negative for educts and positive for products). – Philipp Jul 10 '16 at 21:30  Since the rate of consumption of A is twice the rate of formation of c, to make it equal, the term Δ[A] is divided by 2.

Therefore the rate of reaction is given by = The rate is just a characteristic of a reaction, not how fast one element's concentration is increasing or not. You can simply multiply it by stoichiometric coefficient to find the rate of one element's consumption/creation.

So the answer: "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." is the right one which was shown in a detailed way by philip:

From the stoechiometric coefficients νi

2A+B→C

you can see (as you already correctly stated) that for each molecule of B consumed, νB=−1, two molecules of A must be consumed as well, νA=−2. Similarly for each molecule of B consumed one molecule of C must be formed, νC=+1 (and analoguously for A and C). This will also be reflected in the changes of the amounts of those species, Δni, during the reaction, e.g. if the amount of B changes by 2mol you can expect a change in the amount of A by 4mol. So, the ratios of the {Δni} and the {νi}

will be equal, i.e.

ΔnAΔnB=νAνB ;ΔnAΔnC=νAνC ;ΔnBΔnC=νBνC

Rearranging these equations a bit leads you to:

ΔnAνA=ΔnBνB=ΔnCνC .

Now, divide these equations by the reaction volume V and use ΔniV=ci≡[i]

to introduce concentration changes

Δ[A]νA=Δ[B]νB=Δ[C]νC .

Finally, if you divide now by the reaction time Δt

and use the values for the stoechiometric coefficients, you get the equations from your question:

1νAΔ[A]Δt1−2Δ[A]Δt−12Δ[A]Δt=1νBΔ[B]Δt=1νCΔ[C]Δt=1−1Δ[B]Δt=11Δ[C]Δt=−1Δ[B]Δt=1Δ[C]Δt .