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Consider a gaseous state elementary reaction $$\ce{aA(g) + bB(g)} \overset{k_\mathrm{f}}{\underset{k_{\mathrm{b}}}{\ce{<=>}}}\ce{ cC(g) + dD(g)}$$

I know that for this reaction, $$\Delta G = \Delta G^{\circ} + RT \ln \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) \tag1$$

Now, if $$K = \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) ,$$

then $\Delta G = \Delta G^{\circ} + RT \ln K$.

Now, how can we say that $$K = \frac{k_{\mathrm{b}}}{k_{\mathrm{f}}}$$

and $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{f}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$ $\,$where $r_{\mathrm{f}}$ and $r_{\mathrm{b}}$ are rates of forward and backward reaction as assumed in this answer.

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

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    $\begingroup$ You've neglected the big assumption in the post you quoted---that this reaction is assumed elementary. $\endgroup$ – a-cyclohexane-molecule Aug 8 '17 at 10:52
  • $\begingroup$ @cyclohexane-molecule Edited question. But even though the problem remains. $\endgroup$ – Apoorv Potnis Aug 8 '17 at 11:05
  • $\begingroup$ Khan Academy related video - youtu.be/psLX080RQR8 $\endgroup$ – Apoorv Potnis Sep 5 '17 at 7:18
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I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

I might try to give you some intuition to back up that, for a given (elementary) reaction $\ce{A ->[k] B}$, the reaction rate $r$ can be written as

$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} \propto P_\ce{A}\text{.}$$

(Observe that the second equality above is true due to $P_\ce{A} + P_\ce{B} = \text{constant}$.) First, for an ideal gas, $P_\ce{A} = \frac{n_\ce{A}}{V} RT$. This means that $P_\ce{A} \propto n$ for fixed temperature and volume.

Let's say the reaction happens as a random process. That is to say that, for every time interval $\Delta t$, we have a probability per unit time $p$ of having a single molecule $\ce{A}$ turning into $\ce{B}$. If we wait longer, proportionally more molecules will turn. We'll thus have, for initially $n_\ce{A}$ molecules of $\ce{A}$, after $\Delta t$ seconds,

$$\Delta n_\ce{B} = p n_\ce{A} \Delta t\text{.}$$

This means that, in the time interval $\Delta t$, the population of $\ce{B}$ goes from 0 to $\Delta n_\ce{B}$ (assuming no $\ce{B}$ initially). From the stoichiometry of the reaction, $\Delta n_\ce{B} = -\Delta n_\ce{A}$ (i.e., there's conservation of moles).

Thus,

$$\frac{\Delta n_\ce{B}}{\Delta t} = -\frac{\Delta n_\ce{A}}{\Delta t} = p n_\ce{A}\text{.}$$

After taking the limit of $\Delta t \rightarrow 0$ and multiplying by $\frac{1}{V}$ (i.e., dividing by volume), we get

$$\frac{d \frac{n_\ce{B}}{V}}{dt} = \frac{d [\ce{B}]}{dt} = p \frac{n_\ce{A}}{V} = p [\ce{A}]\text{,}$$

where $[\ce{A}]$ and $[\ce{B}]$ stand for the concentrations of $\ce{A}$ and $\ce{B}$, respectively. Since, again for an ideal gas, $[\ce{A}] = \frac{P_\ce{A}}{RT}$, we get

$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} = p P_\ce{A}\text{,}$$

where the proportionality constant is the probability of reaction per unit time.

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    $\begingroup$ Apologies for accepting the answer late. The answer convinces me fully. $\endgroup$ – Apoorv Potnis Aug 18 '17 at 15:33
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A late answer...

In a molecular approch to chemical reaction rates we asume that the molecules move randomly in solution and have a velocity randomly assigned according to the Boltzmann distribution and that the chance of being in one position in solution is the same as being at any other, i.e. the molecules are in thermal equilibrium at some temperature T.

The reaction rate for the reaction $\mathrm {A+B \rightarrow}$ can be found by considering the number of collisions per second and then this number reduced by the chance that a reaction occurs on collision. To arrive at a rate the chance that a randomly chosen combination of molecules of type A and B at time t will react in the next infinitesimal time interval $t+\delta t$ and has not reacted up to t has to be calculated. As a result the total number of collisions /second between $N_A$ molecules of type A and $N_B$ of type B is found to be $\sim N_AN_B\sigma_{AB}$, where $\sigma_{AB}$ is the cross sectional area for a collision, and this expression produces the rate when multiplied by the probability $p_{AB}$ of A and B reacting, i.e. rate $\sim N_AN_B\sigma_{AB}p_{AB}$. The probability of reaction $p_{AB}$ depends critically on the activation energy.

The important point is, however, that the rate is proportional to the product of the numbers of molecules of type A and B which in constant volume is the same as being proportional to the product of concentrations. That it also is proportional to the collision cross section and the chance of reacting is also physically realistic. The rate constant is proportional just to the probability of reaction and the cross section for collision.

Thus the origin of the product of concentrations in the rate expression is found in the chance that a molecule of type A will collide with that of type B in a small time interval $t, t +\delta t$. Thus if we ignore the fact that there is variation due to changes in experimental conditions, such as temperature, the electronics of the detectors used to measure the reaction etc. perfect kinetics would be measured, i.e $dA/dt$ or $ dB/dt$ vs. time would be noise free.

If we now consider the random nature of the collisions between molecules a slightly different approach has to be taken. If a small volume of the solution is considered and the rate calculated in that region, because the number of species entering and leaving in any given time period varies, the rate expression is also potentially subject to variation. This happens simply because of the random nature of the motion of molecules in a solution or gas. In this case the rate of change of a species X should be written in terms of its average (or mean) value represented as $\langle X(t)\rangle$;

$$ \frac{d\langle X(t)\rangle}{dt} = \langle f\times X(t)\rangle $$

where $\langle f\times X(t)\rangle$ represents some rate expression such as $k_1\langle X(t)\rangle + k_2\langle X(t)\rangle^2$ etc. except that the number of $X$ are represented as their average values. Clearly the $X$ are always positive values. These values have a variance (square of standard deviation) and so we would expect that different experiments give slightly different results because of this. (This is the approach taken by Gillespie, see 'Markov Processes' publ. Academic Press 1992)

In a bimolecular reaction such as $X + X \rightarrow $ with rate constant k the rate expression is

$$ \frac{d\langle X(t)\rangle}{dt} = k\langle \;X(t)(X(t)-1)\;\rangle/2 $$

The term $\langle X(t)(X(t)-1)\rangle /2$ arises becaus of the number of distinct X-X pairs in the reaction volume at time t. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that X is large compared to $1$ and so $\langle X(t)(X(t)-1)\rangle \rightarrow \langle X^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle X^2(t)\rangle \sim \langle X(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $X(t)=\langle X(t)\rangle$ which means that $X$ is a deterministic process, so that the randon nature of the numbers of $X$ entering into the reaction volume is always the same, which is not what was assumed at the outset and seems to fly in the face of what we know about how molecules move in solution, i.e. their random diffusional nature.

However, in any normal kinetic experiment the number of molecules we measure is vast, $\langle X(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the number of molecules is tiny and the approximation made above is valid.

$$ \frac{\sigma_X}{\langle X \rangle} \sim \frac{\sqrt{\langle X \rangle}}{\langle X \rangle} \sim \frac{1}{\sqrt{\langle X \rangle}}\sim 10^{-9}$$

Even if there are as few as 1000 molecules the realtive fluctuation is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.

Notes :

(A) To see that the variance is zero use the definition $\sigma^2_X = \langle X^2 \rangle -\langle X \rangle^2$ where $\sigma$ is the standard deviation)

(B) There are other situations where the rate equation approach breaks down in macroscopic systems and that is when they are sufficiently non-linear, one example is the Schlogl system of equations, which for certain values of concentrations and rate constants becomes bi-stable;

$$B_1+2X \rightleftharpoons 3X; \;\; B_2 \rightleftharpoons X$$

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    $\begingroup$ I understood around half of the answer. I don't know the maths of variance and deviation. But still, this answer could be useful to people who know that or to me in the future. $\endgroup$ – Apoorv Potnis Aug 19 '17 at 17:02
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    $\begingroup$ :) This web page explains variance and standard deviation. mathsisfun.com/data/standard-deviation.html $\endgroup$ – porphyrin Aug 20 '17 at 9:04

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