Let’s consider the reaction, which proceeds through the following mechanism:

$ ClX_{2}kX_{1}kX_{−1} 2ClCl+CHClX_{3}kX_{2} HCl+CClX_{3}Cl+CClX_{3}kX_{3} CClX_{4} (1)(2)(3) $Reactions (1)-(3) are elementary reactions, that is, their rate can be expressed using the rate law with stoichiometric coefficients directly corresponding to the orders.

How might we solve for the overall reaction rate? There are two approaches:

- More conceptual
- More mathematical

First, it’s important to understand what is the overall equation of the chemical reaction. Most of the times it’s known, but even if it isn’t, we can sum reactions (1)+(2)+(3) to obtain:

$ClX_{2}+2Cl+CHClX_{3}+CClX_{3}ClX_{2}+CHClX_{3} 2Cl+HCl+CClX_{3}+CClX_{4} HCl+CClX_{4} (4)(5) $It should be noted that sometimes you have to add some reactions several times (in other words, multiply all coefficients by some number). How could you know? Only if you understand that, for example, $Cl$ and $CClX_{3}$ are unstable particles, and so cannot be isolated, which means that they can’t be initial reactants or final products.

Now, how do we find overall rate?

## Conceptual Approach

In the conceptual approach, we assume that step (2) is the rate-determining step, i.e. it’s the slowest reaction. At least, that’s what the slide says. To be more precise, we have to assume a somewhat stronger statement: the step (2) is slow enough to allow reaction (1) reach an equilibrium.

When (1) reaches an equilibrium, the rate of the forward reaction in step 1 is equal to the rate of the reverse reaction (see notes on equlibrium for derivation of this):

$r_{rev}k_{−1}[Cl]_{2} =r_{fwd}=k_{1}[ClX_{2}] (6)(7) $This will prove to be useful in a moment. We know that the reaction rate is determined by the rate of the slowest step. We’re told (or we assume) that the slowest step is step (2). We can write the rate of this step using rate law:

$r_{2}=k_{2}[Cl][CHClX_{3}](8)$Okay, so we have some expression which depends on the concentration of a reactant $CHCX_{3}$, which is good, but we also have dependence on the concentration of the intermediate particle $Cl$, which is not good. But we can express it from (7) as:

$[Cl]=k_{−1}k_{1} [ClX_{2}] (9)$So the rate of the reaction becomes:

$r_{2}=k_{2}k_{−1}k_{1} [ClX_{2}]_{1/2}[CHClX_{3}](10)$If we redefine

$k_{2}k_{−1}k_{1} =k_{eff}(11)$we get:

$r_{2}=k_{eff}[ClX_{2}]_{1/2}[CHClX_{3}](12)$which looks like the rate law, but with bizarre orders. But this is fine, this is the *apparent* rate law. If we were to measure the dependence of rate on concentrations, we would find the value of $k_{eff}$.

## Mathematical Approach

I guess I have to say it explicitly that this part is **COMPLETELY** optional, but imo is fun and satisfactory.

If we’re interested in the reaction rate, we presumably are interested in the rate at which the product forms:

$dtd[CClX_{4}] (2.1)$$CClX_{4}$ is only formed in step (3), so using the rate law:

$dtd[CClX_{4}] =k_{3}[Cl][CClX_{3}](2.2)$We have the dependence on two unstable particles. To move forward, we introduce the notion of a *steady-state approximation*. Under this approximation, whenever we have an unstable particle, we **assume** that it’s rate of change is almost 0. Conceptually we claim: the unstable particle is unstable, so it cannot accumulate, meaning that as soon as it’s formed, it reacts further. As a result, for some time we have a constant inflow and constant outflow of that particle, and the change in its concentration is negligible (and the concentration itself is pretty small). Within this approximation, we can claim:

Okay, similarly to the derivation of the equilibrium condition, let’s note that $CClX_{3}$ is formed in second reaction and consumed in the third reaction. As a result, the total change in the concentration will be:

$dtd[CClX_{3}] =k_{2}[Cl][CHClX_{3}]−k_{3}[Cl][CClX_{3}](2.4)$Second reaction has a positive sign because it increases the concentration of $CClX_{3}$, and the third reaction has a negative sign because $CClX_{3}$ is consumed in it. As a result of (2.3), from (2.4) it follows:

$k_{2}[Cl][CHClX_{3}]=k_{3}[Cl][CClX_{3}](2.5)$Let’s now repeat the procedure for $Cl$:

$dtd[Cl] =2k_{1}[ClX_{2}]−2k_{−1}[Cl]_{2}−k_{2}[Cl][CHClX_{3}]−k_{3}[Cl][CClX_{3}]=2k_{1}[ClX_{2}]−2k_{−1}[Cl]_{2}−2k_{2}[Cl][CHClX_{3}]=0=k_{1}[ClX_{2}]−k_{−1}[Cl]_{2}−k_{2}[Cl][CHClX_{3}]=0=−k_{−1}[Cl]_{2}−k_{2}[Cl][CHClX_{3}]+k_{1}[ClX_{2}]=0=[Cl]_{2}+k_{−1}k_{2} [CHClX_{3}][Cl]−k_{−1}k_{1} [ClX_{2}]=0 (2.6)(2.7)(2.8)(2.9)(2.10) $The coefficients of $2$ before $k_{1}$ and $k_{−1}$ occur because 2 particles of $Cl$ are produced and consumed in the forward and reverse reaction respectively. To get (2.7) we use (2.5). To get (2.8) we divide by 2. (2.9) is just rearrangement of the terms. We divide (2.9) by $−k_{−1}$ to get (2.10), which looks like a quadratic equation with $a=1$ if we denote $b=k_{−1}k_{2} [CHClX_{3}]$, $c=−k_{−1}k_{1} [ClX_{2}]$, and $x=[Cl]$. That is, we obtain an equation of the form:

$ax_{2}+bx+c=0(2.11)$if you wonder why did I bring it to the form in which $a=1$, well, because at first attempt I didn’t, and I saw that the result was a bit messy, so I tried bringing to $a=1$ and it looked better. Anyway, the equation (2.11) has solutions of the form:

$x_{1/2}=2−b±b_{2}−4c (2.12)$Substituting things back:

$[Cl]_{1/2}=2−k_{−1}k_{2} [CHClX_{3}]±k_{−1}k_{2} [CHClX_{3}]_{2}+4k_{−1}k_{1} [ClX_{2}] (2.13)$The first term in the numerator is negative, the square root is also positive, therefore the only way for the concentration of $[Cl]$ to be positive, only the root with $+$ sign in the numerator is physically meaningful:

$[Cl]=2−k_{−1}k_{2} [CHClX_{3}]+k_{−1}k_{2} [CHClX_{3}]_{2}+4k_{−1}k_{1} [ClX_{2}] (2.14)$Now, lets rewrite (2.2) considering (2.5) and (2.14):

$dtd[CClX_{4}] =k_{3}[Cl][CClX_{3}]=k_{2}[Cl][CHClX_{3}]=k_{2}2−k_{−1}k_{2} [CHClX_{3}]+k_{−1}k_{2} [CHClX_{3}]_{2}+4k_{−1}k_{1} [ClX_{2}] [CHClX_{3}](2.15)$You might say: this looks AWFUL. How does it have ANYTHING to do with what we’ve got in (10)? As in that funny (yes, I insist) joke with the lecturer, “patience, let me explain”.

If we assume step (2) to be faster than step (1), it could mean a few things:

- (A) $k_{2}[Cl][CHClX_{3}]≪k_{1}[ClX_{2}]$ and $k_{2}[Cl][CHClX_{3}]≪k_{−1}[Cl]_{2}$. $≪$ means way smaller. big time smaller (I also think this is funny)
- (B) OR $k_{2}≪k_{1}$ and $k_{2}≪k_{−1}$

these are slightly different conditions. In any case, let’s assume: the relative rates are such that:

$k_{−1}k_{2} [CHClX_{3}]_{2}≪4k_{−1}k_{1} [ClX_{2}](2.16)$which means:

$k_{−1}k_{2} [CHClX_{3}]_{2}+4k_{−1}k_{1} [ClX_{2}] ≈4k_{−1}k_{1} [ClX_{2}] =2k_{−1}k_{1} [ClX_{2}] (2.17)$Which is more likely to be true with the second interpretation (B). Also, when $k_{2}≪k_{−1}$, we might say:

$k_{−1}k_{2} ≈0(2.18)$as a result:

$−k_{−1}k_{2} [CHClX_{3}]≈0(2.19)$and so, (2.15) becomes:

$dtd[CClX_{4}] =k_{2}22k_{−1}k_{1} [ClX_{2}] [CHClX_{3}]=k_{2}k_{−1}k_{1} [ClX_{2}]_{1/2}[CHClX_{3}](2.20)$exactly the result of (10).

If you love being rigorous and you cringed from steps (2.17), (2.18), (2.19) and you said “oh, how convenient,” well yes, that’s exactly the kind of assumptions we make. The result we obtain conceptually is only valid when these conditions are satisfied. In the most strict sense, the (2.15) is the most correct rate law. Well, as long as we’re within the steady-state approximation.

I guess at this point you might wonder why bother with all of this math if we get to the same result as in the conceptual approach, but we spend way more time? We didn’t have to, the only reason why I did it is because this is actually a more general approach that is applicable to any reaction of any complexity. It also may help to see that chemistry, if you want it to be rigorous, can be rigorous.

Also, by the way, we could have said back at (2.6) that the condition “equilibrium establishes faster than reaction proceeds” is equivalent to saying:

$2k_{1}[ClX_{2}]−2k_{−1}[Cl]_{2}≫k_{2}[Cl][CHClX_{3}]+k_{3}[Cl][CClX_{3}]$which meant we could ignore the 3rd and 4th terms in (2.6) and conclude immediately:

$2k_{1}[ClX_{2}]−2k_{−1}[Cl]_{2}=0$which would bring us to the result of (7) and we would have finished way sooner.