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We were given this equation to determine its integrated rate equation:

$$\ce {A + B <=>[$k_1$][$k_2$] C +D ->[$k_3$] E + F}$$

We were given these assumptions to simplify the derivation:

  1. $\displaystyle \frac{\mathrm d[\ce{A}]}{\mathrm dt} = \frac{\mathrm d[\ce{B}]}{\mathrm dt}$, and $\displaystyle \frac{\mathrm d[\ce{C}]}{\mathrm dt} = \frac{\mathrm d[\ce{D}]}{\mathrm dt}$

  2. At any time, $[\ce{A}]=[\ce{B}]$, and $[\ce{C}]=[\ce{D}]$, and $[\ce{E}] = [\ce{A}]+[\ce{B}]$

  3. $\ce{F}$ is continuously removed from the system such that $[\ce{F}] = 0$ at any given time

The answer given to us was:

$$\ln[\ce{A}] = \frac{1}{2} \ln \left( \frac{k_2+k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2 [\ce{A}]\right)$$

However, I refuse to believe that an integrated rate equation can be without the time variable. Our teacher argued that the time variable was removed by using steady state approximation, and that $\frac{\mathrm d[\ce{C}]}{\mathrm dt} =0$. However, I think there is something wrong with using the rate of formation of an intermediate in deriving the integrated rate equation for a certain reaction. Can anyone please help me?

Edit: Here is how our professor got the answer:

$$\begin{align} \frac{\mathrm d[\ce{C}]}{\mathrm dt}=\frac{\mathrm d[\ce{D}]}{\mathrm dt} &= k_1[\ce{A}][\ce{B}] - k_2[\ce{C}][\ce{D}] - k_3[\ce{C}][\ce{D}] \\ \frac{\mathrm d[\ce{C}]}{dt} &= k_1[\ce{A}][\ce{B}] - (k_2 +k_3 )[\ce{C}][\ce{D}] \end{align}$$

$\ce{A}$ and $\ce{B}$ are the only species initially present such that:

$$[\ce{A}]_0 + [\ce{B}]_0 = [\ce{A}] + [\ce{B}] + [\ce{C}] + [\ce{D}] + [\ce{E}] + [\ce{F}] $$

Since $[\ce{C}] = [\ce{D}]$ and $[\ce{A}] = [\ce{B}]$ and $[\ce{F}] = 0$ and $[\ce{E}] = [\ce{A}] + [\ce{B}]$,

$$\begin{align} [\ce{A}]_0 + [\ce{B}]_0 &= 4[\ce{A}] + 2[\ce{C}] \\ [\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}] &= 2[\ce{C}] \\ \frac{[\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}]}{2} &= [\ce{C}] \end{align}$$

$$\begin{align} \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 \\ \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_o}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 \end{align}$$

By steady state approximation (SSA), $\frac{\mathrm d[\ce{C}]}{\mathrm dt} = 0$

$$\begin{align} \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 = 0 \\ k_1[\ce{A}]^2 &= (k_2 +k_3)\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 \end{align}$$

Applying natural logarithms to both sides,

$$\begin{align} 2 \ln(k_1) + 2 \ln[\ce{A}] &= \ln(k_2 +k_3)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\ 2 \ln[\ce{A}] &= \ln(k_2 +k_3 ) - \ln(k_1)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\ \ln[\ce{A}] &= \frac{1}{2}\ln\left(\frac{k_2 +k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \end{align}$$

As you can see from the solution, he was able to eliminate time from the final equation by using the steady state approximation. Is this a valid move to get the integrated rate equation? Please help.

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  • $\begingroup$ The steady state only applies in the 'middle' of the reaction, and not in the initial or final stages. You can see that initially the equilibrium has to be established (assuming only A+B present initially) and that at long times all the species will end up as E & F. In the steady state /equilibrium phase concentrations do not change with time. $\endgroup$ – porphyrin Nov 11 '16 at 8:26
  • $\begingroup$ If we are not to use SSA in the derivation of the integrated rate equation? How then would the rate equation be derived? $\endgroup$ – user37431 Nov 12 '16 at 5:20
  • $\begingroup$ The simple answer is with difficulty :) However, its a lot easier to do this numerically that algebraically. There is lots of code written in free languages such as python with which to do this. To solve any set of differential equations the initial conditions have to be stated. If you want to do this algebraically then this is much harder and you need to consult some standard texts first. Your scheme can be solved; have a look at the equations for pyrene excimers or the Lindemann scheme as its very similar to this. If you have no luck in a day or two I can post an answer . $\endgroup$ – porphyrin Nov 12 '16 at 17:11
  • $\begingroup$ @porphyrin Hello, I have posted what I think is the answer, and I want to know what you think of it, thank you :) $\endgroup$ – user37431 Nov 18 '16 at 3:14
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After consulting the Lindemann mechanism, here is my attempt for an answer:

$$\frac{d[\ce{C}]}{dt}= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 = 0$$ $$[\ce{C}]^2=\frac{k_1}{k_2 +k_3}[\ce{A}]^2$$ $$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +k_2[\ce{C}]^2$$

Substituting $[\ce{C}]^2$,

$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +\frac{k_2k_1}{k_2 +k_3}[\ce{A}]^2$$ $$\frac{d[\ce{A}]}{dt}= \frac{-k_2k_1[\ce{A}]^2 + -k_3k_1[\ce{A}]^2 +k_2k_1[\ce{A}]^2}{k_2 +k_3}$$ $$\frac{d[\ce{A}]}{dt}= \frac{-k_3k_1[\ce{A}]^2}{k_2 +k_3}$$

Let $k'=\frac{-k_3k_1}{k_2+k_3}$, then

$$\frac{d[\ce{A}]}{dt}= k'[\ce{A}]^2$$ $$\frac{d[\ce{A}]}{[\ce{A}]^2}= k'dt$$ $$\int_{[\ce{A}]_o}^{[\ce{A}]} \frac{d[\ce{A}]}{[\ce{A}]^2}= \int_0^t k'dt$$ $$-\left(\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o}\right) = k't$$ $$\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o} = -k't$$ $$\frac{1}{[\ce{A}]}= -k't +\frac{1}{[\ce{A}]_o} $$

I wonder how is it correct when I also used the steady state assumption in my solution. How is it different from the answer posted by my professor? Which is more correct and corresponds to the principles of chemical kinetics? Why would the other one be wrong and the other correct? (That is if my current derivation is correct :) ).

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  • $\begingroup$ Your answer seems to be ok once the equilibrium is established. In your original question the derivation stops at your third equation above and substitutions were made. You now assume that [A] is still changing with time and but you now assume that [C] and [A] are in a fixed ratio. (your second equation) . As you realise some simple chemical kinetics equations are very hard to solve and this is why we often try to use the steady state approximation. This particular scheme does seem to be one that is very difficult to obtain a full time-dependent solution. $\endgroup$ – porphyrin Nov 18 '16 at 16:25
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Your derivation is fine, so feel free to accept your own answer (not sure if this is technically possible, though). Your professor's answer is absurd on more than one level.

Now, when solving differential equations with multiple unknown time-dependent functions, it is not uncommon to stumble upon certain relations involving some of these functions and not involving time (think of conservation laws in celestial mechanics, or look at your second equation which links $[\rm A]$ and $[\rm C]$), so this is not necessarily wrong per se. But your professor's answer is an expression with only one unknown, namely $[\rm A]$. So in fact this is a static equation which defines one particular value of $[\rm A]$. What's the meaning of it? This can't be the equilibrium concentration, for there is no equilibrium. As the second reaction is irreversible, it will inevitably drain the system of all $\rm{[C]+[D]}$, and also of all pathways leading there. The final concentration of $[\rm A]$ can't be anything other than $0$ (that is, unless we have insufficient $[\rm B]$, which is not the case). How this wild expression came about is hardly of any importance. It just can't be right.

Still, if we want to investigate further, we'll find the root of all evil: it is the nonsensical condition $\rm{[E]=[A]+[B]}$ at all times. Just how can that be true? As both reactions progress, $[\rm A]$ and $[\rm B]$ go down, while $[\rm E]$ goes up. To maintain the condition, we must use some sort of Maxwell's demon that would monitor the concentrations of $[\rm A]$ and $[\rm B]$ and manipulate $[\rm E]$ accordingly. But wait, any such manipulation would break the conservation law $\rm{[A]_0+[B]_0=[A]+\dots+[F]}$, which we relied upon, so the whole thing can't be true even then.

So it goes.

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