Determination of the integrated rate equation for a reaction

Question

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.

Answer 1

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 :) ).

Answer 2

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.