Determination of the integrated rate equation for a reaction We were given this equation to determine its integrated rate equation:

$$\ce {A + B &lt;=&gt;[$k_1$][$k_2$] C +D -&gt;[$k_3$] E + F}$$

We were given these assumptions to simplify the derivation:


$\displaystyle \frac{\mathrm d[\ce{A}]}{\mathrm dt} = \frac{\mathrm d[\ce{B}]}{\mathrm dt}$, and $\displaystyle \frac{\mathrm d[\ce{C}]}{\mathrm dt} = \frac{\mathrm d[\ce{D}]}{\mathrm dt}$
At any time, $[\ce{A}]=[\ce{B}]$, and $[\ce{C}]=[\ce{D}]$, and $[\ce{E}] = [\ce{A}]+[\ce{B}]$
$\ce{F}$ 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:

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

$\ce{A}$ 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 $[\ce{E}] = [\ce{A}] + [\ce{B}]$,

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

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

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

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

Applying natural logarithms to both sides,

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

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.
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 :) ).