How do I solve questions pertaining to rate of disappearance and appearance?


> 
> In a reversible reaction $\ce{2NO2 <=>[$k\_1$][$k\_2$] N2O4}$, the rate of disappearance of $\ce{NO2}$ is equal to: 
> 
> 
> 1. $2\frac{k\_1}{k\_2}\ce{[NO2]}^2$
> 2. $2k\_1\ce{[NO2]}^2-2k\_2\ce{[N2O4]}$
> 3. $2k\_1\ce{[NO2]}^2-k\_2\ce{[N2O4]}$
> 4. $(2k\_1-k\_2)\ce{[NO2]}$
> 
> 
> 


The answer, they say, is (2). I just don't understand how they got it. I find it difficult to solve these questions. Is the rate of disappearance the derivative of the concentration of the reactant *divided by its coefficient in the reaction*, or is it simply the derivative? 


If someone could help me with the solution, it would be great. And please, don't assume I'm just picking up a random question from a book and asking it for fun without actually trying to do it. I have worked at it and I don't understand what to do.




In general, if you have a system of elementary reactions, the rate of appearance of a species $\ce{A}$ will be


$$\cfrac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = \sum\limits\_i \nu\_{\ce{A},i} r\_i$$


where


* $i$ is each reaction in the system
* $\nu\_{\ce{A},i}$ is the stoichiometric coefficient of species $\ce{A}$ in reaction $i$ (positive for products, negative for reagents)
* $r\_i$ is the rate for reaction $i$, which in turn will be calculated as a product of concentrations for all reagents $j$ times the kinetic coefficient $k\_i$:


$$r\_i = k\_i \prod\limits\_{j} [j]^{\nu\_{j,i}}$$


In other words, there's a positive contribution to the rate of appearance for each reaction in which $\ce{A}$ is produced, and a negative contribution to the rate of appearance for each reaction in which $\ce{A}$ is consumed, and these contributions are equal to the rate of that reaction times the stoichiometric coefficient.


The rate of disappearance will simply be *minus* the rate of appearance, so the signs of the contributions will be the opposite.


In your example, we have two elementary reactions:


$$\ce{2NO ->[$k\_1$] N2O4} \tag{1}$$


$$\ce{N2O4 ->[$k\_2$] 2NO} \tag{2}$$


So, the rate of *appearance* of $\ce{N2O4}$ would be


$$\cfrac{\mathrm{d}\ce{[N2O4]}}{\mathrm{d}t} = r\_1 - r\_2 $$


Similarly, the rate of *appearance* of $\ce{NO}$ would be


$$\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = - 2 r\_1 + 2 r\_2$$


and the rate of *disappearance* of $\ce{NO}$ would be minus its rate of appearance:


$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 r\_1 - 2 r\_2$$


Since the rates for both reactions would be


$$r\_1 = k\_1 \ce{[NO]}^2$$


$$r\_2 = k\_2 \ce{[N2O4]}$$


the rate of disappearance for $\ce{NO}$ will be


$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 k\_1 \ce{[NO]}^2 - 2 k\_2 \ce{[N2O4]}$$


 7