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

$$\ce{2NO -&gt;[$k_1$] N2O4} \tag{1}$$

$$\ce{N2O4 -&gt;[$k_2$] 2NO} \tag{2}$$

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

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 k_1 \ce{[NO]}^2 - 2 k_2 \ce{[N2O4]}$$