How to write the rates for a bimolecular collisional dissociation?

I have questions concerning two species involved in a reaction, where one species is dissociated completely by collision with another, fairly inert species. The reaction looks like this:


NO + N$\_2$ $\xrightarrow{k\_f}$ N + O + N$\_2$


My understanding is that the reaction *requires* the N$\_2$ molecule in order for the reaction to proceed. So, for the rate:


$\frac{d[NO]}{dt}$ = $-k\_f [NO][N\_2]$  

$\frac{d[N\_2]}{dt}$ = 0


**Is the above expression for the rate of change for NO correct?**


Second, if I want to consider there being a reverse process, what I believe I need is an equilibrium constant such that:


$K\_{eq} = \frac{k\_f}{k\_r} = exp(-\frac{h\_p-h\_r}{RT}) $


Where $h$ is the enthalpy of formation, R is the universal gas constant ($8.314 [\frac{J}{molK}]$, and T is the temperature of the gas. the subscripts $p$ and $r$ refer to the products and reactants, respectively. For "naturally" occuring species, the enthalpy of formation is 0.


For this particular reaction, I looked up the enthalpy of formation for the species involved:


$h\_{N\_2} = 0 [\frac{J}{mol}]$  

$h\_N = 472.68\times 10^3 [\frac{J}{mol}]$  

$h\_O = 249\times 10^3 [\frac{J}{mol}]$  

$h\_{NO} = 90\times 10^3 [\frac{J}{mol}]$


Once the equilbrium constant has been solved for, I determined the reverse reaction rate $k\_r$ by dividing my known rate $k\_f$ by $K\_eq$. The resulting rate then, including reverse reactions would look something like:


$\frac{d[NO]}{dt}$ = $-k\_f [NO][N\_2] + k\_r[N][O][N\_2]$  

$\frac{d[N\_2]}{dt}$ = 0


**Is the above formulation for the reverse reaction rate, and the expression for the time rate of change of the gas correct?**


Thank you


