# 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

• It seems that for one, I am incorrectly calculating the equilibrium constant $K_{eq}$. I should be using the expression $\Delta G = H - T\Delta S$ rather than $h_p, h_r$ in my calculation for the equilibrium constant. Making the change, the equation should instead look like $exp(-\frac{\Delta G_p - \Delta G_r}{RT})$ where the subscripts $p,r$ still mean products and reactants, respectively. – JN3 Mar 25 '18 at 21:35