This is the complete solution to the Zeldovich mechanism of NO formation (albeit neglecting the reverse rxns):
O+N2=NO+N
N+O2=NO+O
N+OH=NO+H

d[NO]/dt=k1[N2][O]+k2[O2][N]+k3[N][OH]
d[N]/dt=k1[N2][O]-k2[O2][N]-k3[N][OH]
d[O]/dt=-k1[N2][O]+k2[O2][N]
d[OH]/dt=-k3[N][OH]
d[N2]/dt=-k1[N2][O]
d[O2]/dt=-k2[O2][N]

temp=2250;
press=10;
k1=1.8 10^8*Exp[-38370/temp]; 
k2=1.8 10^4*temp1*Exp[-4680/temp];
k3=7.1 10^7*Exp[-450/temp];
cNOi=0;

soln=DSolve[{cNO'[t]==k1*cN2[t]*cO[t]+k2*cO2[t]*cN[t]+k3*cN[t]*cOH[t],cNO[0]==cNOi},cNO[t],t]

One ODE is shown above... need to add the other five, plus five initial conditions.

lrr3@psu.edu (updated 10/13/2015)