Intial kinetics of the above reaction and get the following results:

Question

Initial: PNO₂= 1atm P_F2=1 atm to rate of PNO2F formation=0.1 atm/sec

Initial: PNO₂= 0.5 atm P_F2=1 atm to rate of PNO2F formation=0.05 atm/sec

Initial: PNO2= 1 atm PF2=0.25 atm to rate of PNO2F formation=0.025 atm/sec

Rewrite the generalized rate equation from part, now with filled in variables to describe the orders of the reactants.

Answer 1

You did not provide a reaction that you refer to, so let us assume it is the following:

2 NO₂ +F₂ ==> 2NO₂F

When you reduce NO₂ in half from 1 atm to 0.5 atm, but keep F₂ constant at 1 atm, the rate is halved from 0.1 atm/sec to 0.05 atm/sec. Thus, the reaction is 1st order in NO₂

When you reduce F₂ by 4 fold from 1 atm to 0.25 atm but keep NO₂ constant at 1 atm, the rate is reduced 4 times from 0.1 atm/sec to 0.025 atm/sec. Thus, the reaction is 1st order in F₂.

The rate law is...

rate = k(NO₂)¹(F₂)¹

rate = k(P_NO2)¹(P_F2)¹

I think you asked this question earlier.