rate = k(NO)^2(O2)
3.9E-3 = k(5)^2(2)
Solve for k for initial conditions. k stays constant but NO and O2 change.
When 50% O2 has reacted (1 mol), then 2 mols NO have reacted to leave 1 mol O2 and 3 mols NO. Plug those values into the rate equation and solve for rate.
The reaction below is second-order in NO and first order in O2.
2 NO(g) + O2(g) ⟶ 2 NO2(g)
When 5.00 mol of NO and 2.00 mol O2 were reacted in a 1.00 L flask, the initial rate was 3.9×10−3 M s−1.
What is the rate when 50% of the oxygen has reacted?
no idea how to solve this, please help
5 answers
so it would be k=0.0088?
No. It is k = anything. You're looking for the rate.
what do you mean?
The problem gives you k of 0.0039 and you used that to solve for rate under the first conditions. k won't change so I don't understand why you wasted your time calculating a new k. The problem sttes to calculate a new RATE for the new concentrations. I gave you the new concentrations so it should be a slam dunk.