The reaction between NO(g) and O2(g) is second order in NO(g) and first order in O2(g). By what factor will the reaction rate change if the concentrations of both reactants are doubled?
The reaction between NO(g) and O2(g) is second order in NO(g) and first order in O2(g). By what factor will the reaction rate change if the concentrations of both reactants are doubled?
The question threw me off for a second, because in these types of problems, the question could indicate whether or not the reaction was experimentally determined. Since all the choices are whole numbers, we can assume the rate has been experimentally determined. The rate of a reaction is determined by the slow step (eg. the person in the grocery store that takes forever to check out their items, & they keep running back & forth because they either forgot something or duking a billing issue out with the cahsier) anyway. This is a useful fact I thought I would bring to your attention, & is helpful to know.
the rate law would be: r = k ([NO]^2)([O2]) because the NO is second order, & the O2 is first order. So now substitute the amount of increase into the equation & when the smoke clears, you get:
r = k[NO]^2[O2]
r = k(2)^2(2)
r = (4)(2) = 8
r = increased by a factor of 8
Confirm this though, but I'm 99% sure this is the answer. Hope this helps & good luck studying, you'll all pull through, I'm rooting for you guys!
