What would happen to the rate of a reaction with rate law #rate = k[NO]^2[H_2]# if the concentration of #H_2# were halved?

1 Answer
Jan 12, 2016

If we assume that #Delta[NO] = 0#, then the rate is halved. Try making up numbers and performing the calculation.

Let #["NO"] = "0.2 M"# and #["H"_2] = "0.5 M"#, and let #k = "2"# #1/("M"^2*"s")#.

#color(green)(r(t)) = 2*0.2^2*0.5#

#color(green)"= 0.04 M/s"#

Okay, now do the same thing when #[H_2] = "0.25 M"#. Can you predict what you get without writing out any actual math?

What does that tell you about the order of #H_2# in the reaction? What, then, is the order of #NO# in the reaction? How do you know?