If the concentration of #NH_3# (from reaction below) is changing at a rate of #0.25# M/sec, what is the rate of change of #H_2# (in M/sec)?

#N_2# (g) + #3H_2# (g) #\rarr2NH_3# (g)

(I'm probably not understanding this because my brain is sleep deprived, but there is a chance I won't fully get it even when well rested...)

1 Answer
Jun 18, 2018

Here's an intuitive way to understand it...


#"N"_2(g) + 3"H"_2(g) -> 2"NH"_3(g)#

If the change in concentration of #"NH"_3# is

#(Delta["NH"_3])/(Deltat) = "0.25 M/s"#,

then that means for every second, #"0.25 mols"# of it appear in a #"1-L"# container.

There are #"3 mols"# of #"H"_2# consumed (as a reactant) for every #"2 mols"# of #"NH"_3# that get produced (as a product) in a set amount of time.

So, the amount of #"H"_2# that gets consumed is #bb1.5# times the amount of #"NH"_3# produced in a set amount of time.

Lastly, #"H"_2# is consumed as a reactant, so it disappears... with a negative rate #(Delta["H"_2])/(Deltat)#.

Therefore,

#color(blue)((Delta["H"_2])/(Deltat)) = -1 cdot "0.25 M NH"_3/"s" xx ("3 mols H"_2)/("2 mols NH"_3) = color(blue)(-"0.38 M H"_2/"s")#

What about the rate of change of #"N"_2(g)#, and why?
#a)# #"0.50 M/s"#
#b)# #-"0.50 M/s"#
#c)# #"0.13 M/s"#
#d)# #-"0.13 M/s"#