# 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)?

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#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...)

(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

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

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

then that means for every second, **appear** in a

There are *consumed* (as a reactant) **for every** *produced* (as a product) in a set amount of time.

So, the amount of **times** the amount of

Lastly, **consumed** as a reactant, so it ** disappears**... with a

*negative*rate

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