Question

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...)

Answer 1

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

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`"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"`