# 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) + $3 {H}_{2}$ (g) $\setminus \rightarrow 2 N {H}_{3}$ (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...)

Jun 18, 2018

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

${\text{N"_2(g) + 3"H"_2(g) -> 2"NH}}_{3} \left(g\right)$

If the change in concentration of ${\text{NH}}_{3}$ is

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

then that means for every second, $\text{0.25 mols}$ of it appear in a $\text{1-L}$ container.

There are $\text{3 mols}$ of ${\text{H}}_{2}$ consumed (as a reactant) for every $\text{2 mols}$ of ${\text{NH}}_{3}$ that get produced (as a product) in a set amount of time.

So, the amount of ${\text{H}}_{2}$ that gets consumed is $\boldsymbol{1.5}$ times the amount of ${\text{NH}}_{3}$ produced in a set amount of time.

Lastly, ${\text{H}}_{2}$ is consumed as a reactant, so it disappears... with a negative rate $\frac{\Delta \left[{\text{H}}_{2}\right]}{\Delta t}$.

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 ${\text{N}}_{2} \left(g\right)$, and why?
a) $\text{0.50 M/s}$
b) $- \text{0.50 M/s}$
c) $\text{0.13 M/s}$
d) $- \text{0.13 M/s}$