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)?

Md Ariful Islam

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Md Ariful Islam

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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?
$a)$ $"0.50 M/s"$
$b)$ $-"0.50 M/s"$
$c)$ $"0.13 M/s"$
$d)$ $-"0.13 M/s"$

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