# What is the volume after 5 hours in this first order reaction?

## If the rate constant of decomposition for ${\text{AsH}}_{3} \left(g\right)$ is $\text{0.09"/"hour}$, what will be the volume of ${\text{H}}_{2} \left(g\right)$ after $\text{5 hours}$, if the number of moles of ${\text{AsH}}_{3} \left(g\right)$ is $\text{10 mols}$ in the beginning?

Jan 1, 2017

I get about $\text{302.4 L}$, assuming ${\text{H}}_{2} \left(g\right)$ is an ideal gas and that the reaction vessel is large enough to contain this much gas at constant volume.

First, write the reaction:

$2 {\text{AsH"_3(g) -> 2"As"(s) + 3"H}}_{2} \left(g\right)$

Instead of solving for the volume of ${\text{H}}_{2} \left(g\right)$, let's first solve for the $\text{mol}$s. We can assume that the gases are all in the same container, and that the container is big enough, so that the total volume is constant.

Therefore, we can ignore the total volume and look at the $\text{mol}$s. If it's a first-order decomposition, we can approximate the rate law as

$r \left(t\right) = k {n}_{{\text{AsH}}_{3}}$,

where we ignore the volume of the container (even though it would normally give us a concentration if we had $\frac{n}{V}$).

We were given both the rate constant $k$ and the initial $\boldsymbol{\text{mol}}$s, ${n}_{{\text{AsH}}_{3}}$, so we can solve for the initial rate $r \left(t\right)$:

$r \left(t\right) = \left(\text{0.09 hr"^(-1))("10 mols}\right)$

$=$ $\text{0.9 mols/hr}$

That means the rate of reaction is $\text{0.9 mols}$ for every hour that passes. Note that this is not explicitly of ${\text{AsH}}_{3} \left(g\right)$. You need to recall that:

$r \left(t\right) = \text{0.9 mols"/"hr} = - \frac{1}{2} \frac{\Delta {n}_{A s {H}_{3}}}{\Delta t}$

$= \frac{1}{2} \frac{\Delta {n}_{A s}}{\Delta t} = \frac{1}{3} \frac{\Delta {n}_{{H}_{2}}}{\Delta t}$

or that

$\frac{\Delta {n}_{A s {H}_{3}}}{\Delta t} = - \text{1.8 mols/hr}$.

When $\text{5 hrs}$ passes, it means that

("1.8 mols AsH"_3(g))/cancel"hr" xx 5 cancel"hrs" = "9.0 mols AsH"_3(g)

turned into products. Therefore, by stoichiometry:

$\textcolor{g r e e n}{{n}_{{\text{H"_2(g))) = 9.0 cancel("mols AsH"_3(g)) xx ("3 mols H"_2(g))/(2 cancel("mols AsH}}_{3} \left(g\right)}}$

$= \textcolor{g r e e n}{{\text{13.5 mols H}}_{2} \left(g\right)}$

You could also do this by directly using the rate of production of ${\text{H}}_{2} \left(g\right)$:

$\text{0.9 mols"/"hr} = \frac{1}{3} \frac{\Delta {n}_{{H}_{2}}}{\Delta t}$

$\implies \frac{\Delta {n}_{{H}_{2}}}{\Delta t} = \text{2.7 mols/hr}$

=> color(green)(n_("H"_2(g))) = ("2.7 mols H"_2(g))/cancel"hr" xx 5 cancel("hrs") = color(green)("13.5 mols H"_2(g))

If we then assume that the reaction was occurring at $\text{273.15 K}$ and $\text{1 atm}$ pressure, and that ${\text{H}}_{2} \left(g\right)$ is an ideal gas, then we can use a molar volume of $\text{22.4 L}$ for ${\text{H}}_{2} \left(g\right)$ and get:

color(blue)(V_("H"_2(g))) ~~ "22.4 L H"_2(g) xx "13.5 mols H"_2(g) = color(blue)("302.4 L")