# What is the change in volume for the reaction of "16 g CH"_4 and "16 g O"_2 at 700^@ "C" and "1 bar"? The reaction is "CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g).

Apr 11, 2017

About $\text{120 L}$.

For this, since we are asked for the change in volume, it is necessary to solve for the molar volume of an ideal gas at ${700}^{\circ} \text{C}$ (since it is evidently not ${0}^{\circ} \text{C}$ like it would be for STP).

$P V = n R T$

$\implies \frac{V}{n} = \frac{R T}{P}$

= (("0.083145 L"cdot"bar/mol"cdot"K")("700+273.15 K"))/("1 bar")

$=$ $\text{80.91 L/mol}$

Given $\text{16 g}$ of both methane and diatomic oxygen, we can then find their $\text{mol}$s.

${\text{16 g CH"_4 xx ("1 mol CH"_4)/(12.011 + 4 xx 1.0079 "g CH"_4) = "0.9973 mols CH}}_{4}$

${\text{16 g O"_2 xx ("1 mol O"_2)/(15.999 xx 2 "g O"_2) = "0.5000 mols O}}_{2}$

No, it is not necessary to halve anything. You have from the masses the same mol/mol ratio as required in the reaction itself. You needed ${\text{1 mol CH}}_{4} \left(g\right)$ for every ${\text{0.5 mol O}}_{2} \left(g\right)$, and that's what you have, basically.

You do have a bit extra ${\text{O}}_{2}$ than you need, though, so let's just be particular about it... The limiting reactant is ${\text{CH}}_{4}$, since there is slightly less than twice the mols of ${\text{CH}}_{4}$ as ${\text{O}}_{2}$, when the reaction requires exactly twice.

So:

${\text{0.9973 mols CH"_4 harr "0.4987 mols O}}_{2}$

and ${\text{0.0130 mols O}}_{2}$ is in excess. That will remain, but we have to account for that in calculating the mols of the products by using ${\text{CH}}_{4}$ instead of ${\text{O}}_{2}$ to calculate the mols of products.

By the mol/mol ratio given in the reaction:

$\text{mols CH"_4 harr "0.9973 mols CO} \left(g\right)$

${\text{mols CH"_4 xx 2 harr "1.9947 mols H}}_{2} \left(g\right)$

So, really the next thing we can calculate are the final and initial mols:

${n}_{C O} + {n}_{{H}_{2}} = {n}_{2}$

${n}_{C {H}_{4}} + {n}_{{O}_{2}} = {n}_{1}$

$\implies \text{0.9973 mols CO" + "1.9947 mols H"_2 = n_2 = "2.9920 mols gas}$

$\implies \text{0.9973 mols CH"_4 + "(0.4987 + 0.0130) mols O"_2 = n_1 = "1.5090 mols gas}$

That means the ratio of the mols can be related for a big picture using Gay-Lussac's Law:

${V}_{1} / \left({n}_{1}\right) = {V}_{2} / \left({n}_{2}\right)$

${V}_{1} / \left(\text{1.5090 mols") = V_2/("2.9920 mols}\right)$

By the molar volume, we can find each actual volume for the ideal gas combinations.

${V}_{1} / \text{1.5090 mols" = "80.91 L"/"mol}$

$\implies {V}_{1} = \text{122.093 L}$

${V}_{2} / \text{2.9920 mols" = "80.91 L"/"mol}$

$\implies {V}_{2} = \text{242.083 L}$

So, the change in volume is then:

$\boldsymbol{\Delta V = {V}_{2} - {V}_{1}}$

$= \text{242.083 L" - "122.093 L}$

$=$ $\boldsymbol{\text{119.99 L}}$

To two sig figs, $\textcolor{b l u e}{\Delta V = \text{120 L}}$.

As a check, we can see whether the ideal gas law is still satisfied.

PDeltaV stackrel(?" ")(=)DeltanRT

$\left(\text{1 bar")("120 L") stackrel(?" ")(=) ("2.9920 - 1.5090 mols gas")("0.083145 L"cdot"bar/mol"cdot"K")("973.15 K}\right)$

$\text{120 L"cdot"bar" ~~ "119.993 L"cdot"bar}$

Yep, we're good.