What is the change in volume for the reaction of ${16 g CH}_{4}$ $"16 g CH"_4$ and ${16 g O}_{2}$ $"16 g O"_2$ at $700^{\circ} C$ $700^@ "C"$ and $1 bar$ $"1 bar"$ ? The reaction is ${CH}_{4} (g) + \frac{1}{2} O_{2} (g) \to CO (g) + 2 H_{2} (g)$ $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .

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What is the change in volume for the reaction of ${16 g CH}_{4}$ $"16 g CH"_4$ and ${16 g O}_{2}$ $"16 g O"_2$ at $700^{\circ} C$ $700^@ "C"$ and $1 bar$ $"1 bar"$ ? The reaction is ${CH}_{4} (g) + \frac{1}{2} O_{2} (g) \to CO (g) + 2 H_{2} (g)$ $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .

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Answer 1 · 2017-04-11T15:30:54

About $120 L$ $"120 L"$ .

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 $700^{\circ} C$ $700^@ "C"$ (since it is evidently not $0^{\circ} C$ $0^@ "C"$ like it would be for STP).

$P V = n R T$ $PV = nRT$

$\Rightarrow \frac{V}{n} = \frac{R T}{P}$ $=> V/n = (RT)/P$

$= \frac{(0.083145 L \cdot bar/mol \cdot K) (700+273.15 K)}{1 bar}$ $= (("0.083145 L"cdot"bar/mol"cdot"K")("700+273.15 K"))/("1 bar")$

$=$ $=$ $80.91 L/mol$ $"80.91 L/mol"$

Given $16 g$ $"16 g"$ of both methane and diatomic oxygen, we can then find their $mol$ $"mol"$ s.

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

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

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 ${1 mol CH}_{4} (g)$ $"1 mol CH"_4(g)$ for every ${0.5 mol O}_{2} (g)$ $"0.5 mol O"_2(g)$ , and that's what you have, basically.

You do have a bit extra $O_{2}$ $"O"_2$ than you need, though, so let's just be particular about it... The limiting reactant is ${CH}_{4}$ $"CH"_4$ , since there is slightly less than twice the mols of ${CH}_{4}$ $"CH"_4$ as $O_{2}$ $"O"_2$ , when the reaction requires exactly twice.

So:

${0.9973 mols CH}_{4} \leftrightarrow {0.4987 mols O}_{2}$ $"0.9973 mols CH"_4 harr "0.4987 mols O"_2$

and ${0.0130 mols O}_{2}$ $"0.0130 mols O"_2$ is in excess. That will remain, but we have to account for that in calculating the mols of the products by using ${CH}_{4}$ $"CH"_4$ instead of $O_{2}$ $"O"_2$ to calculate the mols of products.

By the mol/mol ratio given in the reaction:

${mols CH}_{4} \leftrightarrow 0.9973 mols CO (g)$ $"mols CH"_4 harr "0.9973 mols CO"(g)$

${mols CH}_{4} \times 2 \leftrightarrow {1.9947 mols H}_{2} (g)$ $"mols CH"_4 xx 2 harr "1.9947 mols H"_2(g)$

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

$n_{C O} + n_{H_{2}} = n_{2}$ $n_(CO) + n_(H_2) = n_2$

$n_{C H_{4}} + n_{O_{2}} = n_{1}$ $n_(CH_4) + n_(O_2) = n_1$

$\Rightarrow 0.9973 mols CO + {1.9947 mols H}_{2} = n_{2} = 2.9920 mols gas$ $=> "0.9973 mols CO" + "1.9947 mols H"_2 = n_2 = "2.9920 mols gas"$

$\Rightarrow {0.9973 mols CH}_{4} + {(0.4987 + 0.0130) mols O}_{2} = n_{1} = 1.5090 mols gas$ $=> "0.9973 mols CH"_4 + "(0.4987 + 0.0130) mols O"_2 = n_1 = "1.5090 mols gas"$

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

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$ $V_1/(n_1) = V_2/(n_2)$

$\frac{V_{1}}{1.5090 mols} = \frac{V_{2}}{2.9920 mols}$ $V_1/("1.5090 mols") = V_2/("2.9920 mols")$

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

$\frac{V_{1}}{1.5090 mols} = \frac{80.91 L}{mol}$ $V_1/"1.5090 mols" = "80.91 L"/"mol"$

$\Rightarrow V_{1} = 122.093 L$ $=> V_1 = "122.093 L"$

$\frac{V_{2}}{2.9920 mols} = \frac{80.91 L}{mol}$ $V_2/"2.9920 mols" = "80.91 L"/"mol"$

$\Rightarrow V_{2} = 242.083 L$ $=> V_2 = "242.083 L"$

So, the change in volume is then:

$bb(DeltaV = V_2 - V_1)$

$= "242.083 L" - "122.093 L"$

$=$ $bb"119.99 L"$

To two sig figs, $color(blue)(DeltaV = "120 L")$ .

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

$PDeltaV stackrel(?" ")(=)DeltanRT$

$("1 bar")("120 L") stackrel(?" ")(=) ("2.9920 - 1.5090 mols gas")("0.083145 L"cdot"bar/mol"cdot"K")("973.15 K")$

$"120 L"cdot"bar" ~~ "119.993 L"cdot"bar"$

Yep, we're good.

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What is the change in volume for the reaction of ${16 g CH}_{4}$ $"16 g CH"_4$ and ${16 g O}_{2}$ $"16 g O"_2$ at $700^{\circ} C$ $700^@ "C"$ and $1 bar$ $"1 bar"$ ? The reaction is ${CH}_{4} (g) + \frac{1}{2} O_{2} (g) \to CO (g) + 2 H_{2} (g)$ $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .

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