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)#.

1 Answer
Apr 11, 2017

About #"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^@ "C"# (since it is evidently not #0^@ "C"# like it would be for STP).

#PV = nRT#

#=> V/n = (RT)/P#

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

#=# #"80.91 L/mol"#

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

#"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 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)# for every #"0.5 mol O"_2(g)#, and that's what you have, basically.

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

So:

#"0.9973 mols CH"_4 harr "0.4987 mols O"_2#

and #"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# instead of #"O"_2# to calculate the mols of products.

By the mol/mol ratio given in the reaction:

#"mols CH"_4 harr "0.9973 mols CO"(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_(CO) + n_(H_2) = n_2#

#n_(CH_4) + n_(O_2) = n_1#

#=> "0.9973 mols CO" + "1.9947 mols H"_2 = n_2 = "2.9920 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:

#V_1/(n_1) = V_2/(n_2)#

#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.

#V_1/"1.5090 mols" = "80.91 L"/"mol"#

#=> V_1 = "122.093 L"#

#V_2/"2.9920 mols" = "80.91 L"/"mol"#

#=> 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.