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
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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
#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 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
You do have a bit extra
So:
#"0.9973 mols CH"_4 harr "0.4987 mols O"_2#
and
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,
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.