Question #a11ba

1 Answer
Jul 6, 2016

Answer:

#DeltaH_"rxn"^@ = -"2776 kJ"#

Explanation:

The idea here is that the standard molar enthalpy of combustion, #DeltaH_"comb"^@#, tells you the enthalpy change associated with the combustion of one mole of a given substance under standard conditions.

In your case, methane, #"CH"_4#, is said to have a standard molar enthalpy of combustion equal to

#DeltaH_"comb"^@ = -"890.8 kJ mol"^(-1) -># the minus sign is used to symbolize heat given off

This tells you that every mole of methane that undergoes combustion under standard conditions gives off #"890.8 kJ"# of heat.

All you have to do here is use the molar mass of methane to see how many moles you have in that #"50.00 g"# sample

#50.00 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.0425color(red)(cancel(color(black)("g")))) = "3.1167 moles CH"_4#

Now use the known value of the molar standard enthalpy of combustion to see how much heat would be given off when #3.1167# moles of methane undergo combustion

#3.1167 color(red)(cancel(color(black)("moles CH"_ 4))) * overbrace("890.8 kJ"/(1color(red)(cancel(color(black)("mole CH"_ 4)))))^(color(blue)(=DeltaH_ "comb"^@)) = "2776.4 kJ"#

Since this represents heat given off, the standard enthalpy change of reaction will carry a minus sign

#DeltaH_"rxn"^@ = color(green)(|bar(ul(color(white)(a/a)color(black)(-"2776 kJ")color(white)(a/a)|))) -># rounded to four sig figs