# Question a11ba

Jul 6, 2016

$\Delta {H}_{\text{rxn"^@ = -"2776 kJ}}$

#### Explanation:

The idea here is that the standard molar enthalpy of combustion, $\Delta {H}_{\text{comb}}^{\circ}$, tells you the enthalpy change associated with the combustion of one mole of a given substance under standard conditions.

In your case, methane, ${\text{CH}}_{4}$, is said to have a standard molar enthalpy of combustion equal to

$\Delta {H}_{\text{comb"^@ = -"890.8 kJ mol}}^{- 1} \to$ 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 $\text{890.8 kJ}$ of heat.

All you have to do here is use the molar mass of methane to see how many moles you have in that $\text{50.00 g}$ 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