Question #0d3d9

Start by making sure that you have clear understanding of what heat capacity means.

As you can se from the units used to express it, heat capacity tells you what the ratio between the amount of heat added, in your case, to the bomb calorimeter, and the subsequent increase in temperature.

SImply put, a heat capacity of #"2.47 kJ/K"# tells you that you need to dd #"2.41 kJ"# of heat to the calorimeter in order to increase its temperature by #"1 K"#.

Mathematically, this is expressed as

#color(blue)(q = C * DeltaT)" "#, where

#q# - the amount of heat added
#C# - the heat capacity of the calorimeter
#DeltaT# - the change in temperature, defined s the final temperature minus the initial temperature

Plug in your values and find #q#, the heat absorbed by the calorimeter

#q = 2.47"kJ"/color(red)(cancel(color(black)("K"))) * 2.22color(red)(cancel(color(black)("K"))) = "5.4834 kJ"#

Now, the idea here is that the heat released by the combustion reaction will be equal to the heat absorbed by the calorimeter.

#-q_"released" = q_"absorbed"#

The minus sign is used here because heat given off carries a negative sign.

So, you know that the combustion of #"0.109 g"# of ethylene, #"C"_2"H"_4#, will give off #"5.4834 kJ"# of heat.

Use ethylene's molar mass to determine how many moles you would get in the sample

#0.109color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_4)/(28.053color(red)(cancel(color(black)("g")))) = "0.003886 moles C"_2"H"_4#

So, if #0.003886# moles would produce this much heat, it follows that one mole would produce

#1color(red)(cancel(color(black)("mole"))) * "5.4834 kJ"/(0.003886color(red)(cancel(color(black)("moles")))) = "1141.1 kJ/mol"#

This means that the enthalpy change of combustion for ethylene will be

#DeltaH_"comb" = -color(green)("1140 kJ/mol") -># rounded to three sig figs

The minus sign here symbolizes that this much heat is being released when one mole of ethylene undergoes combustion.