Question

Question #88aee

Answer 1

Here's what I got.

Explanation:

You know that propane undergoes combustion as described by the balanced chemical equation

`"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))`

You also know that when `"2.5000 g"` of propane undergo combustion, the reaction gives off `"115.75 kJ"` of heat.

Now, in order to find the enthalpy change of combustion of propane, `DeltaH_"comb"`, in kilojoules per mole, you need to figure out how much heat is released when `1` mole of propane undergoes combustion.

Use the molar mass of propane to convert the mass of the sample to moles

`2.5000 color(red)(cancel(color(black)("g"))) * ("1 mole C"_3"H"_8)/(44.1color(red)(cancel(color(black)("g")))) = "0.0566893 moles C"_3"H"_8`

SInce you know that when `0.0566893` moles of propane undergo combustion, `"115.75 kJ"` of heat are being released, you can say that when `1` mole of propane undergoes combustion, the reaction will give off

`1 color(red)(cancel(color(black)("mole C"_3"H"_8))) * "115.75 kJ"/(0.0566893color(red)(cancel(color(black)("moles C"_3"H"_8)))) = "2041.8 kJ"`

This means that the enthalpy change of combustion of propane will be

`color(darkgreen)(ul(color(black)(DeltaH_"comb" = -"2041.8 kJ")))`

The minus sign is used to symbolize heat given off.

SIDE NOTE The difference between this value and what you have at option (E) is most likely caused by the value used for the molar mass of propane.

Now, in order to find the enthalpy change of formation, `DeltaH_f^@`, of propane, you need to find the enthalpy change that accompanies the formation of `1` mole of propane under standard conditions from its constituent elements in their stable form.

`3"C"_ ((s)) + 4"H"_ (2(g)) -> "C"_ 3"H"_ (8(g))" "DeltaH_f^@ = ?`

To do that, you can use standard enthalpy changes of formation of the reactants and of the products for the combustion of propane `->` think Hess' Law here.

`DeltaH_"comb" = sum_i (n * DeltaH_"f products i"^@) - sum_j (m * DeltaH_"f reactants j"^@)`

Here

• `n` represents the number of moles of a given product
• `m` represents the number of moles of a given reactant

The standard enthalpy change of oxygen is `"0 kJ mol"^(-1)` because `"O"_2` is the most stable form of elemental oxygen.

In your case, `1` mole of propane undergoes combustion to produce `3` moles of carbon dioxide and `4` moles of water, so if you take `color(blue)(x)` `"kJ mol"^(-1)` to be the standard enthalpy change of formation of propane

`DeltaH_f^@ = color(blue)(x) color(white)(.)"kJ mol"^(-1)`

you can write

`-"2041.8 kJ"`

`= [3 color(red)(cancel(color(black)("moles CO"_2))) * (-"393.5 kJ"/(1color(red)(cancel(color(black)("mole CO"_2))))) + 4 color(red)(cancel(color(black)("moles H"_2"O"))) * (-"285.5 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))))] - [1 color(red)(cancel(color(black)("mole C"_3"H"_8))) * (color(blue)(x)color(white)(.)"kJ")/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) + 5 color(red)(cancel(color(black)("moles O"_2))) * "0 kJ"/(1color(red)(cancel(color(black)("mole O"_2))))]`

This is equivalent to

`-2041.8 color(red)(cancel(color(black)("kJ"))) = -1180.5 color(red)(cancel(color(black)("kJ"))) - 1142.0 color(red)(cancel(color(black)("kJ"))) - color(blue)(x)color(red)(cancel(color(black)("kJ")))`

Rearrange to solve for `x`

`color(blue)(x) = +2041.8 - 2322.5 = color(blue)(-280.7)`

This means that the standard enthalpy change of formation of propane will be

`color(darkgreen)(ul(color(black)(DeltaH_f^@ = color(blue)(-280.7)color(white)(.)"kJ mol"^(-1))))`

I'll leave the answer rounded to one decimal place.