Question

Question #21cf8

Answer 1

`DeltaH_"rxn" = -"2490 kJ"`

Explanation:

So, the three reactions that you need to use in order to get the target reaction looks like this

`"H"_text(2(g]) + "F"_text(2(g]) -> 2"HF"_text((g])" "color(blue)((1))" "DeltaH_1 = -"537 kJ"`

`"C"_text((s]) + 2"F"_text(2(g]) -> "CF"_text(4(g]) " "color(blue)((2))" "DeltaH_2 = -"680 kJ"`

`2"C"_text((s]) + 2"H"_text(2(]) -> "C"_2"H"_text(4(g]) " "color(blue)((3))" "DeltaH_3 = +"52.3 kJ"`

According to Hess' Law, the enthalpy change of reaction for a particular chemical reaction is independent of the steps needed to get from the reactans to the products.

Your target reaction is

`"C"_2"H"_text(4(g]) + 6"F"_text(2(g]) -> 2"CF"_text(4(g]) + 4"HF"_text((g])`

Notice that you need 2 moles of `"CF"_4` and 4 moles of `"HF"` on the products' side, but that you only have half that amount in equations `color(blue)((1))` and `color(blue)((2))`.

You also need to have `"C"_2"H"_4` on the reactants' side for the target equation, so you're going to have to flip equation `color(blue)((3))` and multiply equations `color(blue)((1))` and `color(blue)((2))` by 2.

This will get you

`2"H"_text(2(g]) + 2"F"_text(2(g]) -> 4"HF"_text((g])" "color(blue)((1^'))"`

`DeltaH_1^' = 2 * DeltaH_1 = 2 * (-"537 kJ") = -"1074 kJ"`

`2"C"_text((s]) + 4"F"_text(2(g]) -> 2"CF"_text(4(g]) " "color(blue)((2^'))"`

`DeltaH_2^' = 2 * DeltaH_2 = 2 * (-"680 kJ") = -"1360 kJ"`

and

`"C"_2"H"_text(4(g]) -> 2"C"_text((s]) + 2"H"_text(2(])" "color(blue)((3^'))`

`DeltaH_3^' = -DeltaH_3 = -"52.3 kJ"`

Now if you add these three reactions, `color(blue)((1^')) + color(blue)((2^')) + color(blue)((3^'))`, you'll get

`color(red)(cancel(color(black)(2"H"_text(2(g])))) + 2"F"_text(2(g]) + color(red)(cancel(color(black)(2"C"_text(2(g])))) + 4"F"_text(2(g]) + "C"_2"H"_text(4(g]) -> 4"HF"_text((g]) + 2"CF"_text(4(g]) + color(red)(cancel(color(black)(2"C"_text((g])))) + color(red)(cancel(color(black)(2"H"_text(2(g]))))`

`"C"_2"H"_text(4(g]) + 6"F"_text(2(g]) -> 2"CF"_text(4(g]) + 4"HF"_text((g])`

The enthalpy change of reaction will thus be

`DeltaH_"rxn" = DeltaH_1^' + DeltaH_2^' + DeltaH_3^'`

`DeltaH_"rxn" = -1074 - 1360 + (-52.3) = color(green)(-"2490 kJ")`

I'll leave the answer rounded to three sig figs.