Question #21cf8

1 Answer
Aug 26, 2015

Answer:

#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.