# Question 21cf8

Aug 26, 2015

$\Delta {H}_{\text{rxn" = -"2490 kJ}}$

#### Explanation:

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

$\text{H"_text(2(g]) + "F"_text(2(g]) -> 2"HF"_text((g])" "color(blue)((1))" "DeltaH_1 = -"537 kJ}$

$\text{C"_text((s]) + 2"F"_text(2(g]) -> "CF"_text(4(g]) " "color(blue)((2))" "DeltaH_2 = -"680 kJ}$

$2 \text{C"_text((s]) + 2"H"_text(2(]) -> "C"_2"H"_text(4(g]) " "color(blue)((3))" "DeltaH_3 = +"52.3 kJ}$

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.

${\text{C"_2"H"_text(4(g]) + 6"F"_text(2(g]) -> 2"CF"_text(4(g]) + 4"HF}}_{\textrm{\left(g\right]}}$

Notice that you need 2 moles of ${\text{CF}}_{4}$ and 4 moles of $\text{HF}$ on the products' side, but that you only have half that amount in equations $\textcolor{b l u e}{\left(1\right)}$ and $\textcolor{b l u e}{\left(2\right)}$.

You also need to have ${\text{C"_2"H}}_{4}$ on the reactants' side for the target equation, so you're going to have to flip equation $\textcolor{b l u e}{\left(3\right)}$ and multiply equations $\textcolor{b l u e}{\left(1\right)}$ and $\textcolor{b l u e}{\left(2\right)}$ by 2.

This will get you

$2 \text{H"_text(2(g]) + 2"F"_text(2(g]) -> 4"HF"_text((g])" "color(blue)((1^'))}$

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

$2 \text{C"_text((s]) + 4"F"_text(2(g]) -> 2"CF"_text(4(g]) " "color(blue)((2^'))}$

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

and

$\text{C"_2"H"_text(4(g]) -> 2"C"_text((s]) + 2"H"_text(2(])" } \textcolor{b l u e}{\left({3}^{'}\right)}$

$\Delta {H}_{3}^{'} = - \Delta {H}_{3} = - \text{52.3 kJ}$

Now if you add these three reactions, $\textcolor{b l u e}{\left({1}^{'}\right)} + \textcolor{b l u e}{\left({2}^{'}\right)} + \textcolor{b l u e}{\left({3}^{'}\right)}$, you'll get

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {\text{H"_text(2(g])))) + 2"F"_text(2(g]) + color(red)(cancel(color(black)(2"C"_text(2(g])))) + 4"F"_text(2(g]) + "C"_2"H"_text(4(g]) -> 4"HF"_text((g]) + 2"CF"_text(4(g]) + color(red)(cancel(color(black)(2"C"_text((g])))) + color(red)(cancel(color(black)(2"H}}_{\textrm{2 \left(g\right]}}}}}$

${\text{C"_2"H"_text(4(g]) + 6"F"_text(2(g]) -> 2"CF"_text(4(g]) + 4"HF}}_{\textrm{\left(g\right]}}$

The enthalpy change of reaction will thus be

$\Delta {H}_{\text{rxn}} = \Delta {H}_{1}^{'} + \Delta {H}_{2}^{'} + \Delta {H}_{3}^{'}$

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

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