# Question 5a15a

Apr 5, 2015

-390.3 kJ

#### Explanation:

Once again, to get to the reaction of interest, all you have to do is manipulate the three given reactions.

$\textcolor{b l u e}{\left(1\right)}$: ${C}_{\left(s\right)} + 2 {H}_{2 \left(g\right)} \to C {H}_{4 \left(g\right)}$, $\Delta {H}_{1} = \text{-74.6 kJ}$

$\textcolor{b l u e}{\left(2\right)}$: ${C}_{\left(s\right)} + 2 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)}$, $\Delta {H}_{2} = \text{-95.7 kJ}$

$\textcolor{b l u e}{\left(3\right)}$: ${H}_{2 \left(g\right)} + C {l}_{2 \left(g\right)} \to 2 H C {l}_{\left(g\right)}$, $\Delta {H}_{3} = \text{-184.6 kJ}$

Now look at your main reaction and try to determine what you have/need from the above three reactions

$C {H}_{4 \left(g\right)} + 4 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)} + 4 H C {l}_{\left(g\right)}$

Notice that the main reaction has $C {H}_{4}$ on the reactants' side, while reaction $\textcolor{b l u e}{\left(1\right)}$ has it on the products' side $\to$ flip reaction $\textcolor{b l u e}{\left(1\right)}$ to get $C {H}_{4}$ on the reactants' side

$\textcolor{b l u e}{\left(1\right) \text{reversed}} \implies C {H}_{4 \left(g\right)} \to {C}_{\left(s\right)} + 2 {H}_{2 \left(g\right)}$

DeltaH_("1 reversed") = "+74.6 kJ"

Now notice that you need 4 moles of $H C l$ for the main reaction, but only get 2 moles from reaction $\textcolor{b l u e}{\left(3\right)}$ $\to$ multiply reaction $\textcolor{b l u e}{\left(3\right)}$ by 2 to get the correct number of $H C l$ moles on the products' side

$\textcolor{b l u e}{\left(3\right)} \cdot 2 \implies 2 {H}_{2 \left(g\right)} + 2 C {l}_{2 \left(g\right)} \to 4 H C {l}_{\left(g\right)}$

DeltaH_("3 doubled") = 2 * DeltaH_3 = 2 * ("-184.6 kJ") = "-369.2 kJ"

Now add all the three resulting reactions to get your main one,

color(blue)("(1) reversed") + color(blue)((2)) + color(blue)(("3 doubled"))#

$C {H}_{4 \left(g\right)} \to \cancel{{C}_{\left(s\right)}} + \cancel{2 {H}_{2 \left(g\right)}}$
$\cancel{{C}_{\left(s\right)}} + 2 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)}$
$\cancel{2 {H}_{2 \left(g\right)}} + 2 C {l}_{2 \left(g\right)} \to 4 H C {l}_{\left(g\right)}$

$C {H}_{4 \left(g\right)} + 2 C {l}_{2 \left(g\right)} + 2 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)} + 4 H C {l}_{\left(g\right)}$, which is equivalent to

$C {H}_{4 \left(g\right)} + 4 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)} + 4 H C {l}_{\left(g\right)}$

Do the same for the $\Delta H$s to get

$\Delta {H}_{\text{rxn") = DeltaH_("1 reversed") + DeltaH_2 + DeltaH_("3 doubled}}$

$\Delta {H}_{\text{rxn") = "+74.6 kJ" + ("-95.7 kJ") + ("-369.2 kJ}}$

$\Delta {H}_{\text{rxn") = color(green)("-390.3 kJ}}$