# Question 342e1

Mar 24, 2015

The enthalpy of reaction in this case will be DeltaH_("rxn") = "-390. kJ".

The enthalpy of reaction is defined as the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants.

$\Delta {H}_{\text{rxn")^(0) = Sigma(DeltaH_f^(0) "products") - Sigma(DeltaH_f^(0)"rectants}}$

The reaction you're interested in is

$C {H}_{4 \left(g\right)} + 4 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)} + \textcolor{red}{4} H C {l}_{\left(g\right)}$ (1)

You know that you must use the following reactions

${C}_{\left(s\right)} + 2 {H}_{2 \left(g\right)} \to C {H}_{4 \left(g\right)}$, DeltaH_("rxn 1") = "-74.6 kJ"
${C}_{\left(s\right)} + 2 C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)}$, DeltaH_("rxn 2") = "-95.7 kJ"
${H}_{2 \left(g\right)} + C {l}_{2 \left(g\right)} \to 2 H C {l}_{\left(g\right)}$, DeltaH_("rxn 3") = "-184.6 kJ" (2)

Because the standard enthalpy of formation for elements in their most stable form is zero, the enthalpies of reaction given to you for the above three reactions are actually the standard enthalpies of formation of $C {H}_{4}$, $C C {l}_{4}$, and $H C l$.

This means that you can use them in equation (1) to solve for $\Delta {H}_{\text{rxn}}$. Keep in mind, however, that you have $\textcolor{red}{4}$ moles of $H C l$, not 2 like equation (2) suggests.

Since equation (2) gives you an enthalpy value for 2 moles of $H C l$, you'll have to multiply this by $\textcolor{b l u e}{2}$ to get the proper enthalpy for your reaction of interest.

So,

$\Delta {H}_{\text{rxn") = (-"95.7 kJ" + color(blue)(2) * (-"184.6 kJ")) - (-"74.6 kJ}}$

DeltaH_("rxn") = -"464.9 kJ" + "74.6 kJ" = "-390.3 kJ"

Rounded to three sig figs, the answer will be

$\Delta {H}_{\text{rxn") = color(red)(-"390. kJ}}$

SIDE NOTE Enthalpies of formation are usually given for the formation of 1 mole of a substance, which would make equation (2) actually look like this

$\text{1/2"H_(2(g)) + "1/2} C {l}_{2 \left(g\right)} \to H C {l}_{\left(g\right)}$

DeltaH_("actual 3") = 1/2 * DeltaH_("rxn 3") = -"92.3 kJ"#

In this case, you'll have to multiply this value by 4 in order to get the correct enthalpy needed for your reaction because, like I've said, you have 4 moles of HCl that are produced.