# Question 59ed6

Apr 5, 2015

You can easily solve for the enthalpy change of reaction for your particlar reaction by manipulating the two given reactions.

$\textcolor{b l u e}{\left(1\right)}$: $C {a}_{\left(s\right)} + C {O}_{2 \left(g\right)} + \text{1/2} {O}_{2 \left(g\right)} \to C a C {O}_{3 \left(s\right)}$, $\Delta {H}_{1} = \text{-812.8 kJ}$

$\textcolor{b l u e}{\left(2\right)}$: $2 C {a}_{\left(s\right)} + {O}_{2 \left(g\right)} \to 2 C a {O}_{\left(s\right)}$, $\Delta {H}_{2} = \text{-1269.8 kJ}$

Now, think of how you can manipulate these two reactions to get to

$C a {O}_{\left(s\right)} + C {O}_{2 \left(g\right)} \to C a C {O}_{3 \left(s\right)}$

Notice that your target reaction has 1 mole of $C a O$ reacting, while reaction $\textcolor{b l u e}{\left(2\right)}$ has 2 moles of $C a O$ being produced.

If you flip equation $\textcolor{b l u e}{\left(2\right)}$ and divide it by 2, you'll get

$\frac{\textcolor{b l u e}{\left(2\right)}}{2} \implies C {a}_{\left(s\right)} + \text{1/2} {O}_{2 \left(g\right)} \to C a {O}_{\left(s\right)}$

(DeltaH_2)/2 = ("-1269.8 kJ")/2 = "-634.9 kJ"

If you flip it, the reverse reaction will be

color(blue)((2) "reversed") => CaO_((s)) -> Ca_((s)) + "1/2"O_(2(g))

DeltaH_("2 reversed") = "+634.9 kJ"

Now just add reaction $\textcolor{b l u e}{\left(1\right)}$ to reaction $\textcolor{b l u e}{\left(2\right) \text{reversed}}$ to get

$\cancel{C {a}_{\left(s\right)}} + C {O}_{2 \left(g\right)} + \cancel{\text{1/2} {O}_{2 \left(g\right)}} \to C a C {O}_{3 \left(s\right)}$
$C a {O}_{\left(s\right)} \to \cancel{C {a}_{\left(s\right)}} + \cancel{\text{1/2} {O}_{2 \left(g\right)}}$

$C a {O}_{\left(s\right)} + C {O}_{2 \left(g\right)} \to C a C {O}_{3 \left(s\right)}$

Now do the same for the $\Delta H$s

DeltaH_"rxn" = DeltaH_1 + DeltaH_("2 reversed")#

$\Delta {H}_{\text{rxn") = "-812.8 kJ" + "634.9 kJ" = color(green)("-177.9 kJ}}$