# Question 079ac

Jul 2, 2016

I got $- \text{414 kJ/mol}$, and then $- \text{11 kJ/mol}$.

The basic idea is that since enthalpy is a state function, you can focus on the initial and final states of the system. It does not depend on the path you take.

Hess's Law is based off of this property of enthalpy, and states that you can:

• Reverse a reaction --- this switches the sign of $\Delta H$ for that reaction.
• Scale a reaction by a constant (e.g. $\frac{1}{2}$, $3$, etc) --- this multiplies $\Delta H$ for that reaction by the number you used.

So, what you should do for these problems is try to make your intermediates and/or catalysts cancel.

$\to$ Anything that appears in both the products and reactants sides cancel accordingly.

PROBLEM 4

Overall reaction:

${\text{Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(s) + "H}}_{2} \left(g\right)$

Unmodified component reactions:

(1): $\text{ H"_2(g) + 1/2"O"_2(g) -> "H"_2"O} \left(l\right)$
$\Delta H = - \text{285 kJ/mol}$

(2): " CaO"(s) + "H"_2"O"(l) -> "Ca"("OH")_2(s)
$\Delta H = - \text{64 kJ/mol}$

(3): $\text{ Ca"(s) + 1/2"O"_2(g) -> "CaO} \left(s\right)$
$\Delta H = - \text{635 kJ/mol}$

• Since $\text{Ca} \left(s\right)$ is a reactant in the overall reaction with a stoichiometric coefficient of $1$, we can use (3) as-is.
• Since ${\text{H}}_{2} \left(g\right)$ is a product in the overall reaction with a stoichiometric coefficient of $1$, we can reverse (1) and then use it like that.
• Since "Ca"("OH")_2(s) is a product in the overall reaction with a stoichiometric coefficient of $1$, we can use (2) as-is.

Everything else cancels out accordingly.

Modified component reactions:

$\setminus m a t h b f \left(-\right)$(1): " H"_2"O"(l) -> "H"_2(g) + cancel(1/2"O"_2(g)), $\Delta H = - \left(- \text{285 kJ/mol") = color(red)(+"285 kJ/mol}\right)$

(2):   ${\cancel{\text{CaO"(s)) + "H"_2"O"(l) -> "Ca"("OH}}}_{2} \left(s\right)$, $\Delta H = - \text{64 kJ/mol}$

(3): " Ca"(s) + cancel(1/2"O"_2(g)) -> cancel("CaO"(s)), $\Delta H = - \text{635 kJ/mol}$

$\text{---------------------------------------------------------}$
$\setminus m a t h b f \left({\text{Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(s) + "H}}_{2} \left(g\right)\right)$

$\implies \textcolor{b l u e}{\Delta {H}_{\text{overall}}}$

$= \left(+ \text{285 kJ/mol") + (-"64 kJ/mol") + (-"635 kJ/mol}\right)$

$= \textcolor{b l u e}{- \text{414 kJ/mol}}$

PROBLEM 5 (the units of $\Delta H$ here are NOT $\text{kJ}$. They are $\setminus m a t h b f \left(\underline{\underline{\text{kJ/mol}}}\right)$.)

Overall reaction:

${\text{FeO"(s) + "CO"(g) -> "Fe"(s) + "CO}}_{2} \left(g\right)$

Unmodified component reactions:

(1): ${\text{ 3Fe"_2"O"_3(s) + "CO"(g) -> 2"Fe"_3"O"_4(s) + "CO}}_{2} \left(g\right)$
$\Delta H = - \text{47 kJ/mol}$
(2): ${\text{ Fe"_2"O"_3(s) + 3"CO"(g) -> 2"Fe"(s) + 3"CO}}_{2} \left(g\right)$
$\Delta H = - \text{25 kJ/mol}$
(3): ${\text{ Fe"_3"O"_4(s) + "CO"(g) -> 3"FeO"(s) + "CO}}_{2} \left(g\right)$
$\Delta H = + \text{19 kJ/mol}$

• Since $\text{FeO} \left(s\right)$ is a reactant in the overall reaction with a stoichiometric coefficient of $1$, but it is a product in (3) with a stoichiometric coefficient of $3$, we must scale (3) by $\frac{1}{3}$ and reverse it.
• Since $\text{Fe} \left(s\right)$ is a product in the overall reaction with a stoichiometric coefficient of $1$, but it is a product in (2) with a stoichiometric coefficient of $2$, we have to scale (2) by $\frac{1}{2}$.
• For ${\text{CO}}_{2}$ after modifying (2) and (3), we will have $\frac{1}{3} {\text{CO}}_{2}$ as reactants and $\frac{3}{2} {\text{CO}}_{2}$ as products, and in the overall reaction we have $1 {\text{CO}}_{2}$ as products. Right now, this will cancel as $\frac{3}{2} - \frac{1}{3} = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} {\text{CO}}_{2}$ as products. Since in the overall reaction, there is only one ${\text{CO}}_{2}$, we need to scale (1) by $\frac{1}{6}$ and reverse it so that $\frac{7}{6} - \frac{1}{6} = \frac{6}{6} = 1 {\text{CO}}_{2}$ on the reactants side.

Everything else cancels out accordingly. For each reactant, I've converted the fractions to common denominators for you, but you will have to do it yourself on a test.

Modified component reactions:

$\setminus m a t h b f \left(- \frac{1}{6}\right)$(1): cancel(1/3"Fe"_3"O"_4(s)) + 1/6"CO"_2(g) -> cancel(1/2"Fe"_2"O"_3(s)) + 1/6"CO"(g), $\Delta H = - \frac{1}{6} \left(- \text{47 kJ/mol}\right) = \textcolor{red}{+ 7.8 \overline{33}}$ $\textcolor{red}{\text{kJ/mol}}$

$\setminus m a t h b f \left(\frac{1}{2}\right)$(2):   cancel(1/2"Fe"_2"O"_3(s)) + 9/6"CO"(g) -> "Fe"(s) + 9/6"CO"_2(g), $\Delta H = \frac{1}{2} \left(- \text{25 kJ/mol") = -color(red)("12.5}\right)$ $\textcolor{red}{\text{kJ/mol}}$

$\setminus m a t h b f \left(- \frac{1}{3}\right)$(3): $\text{FeO"(s) + 2/6"CO"_2(g) -> cancel(1/3"Fe"_3"O"_4(s)) + 2/6"CO} \left(g\right)$, $\Delta H = - \frac{1}{3} \left(+ \text{19 kJ/mol}\right) = \textcolor{red}{- 6. \overline{33}}$ $\textcolor{red}{\text{kJ/mol}}$
$\text{---------------------------------------------------------}$
$\setminus m a t h b f \left({\text{FeO"(s) + "CO"(g) -> "Fe"(s) + "CO}}_{2} \left(g\right)\right)$

$\implies \textcolor{b l u e}{\Delta {H}_{\text{overall}}}$

= (+"7.8"bar(33) "kJ/mol") + (-"12.5 kJ/mol") + (-6.bar(33) "kJ/mol")#

$= \textcolor{b l u e}{- \text{11 kJ/mol}}$