# What is the enthalpy of reaction in "kJ/mol" for when "0.254 g" of "Mg" is dropped into a coffee-cup calorimeter, if the temperature of "200. mL" of water rose from 25.00^@ "C" to 32.20^@ "C"?

Dec 15, 2016

I got $\Delta {H}_{\text{rxn" = -"577 kJ/mol}}$.

Well, the first thing you should recognize is that $\Delta {H}_{\text{rxn}}$ in a constant pressure scenario is equal to $q$ of the reaction, ${q}_{\text{rxn}}$, divided by the $\text{mol}$s of $\text{Mg} \left(s\right)$.

Furthermore, by conservation of energy, we assume that $- {q}_{\text{cal" = q_"rxn}}$, i.e., that the heat released by the reaction is completely absorbed into the calorimeter solution to increase its temperature.

(It's not perfect, as the calorimeter is open to the atmosphere to manage a constant pressure, but it's a pretty decent assumption.)

From that, we have the main equation:

$\boldsymbol{{q}_{\text{cal}} = {m}_{w} {c}_{w} \Delta T}$

where:

• ${m}_{w}$ is the mass of the solution, obtained by using the volume of the product solution and the density of the solution (which we assume is that of pure water).
• ${c}_{w} = \text{4.184 J/g"^@ "C}$ is the specific heat capacity of the solution, again assumed to be equal to that for pure water.
• $\Delta T$ is the change in temperature of the solution due to the exothermic reaction.

So, the main steps would be:

1. Find ${q}_{\text{cal}}$, in $\text{kJ}$.
2. Solve for ${q}_{\text{rxn}}$ via the sign difference, in $\text{kJ}$.
3. Solve for $\Delta {H}_{\text{rxn}}$ in "kJ""/"ul(ul("mol")) by utilizing the given mass of $\text{Mg} \left(s\right)$.

FINDING HEAT EVOLVED IN THE CALORIMETER

$\textcolor{g r e e n}{{q}_{\text{cal") = ("200. mL" xx "1.00 g"/"mL")("4.184 J/g"^@ "C")(32.20^@ "C" - 25.00^@ "C}}}$

$=$ ${6.02}_{496} \times {10}^{3}$ $\text{J}$

$= \textcolor{g r e e n}{{6.02}_{496}}$ $\textcolor{g r e e n}{\text{kJ}}$

where subscripts indicate the uncertain digits (the digits past the last significant figure).

FINDING HEAT EVOLVED DUE TO THE REACTION

This is easy. We stated that $- {q}_{\text{cal" = q_"rxn}}$, so

$\textcolor{g r e e n}{{q}_{\text{rxn}} = - {6.02}_{496}}$ $\textcolor{g r e e n}{\text{kJ}}$

FINDING ENTHALPY OF THE REACTION

This part is where you really have to think. The reaction is based on the given mass of $\text{Mg} \left(s\right)$, and you are asked to report your answer in "kJ"/("mol Mg"(s)).

The ${q}_{\text{rxn}}$ we just found is numerically equal to $\Delta {H}_{\text{rxn}}$ in $\text{kJ}$.... but not in $\text{kJ/mol}$. So, we first utilize the mass of $\text{Mg} \left(s\right)$ to get its $\text{mol}$s:

color(green)(n_("Mg"(s))) = "0.254 g Mg"(s) xx ("1 mol Mg"(s))/("24.305 g Mg"(s))

$= \textcolor{g r e e n}{{0.0104}_{505}}$ $\textcolor{g r e e n}{\text{mols Mg} \left(s\right)}$

Finally, we note the unit conversion to get:

color(blue)(DeltaH_"rxn") = q_"rxn"/(n_("Mg"(s)))

$= - \left({6.02}_{496} \text{kJ")/(0.0104_(505) "mols Mg} \left(s\right)\right)$

$= \textcolor{b l u e}{- \text{577 kJ/mol}}$ (rounded to three sig figs)