# For the aqueous reaction of "100 mL" of "0.50 M" ammonia with "300 mL" of "0.50 M" hydrochloric acid to form ammonium chloride, if "HCl" is in excess, and the solution heated up by "1.6 K", what is the enthalpy of reaction?

Jul 10, 2017

$\Delta {H}_{r x n} = - \text{53.56 kJ/mol}$.

The reaction is an acid-base neutralization:

$\text{NH"_3(aq) + "HCl"(aq) -> "NH"_4"Cl} \left(a q\right)$

(the ${\text{NH}}_{3} \left(a q\right)$ is introduced as $\text{NH"_4"OH} \left(a q\right)$, which equilibrates to be primarily ${\text{NH}}_{3}$.)

This reaction generates heat, and the heat transferred out into the solution from the reaction at constant pressure is given by

${q}_{P} = m {C}_{P} \Delta T$,

where:

• ${q}_{P}$ is the heat flow with respect to the solution at constant pressure, i.e. lab bench conditions.
• $m$ is the mass of the solution in $\text{g}$.
• ${C}_{P}$ is the specific heat capacity of water at this $T$ and $P$.
• $\Delta T$ is the change in temperature in either $\text{^@ "C}$ or $\text{K}$ (why can we say that?).

We know that the $\text{HCl}$ is in excess because we were told that, so the ${\text{NH}}_{3} \left(a q\right)$ is clearly the limiting reagent. That means it must be equimolar with $\text{NH"_4"Cl}$ as per the reaction stoichiometry above.

"0.50 mol NH"_3/cancel"L soln" xx 100 cancel"mL soln" xx cancel"1 L"/(1000 cancel"mL")

$= \text{0.050 mols NH"_3 = "0.050 mols NH"_4"Cl}$

The total volume of the solution is

$\text{100 mL NH"_3 + "300 mL HCl" ~~ "400 mL}$,

if the solution volumes are assumed to be truly additive.

We can also assume the density and specific heat capacity of the solution is the same as for water at ${25}^{\circ} \text{C}$. For simplicity, we can use $\text{1.00 g/mL}$ and $\text{4.184 J/g" cdot "K}$, respectively.

Thus, the approximate mass of the solution is

$400 \cancel{\text{mL" xx "1.00 g"/cancel"mL" ~~ "400 g}}$,

So, for the heat transferred into the solution (coming out from the reaction!) at constant pressure to increase its temperature is given by:

${q}_{P} = \left(400 \cancel{\text{g")("4.184 J/"cancel"g"cdot cancel"K")(1.6 cancel"K}}\right)$

$=$ $\text{2677.76 J}$

But for the REACTION, we have, from conservation of thermal energy...

${q}_{r x n} + {q}_{P} = 0$.

So, at constant pressure (there's a reason why I keep saying this!), where the heat flow is DEFINED to be the change in enthalpy in the same units, we actually have:

$\Delta {H}_{r x n} = {q}_{r x n} = - {q}_{P}$

And thus, with just ${q}_{P}$, we currently have the negative change in enthalpy of reaction in $\text{J}$. But traditionally we report $\Delta {H}_{r x n}$ in $\text{kJ/mol}$ for the reaction.

Now hopefully you recognize that this is why we calculated the mols of $\text{NH"_4"Cl}$ before:

$\textcolor{b l u e}{\Delta {H}_{r x n}} \equiv {q}_{r x n} / \left({n}_{\text{NH"_4"Cl}}\right)$

$= - {q}_{P} / \left({n}_{\text{NH"_4"Cl}}\right)$

$= - \left(2677.76 \cancel{\text{J")/("0.050 mols NH"_4"Cl") xx "1 kJ"/(1000 cancel"J}}\right)$

$=$ $\textcolor{b l u e}{- \text{53.56 kJ/mol}}$