# Question 97a5e

Jul 22, 2016

I got $\text{6.2 kJ}$, though it's an unrealistic temperature change.

The first thing you should recognize is that the heat was generated through the reaction, and released into the water due to the neutralization of the staggering $\text{0.75 mols"/"0.075 L" = "10 M}$ $\text{HCl}$ by $\text{10 M}$ $\text{NaOH}$ (which, mind you, is quite dangerous).

It wasn't released into the atmosphere; if it was, the water would get colder!

Next, try writing the chemical reaction:

$\text{NaOH"(aq) + "HCl"(aq) -> "H"_2"O"(l) + "NaCl} \left(a q\right)$

We assume that:

• The density of the solution is similar to that of regular water at ${25}^{\circ} \text{C}$.
• The density changes minimally at ${35}^{\circ} \text{C}$. The density would change from about $0.997$ to about $0.994$, so we could just use $\text{0.995 g/mL}$.
• Adding the two solutions together is completely additive - it doesn't generate a solution of more than $75 + 75 = \text{150 mL}$. Also, the solutions themselves are already made, so they do both start at $\text{75 mL}$ each.
• All the heat released actually goes into heating the water, and none of it is released into the atmosphere like it would be in real life.

With that, we can calculate the mass of the solution:

$\text{150 mL" xx "0.995 g/mL" ~~ "149.25 g}$

Then, we can use the heat flow equation to determine the heat $q$ released into the water, from the reaction. So by convention we would write it to be negative, relative to the reaction, to emphasize it is exothermic.

color(blue)(q_"rxn") = -m_"soln"c_"soln"DeltaT_"soln"

= -m_"soln"c_"soln"(T_f^"soln" - T_i^"soln")

$= - \left(\text{149.25 g")("4.18 J/g"cdot""^@ "C")(35^@ "C" - 25^@ "C}\right)$

$= - 6238.65$

$\approx - 6.2 \times {10}^{3}$ $\text{J}$

$\implies$ $\textcolor{b l u e}{- \text{6.2 kJ}}$

So you can say $\text{6.2 kJ}$ was absorbed by the water, or released by the reaction.

As for the $\text{mol}$s you were given, since:

• $\text{HCl}$ and $\text{NaOH}$ have equimolar quantities of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ (they're 1:1 with each other)
• Their concentrations are identical (due to the same number of $\text{mol}$s and the same volumes)

You can use the $\text{mol}$s of either one to determine the enthalpy of reaction. You have $\text{0.75 mols}$ of $\text{HCl}$, so we can use that. Relative to the reaction, which was exothermic, $\Delta {H}_{\text{rxn}} < 0$. So:

$\textcolor{red}{\Delta {H}_{\text{rxn") = -"6.23865 kJ"/("0.75 mols HCl}}}$

$\approx$ $\textcolor{red}{- \text{8.3 kJ/mol}}$

Keep in mind that this number is not realistic; it really should be closer to $- \text{56 kJ/mol}$, so the temperature change was unrealistically low.

In real life, the final temperature should be more like $\setminus m a t h b f \left({92}^{\circ} \text{C}\right)$, because the concentrations are so high ($\text{10 M}$ is about 30% $\text{w/w}$ $\text{HCl}$, and labs generally buy 37%# stock $\text{HCl}$, which is about $\text{12 M}$).