# Adding nitric acid to water is quite exothermic. What is the temperature change of 100 mL of water (d = 1.00g/mL) at 20.0°C [cp = 75.3 J/(mol·°C)] after adding 10.0 mL of concentrated HNO3(14.0M , ΔH°soln = –33.3 kJ/mol) ?

Jul 28, 2017

Well... this is teaching you a safe practice: add acid to water!

The temperature rise should be about ${10.05}^{\circ} \text{C}$.

INITIAL SETUP

Well, assuming a density of $\text{1.00 g/mL}$ for water, the combined mass of the starting water and the added ${\text{HNO}}_{3}$ is approximately:

$\text{100 g" + "10.0 g" ~~ "110.0 g}$

The initial state is the created mixture, and the final state is the mixture after the initial temperature change. We focus on the initial temperature rise, assumed instantaneous, and ignore the slow temperature drop that follows. Treat time as the horizontal axis and temperature as the vertical axis here.

In the above image (ignore the scale), the temperature rise would be the vertical increase starting from the horizontal line to an imaginary extrapolated point at roughly $\left(50 , 38.5\right)$ or so.

The amount heat flow out of the nitric acid into the mixture (the process being exothermic with respect to the nitric acid, endothermic with respect to the solution!) is governed by the specific heat capacity of the mixture and its mass.

FINDING THE HEAT FLOW IN GENERAL

In this case, you were given the molar specific heat capacity of water, ${C}_{P}$, and the heat flow associated with that is:

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

where ${q}_{P}$ is the heat flow at constant pressure, $m$ is the mass of the mixture in $\text{g}$, and $\Delta T$ is its temperature change.

Assume that your ${C}_{P}$ applies to the mixture as well.

${q}_{P} = \left(\text{110.0 g")("75.3 J/mol"^@ "C" xx "1 mol"/"18.015 g water}\right) \Delta T$

$= 459.78 \Delta T$ $\text{J}$

RELATING TO THE ENTHALPY OF THE DISSOLUTION

Now, since the ${\text{HNO}}_{3}$ was $\text{14.0 M}$, this many mols were involved:

${\text{14.0 mols"/cancel"L" xx 0.010 cancel("L HNO"_3(l)) = "0.14 mols HNO}}_{3} \left(l\right)$

At constant pressure (this is important!), we have that

${q}_{P} = \Delta H$,

the enthalpy of the dissolution process (again, exothermic with respect to the nitric acid, not the solution; bad notation).

So, the heat released into the solution due to dissolving the nitric acid into water is equal in magnitude to the enthalpy of dissolving the nitric acid in solution:

${\overbrace{0.14 \cancel{{\text{mols HNO"_3(l)))^(n_(HNO_3)) xx overbrace(|"33.0 kJ"/cancel("mol HNO}}_{3} \left(l\right)} |}}^{| \Delta {H}_{s o l n} |}$

= overbrace(459.78DeltaT" "cancel"J" xx "1 kJ"/(1000 cancel"J"))^(q_P)

$\implies \textcolor{b l u e}{| \Delta T | = {10.05}^{\circ} \text{C}}$

So, the temperature change was by ${10.05}^{\circ} \text{C}$, and after thinking about it, the solution became ${10.05}^{\circ} \text{C}$ hotter.