# If "85 mL" of "0.900 M" "HCl" is combined with "85 mL" of "0.900 M" "KOH" in a coffee-cup calorimeter, and the heat capacity of the calorimeter is "0.325 kJ/"^@ "C", what is the final temperature if the initial temperature was 18.24^@ "C"?

Jan 4, 2017

I got $\text{31.80"^@ "C}$.

The reaction occurring is:

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

The final solution has a volume assumed equal to the combined initial volumes:

$V = \left(85 + 85\right) \cancel{\text{mL" xx "1 L"/(1000 cancel"mL") = "0.170 L}}$

and a common error is to assume the concentration of $\text{KCl}$ is $\text{0.900 M}$ (when it's really not).

But we don't even know what the enthalpy of reaction is. We just have to look it up, I guess (you shouldn't have to calculate it in this context).

$\Delta {H}_{\text{rxn" = -"57.6 kJ/mol}}$

At constant pressure,

$\Delta {H}_{\text{rxn" = q_"rxn}}$

whereas given a calorimeter's heat capacity:

q_"cal" = C_"cal"DeltaT = ulul(-q_"rxn"),

where ${C}_{\text{cal}}$ is the heat capacity of the calorimeter in $\text{J/"^@ "C}$, and $\Delta H$ and $q$ are both in units of $\text{J}$.

The reaction is exothermic since $\Delta {H}_{\text{rxn}} < 0$, so the solution absorbs the heat. Thus, we expect it to increase in temperature, so that ${T}_{f} > {T}_{i}$ (i.e. $\Delta T > 0$).

So, to solve for the final temperature, we first need to figure out the $\text{mol}$s of $\text{KCl}$ product. Pick one reactant and go with it, since the reactants are both going to react 1:1 (one ${\text{H}}^{+}$ to one ${\text{OH}}^{-}$):

$\text{0.900 mol/L" xx "0.085 L" = n_("KOH") = n_("HCl") = n_"KCl}$

$=$ $\text{0.0765 mols}$

So, the heat absorbed by the solution is (where the units are all made to be in $\text{kJ}$):

${q}_{\text{cal" = -q_"rxn" = -n_"KCl"DeltaH_"rxn}}$

$= - \left(- \text{57.6 kJ"/"mol")("0.0765 mols}\right)$

$=$ $\text{4.406 kJ}$

Since ${q}_{\text{cal" = C_"cal}} \left({T}_{f} - {T}_{i}\right)$, the final temperature is:

$\textcolor{b l u e}{{T}_{f}} = \left({q}_{\text{cal")/(C_"cal}}\right) + {T}_{i}$

= ("4.406 kJ")/("0.325 kJ/"^@ "C") + "18.24"^@ "C"

$=$ $\textcolor{b l u e}{\text{31.80"^@ "C}}$