Question

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"`?

Answer 1

I got `"31.80"^@ "C"`.

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The reaction occurring is:

`"HCl"(aq) + "KOH"(aq) -> "H"_2"O"(l) + "KCl"(aq)`

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

`V = (85 + 85) cancel"mL" xx "1 L"/(1000 cancel"mL") = "0.170 L"`

and a common error is to assume the concentration of `"KCl"` is `"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).

`DeltaH_"rxn" = -"57.6 kJ/mol"`

At constant pressure,

`DeltaH_"rxn" = q_"rxn"`

whereas given a calorimeter's heat capacity:

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

where `C_"cal"` is the heat capacity of the calorimeter in `"J/"^@ "C"`, and `DeltaH` and `q` are both in units of `"J"`.

The reaction is exothermic since `DeltaH_"rxn" < 0`, so the solution absorbs the heat. Thus, we expect it to increase in temperature, so that `T_f > T_i` (i.e. `DeltaT > 0`).

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

`"0.900 mol/L" xx "0.085 L" = n_("KOH") = n_("HCl") = n_"KCl"`

`=` `"0.0765 mols"`

So, the heat absorbed by the solution is (where the units are all made to be in `"kJ"`):

`q_"cal" = -q_"rxn" = -n_"KCl"DeltaH_"rxn"`

`= -(-"57.6 kJ"/"mol")("0.0765 mols")`

`=` `"4.406 kJ"`

Since `q_"cal" = C_"cal"(T_f - T_i)`, the final temperature is:

`color(blue)(T_f) = (q_"cal")/(C_"cal") + T_i`

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

`=` `color(blue)("31.80"^@ "C")`