# 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"#?

##### 1 Answer

I got

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

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

So, to solve for the final temperature, we first need to figure out the

#"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

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

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

#=# #"4.406 kJ"#

Since

#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")#