Question

A 100.0 mL of 0.50 M HCl at is reacted with 125.0mL of 0.50 M NaOH in a calorimeter whose heat capacity is 500 J/C. What will the maximum temperature reached by the resulting solution if the initial temperature for the solution is 20.0°C? (Assume that the specific heat of the final solution is 4.18 J/g·°C, and that the density of the final solution is that of water.) The enthalpy change of HCl acid is neutralized by NaOH is –56 kJ/mol.

Answer 1

moles of Hal in 18000 ml of 0.50m Hl. = 0.50 imol x 198 my = 0.0 Samol inoles of Naoth in 125.0m of o.som Noott - 0.50 mo & tex 125 m = 0·0625 mol von reaction of all & Nach: Het acht Noot (2) H2O(l) & Nach(ar) According to stoichiometric ratio of this reaction mol HCl react with no woott : 0.05" & a & 0.05 . u. out here given Naotte 0. 0625 mol So, here, noot is excess reagent of the is limiting reagent & heat will be released according to limiting reaction & reaction will occur according to limiting reagent.

-0 i Enthalpy change of neutralisation reaction of lund ta & imol Noot is – 56 kJ/mol Enthalpy change heren) of neutralisation reaction of 0.05 mol 40 40.05 mot Noon awill be – 568 0.05 kI - 2.8kJ – 2800J here, attorne – 28007 Here heat released by the reaction. A ren) = heat gained bxsolution (Vrol) & heat gained by the * Calorimeter Pear).-0) Now, or sol sol X AT where, was mans of solm . - volume of solar & density of = (10 + 125 m x 18 Edelmeiben - 225g = denuto ofarter el & Specific heat of solution = 4.18 2 AT - Temperature charge of solution Ti-Initial 7T Carl = (t – 20) 2 = 200 Ana = 225 X 4+18X(7-2019 Fires temp q cara CarX AT T= ?) [co spenfie = 500X (T-20) heat of calorimeter - 500 5/6 AT = same as non = (-20) 9°C

u using equation (3) 280 = 225x 4•18 x (74 – 20) + 500 x(7-26) =) 280 = (71 -20 ) [2254498)+507 -) 2500 = (2-20) x 7490.5 =) (74 -20) = 269.0.5 = 1.94 in T & 20+ 1942 21.94 ~ 22.6. Final maximum temperature can be reached to 22.0