Question

A quantity of 2.00 x 102 mL of 0.745 M HCl is mixed with 2.00 102 mL of 0.373 M Ba(OH), in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.02°C. For the process ht(aq) + OH(aq) + H2O(1) the heat of neutralization is -56.2 kJ/mol. What is the final temperature of the mixed solution? 1 °C

Answer 1

Sol: the heat of neutralization of reaction is used to bind heat of Peachon. from this beat of solution is Calculated finally we use relation between heat of solution and temperature change to bind the binal temperature of solutions. I woles are related to volume & Conc.as shown below Nooo noles = conc. x volume Calculate moles of H+ present in HCL is (2x6 mi)/h * 0.145 mol HCl & Imotht loomi Il Solution imot Hal = 0.149 moi. moles of ot present in BaoH) are as follows: ( xs mi) ( IC ) x 0.313 mol Tab W) x 2.7 | bị IL solution Imol B4OH) LOCA = 0.149 mol OH Heald neutralizam o lachon is -56.2 Ismoi Htas & Ott lan) . H₂O(l) AH D -56.2kJ/mol As pressure Constant in calorimeter heat of reaction will be equal to real enthalpy of reaction (Altexn) & vxn -56.2 kJ/mol axy=-56.2 KJ x 0.149 mol Imol a - 8. 37K5 heat of solution is usor - Unn. (as no chance takes place. q sol = 8.37 KJ

heat of solution. Usol = mols to DT 6 msol mass of solution, ssol Specific heat of solution o change in temperature mso vol.of solution x density = (2x tomi + 2x10"mi) Gime solution) 19 = 4x10? G. Ssol = 4.189 J/g.cc Vsol = msal Ssol AT 8.37KJ * 10005 = (4x10'9) x 4.184 1190 x (T- 20.02) IKI T= 8.37 xlo J + 20.02°C (4x109) (4.184 I/9..) T = 20.02°c + 5.001°c Tz 25.021c ~ 250 T= 25.0c therefore, the final temperature is 25°c of solutions