Concentration of NaoH Trial 4 Part C: Heat of Neutralization Trial 3 (7 marks) T

Question

Concentration of NaoH Trial 4 Part C: Heat of Neutralization Trial 3 (7 marks) T Trial 2 concentration of HCI Trial 1 34.49 A. Mass of calorimeter (g) Mass HCl and Calorimeter (g) C. Mass calorimeter, HCI and NaoH (g) D1. Mass HCI (g) (B-A) D2. Mass NaoH (g) (C-B) Mass of solution (D1-D2) Temperature Hci measured c) 55.640 55.42 54.06 l.oug 1.32 24.23 &l23; O thermometer 1) itial Temperature HCI corrected C) 240 240 230' 23.0 itial Temperature NaoH measured (C) tial Temperature NaoH corrected (C) 2133 23.3 23.33 ermometer 2) Average Initial Temperature corrected C) 23.620 2 2 al Temperature measured C) (thermometer 1) Final Temperature corrected C) Temperature change corrected C) (F E)

Explanation / Answer

Calorimeter

Heat capacity of calorimeter

Trial 1:

q = mCp (m = mass od HCl + NaOH ; Cp = specific heat of water)

   = 40.44 g x 4.184 J/g.oC

   = 169.201 kJ/mol

Similarly for other trials can be calculated.

molar heat of neutralization = mCpdT/n

with,

m = total mass of HCl + NaOH

Cp = specific heat of water

dT = change n temperature

n = moles of HCl = molarity x volume

Molar Heat if neutralization (dH) calculation

Trial 1 :

dH = mCpdT/n

     = 40.44 g x 4.184 J/g.oC x 4.69 oC/1.0 M x 0.01991 L

     = 39.857 kJ/mol

similarly for other trial runs,

Trial 2 :

dH = mCpdT/n

     = 40.45 g x 4.184 J/g.oC x 3.26 oC/1.0 M x 0.01976 L

     = 27.922 kJ/mol

Trial 3 :

dH = mCpdT/n

     = 39.98 g x 4.184 J/g.oC x 3.58 oC/1.0 M x 0.01947 L

     = 30.843 kJ/mol

Trial 4 :

dH = mCpdT/n

     = 40.36 g x 4.184 J/g.oC x 4.60 oC/1.0 M x 0.01952 L

     = 39.794 kJ/mol

average molar heat of neutralization = 34.604 kJ/mol

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