Q19 / 60 · JEE Advanced 2015 (Paper 2) · Chemistry

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1) , this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka=2.0×10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC was measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1 )
Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 is

  • Correct answer
    A
    1.0
  • B
    10.0
  • C
    24.5
  • D
    51.4

Solution

Let the heat capacity of insulated beaker be C.
Mass of aqueous content in Expt. 1
=100+100×1=200 g
⇒Total heat capacity=C+200×4.2 J/K
Moles of acid, base neutralized in Expt. 1 =1×0.1=0.1
⇒ Heat absorbed in Expt. 1 =0.1×57=5.7 kJ
⇒5.7×1000=C+200×4.2×ΔT
⇒5.7×1000=C+200+4.2×5.7
⇒C+200×4.2=1000
In Second experiment
nCH3COOH=0.2 , nNaOH=0.1
Total mass of aqueous content = 200 g
⇒ Total heat capacity =C+200×4.2=1000
⇒ Heat released =1000×5.6=5600 J
Overall, only 0.1 mol of CH3COOH undergo neutralization.
⇒ΔHneutralization of CH3COOH=-56000.1
=-56,000 J/mol=-56 KJ/mol
⇒ΔHionization of CH3COOH=57-56=1KJ/mol

Hence, the correct answer is option: 'A'.

Relevant Questions