ExamBro
ExamBro
JEE Advanced · Chemistry · 9. Redox Reactions

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

  1. A 1.0
  2. B 10.0
  3. C 24.5
  4. D 51.4
Verified Solution

Answer & Solution

Correct Answer

(A) 1.0

Step-by-step Solution

Detailed explanation

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
\(\Rightarrow \Delta \text H _{\text {ionization }}\) of \(\text{CH} _3 \text{COOH} =57-56\) \(=1 \text{KJ} / \text{mol}\)
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app