Verified Solution • 2025–26 Session moderate

QJEEAdv15Chc92f: JEE Advanced 2015 Chemistry - Class 11 JEEAdv15Chc92f Solution: Paragraph When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 …

Paragraph
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 5.7o C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is constant (-57.0 kJ/mol), 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 ×\times 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
Enthalpy of dissociation (in kJ/mol) of acetic acid obtained from the Expt. 2 is

Multiple Choice Options

  • A 1
  • B 10
  • C 24.5
  • D 51.4

Step-by-Step Verified Solution

Energy evolved on neutralization of HCl and NaOH is

0.1 ×\times 57 = 5.7 kJ = 5700 J

Energy utilized to rise the temperature of the solution is

ms . Δ\Delta T = 200 ×\times 1 ×\times 4.2 ×\times 5.7 = 4788 J

Energy used to increase temperature of calorimeter is

= 5700 −- 4788 = 912 J

ms . Δ\Delta T = 912

m ×\times s ×\times 5.7 = 912 →\rightarrow ms = 160 J ∘^\circ C −- 1 [Calorimeter constant]

Energy evolved by neutralization of CH3COOH and NaOH is

= 200 ×\times 4.2 ×\times 5.6 + 160 ×\times 5.6 = 5600 J

So, the energy used in dissociation of 0.1 mol CH3COOH is

= 5700 −- 5600 = 100 J

Thus, enthalpy of dissociation is 1 kJ mol −- 1.

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