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← JEE Advanced 2015 · Paper 2
ChemistryMultiple correct+4 / -2Question 39 of 60

JEE Advanced 2015 · Paper 2, Chemistry question 39

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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 5.7 °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 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=2.0×10−5K_a=2.0\times10^{-5}) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C 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

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Answer and worked solution

Official answer: A

Calorimeter constant (Expt. 1). 0.1 mol is neutralised, releasing 5.7 kJ, for a rise of 5.7 °C. The total heat capacity is 1.0 kJ K⁻¹.

Expt. 2. 0.1 mol of NaOH neutralises 0.1 mol of acetic acid. The rise of 5.6 °C means 5.6 kJ released, i.e. −56.0 kJ mol⁻¹.

Dissociation enthalpy. The 1.0 kJ mol⁻¹ shortfall went into dissociating the acid:
ΔHdiss=57.0−56.0=1.0 kJ mol−1.\Delta H_{diss}=57.0-56.0=1.0\text{ kJ mol}^{-1}.

(A)

Topics:CalorimetryHess's law and enthalpies of reaction
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