JEE Challenger
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ChemMedAdv 2015 • Paper 2

Enthalpy of Dissociation of Acetic Acid from Calorimetric Experiment

Comprehension Passage

PARAGRAPH 2

When 100 mL100\text{ mL} of 1.0 M HCl1.0\text{ M HCl} solution was mixed with 100 mL100\text{ mL} of 1.0 M NaOH1.0\text{ M NaOH} inside an insulated beaker under constant pressure, a temperature rise of 5.7∘C5.7^\circ\text{C} was recorded for the beaker along with its contents (Expt. 1). Since the enthalpy of neutralization for a strong acid with a strong base remains a constant (−57.0 kJ mol−1-57.0\text{ kJ mol}^{-1}), this setup was utilized to determine the calorimeter constant. In a subsequent trial (Expt. 2), 100 mL100\text{ mL} of 2.0 M2.0\text{ M} acetic acid (Ka=2.0×10−5K_a = 2.0 \times 10^{-5}) was mixed with 100 mL100\text{ mL} of 1.0 M NaOH1.0\text{ M NaOH} under identical conditions as Expt. 1, resulting in a temperature increase of 5.6∘C5.6^\circ\text{C}. (Assume heat capacity for all solutions to be 4.2 J g−1 K−14.2\text{ J g}^{-1}\text{ K}^{-1} and density for all solutions to be 1.0 g mL−11.0\text{ g mL}^{-1})

Calculate the enthalpy of dissociation (expressed in kJ mol−1\text{kJ mol}^{-1}) of acetic acid derived from Expt. 2.

Options

A

1.0

Correct
B

10.0

C

24.5

D

51.4

Topics & Concepts

ThermodynamicsEnthalpyAcid-Base Equilibrium

Step-by-Step Solution

To find the enthalpy of dissociation of acetic acid (ΔHdiss\Delta H_{\text{diss}}), we analyze the heat changes in both Experiment 1 and Experiment 2.

Step 1: Determine the total heat capacity of the calorimeter setup from Expt. 1

In Expt. 1, a strong acid (HCl\text{HCl}) is mixed with a strong base (NaOH\text{NaOH}):

  • Volume of HCl\text{HCl} solution = 100 mL100\text{ mL}
  • Volume of NaOH\text{NaOH} solution = 100 mL100\text{ mL}
  • Total volume of mixture, V=200 mLV = 200\text{ mL}

Given:

  • Density of solution, d=1.0 g mL−1  ⟹  Total mass m=200 gd = 1.0\text{ g mL}^{-1} \implies \text{Total mass } m = 200\text{ g}
  • Specific heat capacity of solution, s=4.2 J g−1 K−1s = 4.2\text{ J g}^{-1}\text{ K}^{-1}
  • Moles of HCl=0.100 L×1.0 M=0.1 mol\text{HCl} = 0.100\text{ L} \times 1.0\text{ M} = 0.1\text{ mol}
  • Moles of NaOH=0.100 L×1.0 M=0.1 mol\text{NaOH} = 0.100\text{ L} \times 1.0\text{ M} = 0.1\text{ mol}
  • Temperature rise, ΔT1=5.7 K\Delta T_1 = 5.7\text{ K}
  • Enthalpy of neutralization for strong acid-strong base, ΔHneut=−57.0 kJ mol−1\Delta H_{\text{neut}} = -57.0\text{ kJ mol}^{-1}

The heat released during neutralization (q1q_1) is given by: q1=n×(−ΔHneut)=0.1 mol×57.0 kJ mol−1=5.7 kJ=5700 Jq_1 = n \times (-\Delta H_{\text{neut}}) = 0.1\text{ mol} \times 57.0\text{ kJ mol}^{-1} = 5.7\text{ kJ} = 5700\text{ J}

This heat is absorbed by both the solution and the calorimeter (beaker): q1=Ctotal×ΔT1q_1 = C_{\text{total}} \times \Delta T_1 5700 J=Ctotal×5.7 K5700\text{ J} = C_{\text{total}} \times 5.7\text{ K} Ctotal=5700 J5.7 K=1000 J K−1C_{\text{total}} = \frac{5700\text{ J}}{5.7\text{ K}} = 1000\text{ J K}^{-1}

Step 2: Determine the reaction enthalpy in Expt. 2

In Expt. 2:

  • Moles of CH3COOH=0.100 L×2.0 M=0.2 mol\text{CH}_3\text{COOH} = 0.100\text{ L} \times 2.0\text{ M} = 0.2\text{ mol}
  • Moles of NaOH=0.100 L×1.0 M=0.1 mol\text{NaOH} = 0.100\text{ L} \times 1.0\text{ M} = 0.1\text{ mol}
  • Temperature rise, ΔT2=5.6 K\Delta T_2 = 5.6\text{ K}

Since NaOH\text{NaOH} is the limiting reagent, 0.1 mol0.1\text{ mol} of CH3COOH\text{CH}_3\text{COOH} reacts completely with 0.1 mol0.1\text{ mol} of OH−\text{OH}^-.

The total heat evolved (q2q_2) in Expt. 2 is: q2=Ctotal×ΔT2=1000 J K−1×5.6 K=5600 J=5.6 kJq_2 = C_{\text{total}} \times \Delta T_2 = 1000\text{ J K}^{-1} \times 5.6\text{ K} = 5600\text{ J} = 5.6\text{ kJ}

The enthalpy change of neutralization per mole of weak acid (ΔHrxn\Delta H_{\text{rxn}}) is: ΔHrxn=−q2nreacted=−5.6 kJ0.1 mol=−56.0 kJ mol−1\Delta H_{\text{rxn}} = -\frac{q_2}{n_{\text{reacted}}} = -\frac{5.6\text{ kJ}}{0.1\text{ mol}} = -56.0\text{ kJ mol}^{-1}

Step 3: Calculate the enthalpy of dissociation of acetic acid

The neutralization of a weak acid (CH3COOH\text{CH}_3\text{COOH}) by a strong base (OH−\text{OH}^-) involves two processes:

  1. Dissociation of acetic acid: CH3COOH(aq)⟶CH3COO−(aq)+H+(aq)(ΔHdiss)\text{CH}_3\text{COOH}(aq) \longrightarrow \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \quad (\Delta H_{\text{diss}})
  2. Neutralization of H+\text{H}^+ by OH−\text{OH}^-: H+(aq)+OH−(aq)⟶H2O(l)(ΔHneut=−57.0 kJ mol−1)\text{H}^+(aq) + \text{OH}^-(aq) \longrightarrow \text{H}_2\text{O}(l) \quad (\Delta H_{\text{neut}} = -57.0\text{ kJ mol}^{-1})

According to Hess's Law: ΔHrxn=ΔHdiss+ΔHneut\Delta H_{\text{rxn}} = \Delta H_{\text{diss}} + \Delta H_{\text{neut}} −56.0 kJ mol−1=ΔHdiss+(−57.0 kJ mol−1)-56.0\text{ kJ mol}^{-1} = \Delta H_{\text{diss}} + (-57.0\text{ kJ mol}^{-1}) ΔHdiss=−56.0+57.0=1.0 kJ mol−1\Delta H_{\text{diss}} = -56.0 + 57.0 = 1.0\text{ kJ mol}^{-1}

Correct Answer: A (1.0)

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