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Thermodynamics question

2022 · 25 Jul · Shift 2 · Q16
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Thermodynamics question

2022 · 25 Jul · Shift 2 · Q16

JEE MainChemistryThermodynamicsNumerical+4 / −1
While performing a thermodynamics experiment, a student made the following observations. HClHClHCl + NaOHNaOHNaOH →\to→ NaClNaClNaCl + H2OH_2OH2​O Δ\DeltaΔ H = −-− 57.3 kJ mol −-− 1 CH3COOHCH_3COOHCH3​COOH + NaOHNaOHNaOH →\to→ CH3COONaCH_3COONaCH3​COONa + H2OH_2OH2​O Δ\DeltaΔ H = −-− 55.3 kJ mol −-− 1 The enthalpy of ionization of CH3COOHCH_3COOHCH3​COOH as calculated by the student is ‾\underline{\hspace{2cm}}​ kJ mol −-− 1. (nearest integer)
Numerical answer
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Correct answer: 2

  1. Identify the reactions

For a strong acid–strong base neutralization: HCl+NaOH→NaCl+H2O,ΔH=−57.3 kJ mol−1\mathrm{HCl + NaOH \to NaCl + H_2O}, \qquad \Delta H = -57.3\ \text{kJ mol}^{-1}HCl+NaOH→NaCl+H2​O,ΔH=−57.3 kJ mol−1

Since HCl\mathrm{HCl}HCl, NaOH\mathrm{NaOH}NaOH, and NaCl\mathrm{NaCl}NaCl are strong electrolytes, the net ionic reaction is: H++OH−→H2O,ΔH=−57.3 kJ mol−1\mathrm{H^+ + OH^- \to H_2O}, \qquad \Delta H = -57.3\ \text{kJ mol}^{-1}H++OH−→H2​O,ΔH=−57.3 kJ mol−1

  1. For acetic acid neutralization

CH3COOH+NaOH→CH3COONa+H2O,ΔH=−55.3 kJ mol−1\mathrm{CH_3COOH + NaOH \to CH_3COONa + H_2O}, \qquad \Delta H = -55.3\ \text{kJ mol}^{-1}CH3​COOH+NaOH→CH3​COONa+H2​O,ΔH=−55.3 kJ mol−1

Here, CH3COOH\mathrm{CH_3COOH}CH3​COOH is a weak acid, so before neutralization it must ionize: CH3COOH→CH3COO−+H+\mathrm{CH_3COOH \to CH_3COO^- + H^+}CH3​COOH→CH3​COO−+H+

Let the enthalpy of ionization of acetic acid be ΔHion\Delta H_{\text{ion}}ΔHion​.

Then the overall process is:

  • Ionization of acetic acid: CH3COOH→CH3COO−+H+ΔH=ΔHion\mathrm{CH_3COOH \to CH_3COO^- + H^+} \qquad \Delta H = \Delta H_{\text{ion}}CH3​COOH→CH3​COO−+H+ΔH=ΔHion​

  • Neutralization of H+\mathrm{H^+}H+ by OH−\mathrm{OH^-}OH−: H++OH−→H2OΔH=−57.3 kJ mol−1\mathrm{H^+ + OH^- \to H_2O} \qquad \Delta H = -57.3\ \text{kJ mol}^{-1}H++OH−→H2​OΔH=−57.3 kJ mol−1

So, by Hess's law, ΔHoverall=ΔHion+(−57.3)\Delta H_{\text{overall}} = \Delta H_{\text{ion}} + (-57.3)ΔHoverall​=ΔHion​+(−57.3)

Given: −55.3=ΔHion−57.3-55.3 = \Delta H_{\text{ion}} - 57.3−55.3=ΔHion​−57.3

  1. Calculate ΔHion\Delta H_{\text{ion}}ΔHion​

ΔHion=−55.3+57.3=2.0 kJ mol−1\Delta H_{\text{ion}} = -55.3 + 57.3 = 2.0\ \text{kJ mol}^{-1}ΔHion​=−55.3+57.3=2.0 kJ mol−1

  1. Nearest integer

2\boxed{2}2​

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