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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 2 April, Morning Shift — Question 43

Consider the following data (i) 2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2( g)+1200 kJ/mol2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{Al}_{2} \mathrm{Cl}_{6}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g})+ 1200 \mathrm{~kJ} / \mathrm{mol}. (ii) H2( g)+Cl2( g)→2HCl(g)+164 kJ/mol\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+164 \mathrm{~kJ} / \mathrm{mol}. (iii) HCl(g)+aq→HCl(aq)+83 kJ/mol\mathrm{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow \mathrm{HCl}(\mathrm{aq})+83 \mathrm{~kJ} / \mathrm{mol}. (iv) Al2Cl6( s)+aq→Al2Cl6(aq)+663 kJ/mol\mathrm{Al}_{2} \mathrm{Cl}_{6}(\mathrm{~s})+\mathrm{aq} \rightarrow \mathrm{Al}_{2} \mathrm{Cl}_{6}(\mathrm{aq})+663 \mathrm{~kJ} / \mathrm{mol}.

The enthalpy of formation of anhydrous solid Al2Cl6\mathrm{Al}_{2} \mathrm{Cl}_{6} is :

  1. Option A:

    −648 kJ mol−1-648 \mathrm{~kJ} \mathrm{~mol}^{-1}

  2. Option B:

    −1350 kJ mol−1-1350 \mathrm{~kJ} \mathrm{~mol}^{-1}

  3. Option C:

    −2002 kJ mol−1-2002 \mathrm{~kJ} \mathrm{~mol}^{-1}

  4. Option D:

    −1527 kJ mol−1-1527 \mathrm{~kJ} \mathrm{~mol}^{-1}

    Correct

Answer: D

Step-by-step solution

2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)2\mathrm{Al}(\mathrm{s}) + 6\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(\mathrm{aq}) + 3\mathrm{H}_2(\mathrm{g})

ΔrH1=−1200 kJ/mol[H2(g)+Cl2(g)→2HCl(g)]×3ΔrH2=−163×3 kJ/mol[HCl(g)+aq.→HCl(aq)]×6ΔrH3=−83×6 kJ/molAl2Cl6(aq)→Al2Cl6(s)+aq.ΔrH4=+663 kJ/mol2Al(s)+3Cl2→Al2Cl6(s)\begin{aligned} &\Delta_{\mathrm{r}}H_1 = -1200\,\mathrm{kJ/mol} \\[4pt] &\left[\mathrm{H}_2(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \rightarrow 2\mathrm{HCl}(\mathrm{g})\right]\times 3 \\ &\Delta_{\mathrm{r}}H_2 = -163\times 3\,\mathrm{kJ/mol} \\[4pt] &\left[\mathrm{HCl}(\mathrm{g}) + \mathrm{aq.} \rightarrow \mathrm{HCl}(\mathrm{aq})\right]\times 6 \\ &\Delta_{\mathrm{r}}H_3 = -83\times 6\,\mathrm{kJ/mol} \\[4pt] &\mathrm{Al}_2\mathrm{Cl}_6(\mathrm{aq}) \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(\mathrm{s}) + \mathrm{aq.} \\ &\Delta_{\mathrm{r}}H_4 = +663\,\mathrm{kJ/mol} \\[4pt] &2\mathrm{Al}(\mathrm{s}) + 3\mathrm{Cl}_2 \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(\mathrm{s}) \end{aligned} ΔfH=ΔrH1+ΔrH2+ΔrH3+ΔrH4=−1200−163×3−83×6+663=−1527 kJ/mol\begin{aligned} \Delta_{\mathrm{f}}H &= \Delta_{\mathrm{r}}H_1 + \Delta_{\mathrm{r}}H_2 + \Delta_{\mathrm{r}}H_3 + \Delta_{\mathrm{r}}H_4 \\ &= -1200 - 163\times 3 - 83\times 6 + 663 \\ &= -1527\,\mathrm{kJ/mol} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes

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