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ChemMedMain 2026 • 2nd Apr - Shift 1

Calculate Enthalpy of Formation of Anhydrous Aluminum Chloride

Consider the following data.

(i) 2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)+1200 kJ/mol2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)} + 1200\text{ kJ/mol}
(ii) H2(g)+Cl2(g)→2HCl(g)+164 kJ/mol\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)} + 164\text{ kJ/mol}
(iii) HCl(g)+aq→HCl(aq)+83 kJ/mol\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)} + 83\text{ kJ/mol}
(iv) Al2Cl6(s)+aq→Al2Cl6(aq)+663 kJ/mol\text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 663\text{ kJ/mol}

The enthalpy of formation of anhydrous solid Al2Cl6\text{Al}_2\text{Cl}_6 is :

Options

A

−648 kJ mol−1-648\text{ kJ mol}^{-1}

B

−1350 kJ mol−1-1350\text{ kJ mol}^{-1}

C

−2002 kJ mol−1-2002\text{ kJ mol}^{-1}

D

−1527 kJ mol−1-1527\text{ kJ mol}^{-1}

Correct

Topics & Concepts

ThermodynamicsEnthalpy

Step-by-Step Solution

To find the enthalpy of formation of anhydrous solid Al2Cl6(s)\text{Al}_2\text{Cl}_6\text{(s)}, we need to determine the enthalpy change for the target reaction: 2Al(s)+3Cl2(g)→Al2Cl6(s)2\text{Al(s)} + 3\text{Cl}_2\text{(g)} \rightarrow \text{Al}_2\text{Cl}_6\text{(s)}

The given thermochemical equations and their corresponding enthalpy changes (ΔH\Delta H) are:

  1. 2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)}
    ΔH1=−1200 kJ/mol\Delta H_1 = -1200\text{ kJ/mol}

  2. H2(g)+Cl2(g)→2HCl(g)\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}
    ΔH2=−164 kJ/mol\Delta H_2 = -164\text{ kJ/mol}

  3. HCl(g)+aq→HCl(aq)\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)}
    ΔH3=−83 kJ/mol\Delta H_3 = -83\text{ kJ/mol}

  4. Al2Cl6(s)+aq→Al2Cl6(aq)\text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)}
    ΔH4=−663 kJ/mol\Delta H_4 = -663\text{ kJ/mol}

Step-by-Step Combination of Reactions:

To obtain the target reaction, we can combine the given reactions using Hess's Law of Constant Heat Summation:

  • Keep reaction (i) as it is:
    2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)(ΔH1=−1200 kJ)2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)} \quad (\Delta H_1 = -1200\text{ kJ})

  • Multiply reaction (ii) by 33:
    3H2(g)+3Cl2(g)→6HCl(g)(3ΔH2=3×(−164)=−492 kJ)3\text{H}_2\text{(g)} + 3\text{Cl}_2\text{(g)} \rightarrow 6\text{HCl(g)} \quad (3\Delta H_2 = 3 \times (-164) = -492\text{ kJ})

  • Multiply reaction (iii) by 66:
    6HCl(g)+aq→6HCl(aq)(6ΔH3=6×(−83)=−498 kJ)6\text{HCl(g)} + \text{aq} \rightarrow 6\text{HCl(aq)} \quad (6\Delta H_3 = 6 \times (-83) = -498\text{ kJ})

  • Reverse reaction (iv):
    Al2Cl6(aq)→Al2Cl6(s)+aq(−ΔH4=+663 kJ)\text{Al}_2\text{Cl}_6\text{(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \quad (-\Delta H_4 = +663\text{ kJ})

Adding the Reactions:

Adding the four modified equations cancels out all intermediate species (HCl(aq)\text{HCl(aq)}, HCl(g)\text{HCl(g)}, H2(g)\text{H}_2\text{(g)}, Al2Cl6(aq)\text{Al}_2\text{Cl}_6\text{(aq)}, and aq\text{aq}):

2Al(s)+3Cl2(g)→Al2Cl6(s)2\text{Al(s)} + 3\text{Cl}_2\text{(g)} \rightarrow \text{Al}_2\text{Cl}_6\text{(s)}

Enthalpy Calculation:

ΔHf∘=ΔH1+3ΔH2+6ΔH3−ΔH4\Delta H_f^\circ = \Delta H_1 + 3\Delta H_2 + 6\Delta H_3 - \Delta H_4

ΔHf∘=(−1200)+3(−164)+6(−83)−(−663)\Delta H_f^\circ = (-1200) + 3(-164) + 6(-83) - (-663)

ΔHf∘=−1200−492−498+663\Delta H_f^\circ = -1200 - 492 - 498 + 663

ΔHf∘=−2190+663=−1527 kJ mol−1\Delta H_f^\circ = -2190 + 663 = -1527\text{ kJ mol}^{-1}

Thus, the enthalpy of formation of anhydrous solid Al2Cl6\text{Al}_2\text{Cl}_6 is −1527 kJ mol−1-1527\text{ kJ mol}^{-1}.

Correct Option: D

Calculate Enthalpy of Formation of Anhydrous Aluminum Chloride | Chemistry PYQ Solution - JEE Challenger