Consider the following data. (i) 2Al(s) + 6HCl(aq) -> Al2Cl6(aq) + 3H2(g) + 1200 kJ/mol (ii) H2(g) + Cl2(g) -> 2HCl(g) + 164 kJ/mol (iii) HCl(g) + aq -> HCl(aq) + 83 kJ/mol (iv) Al2Cl6(s) + aq -> Al2Cl6(aq) + 663 kJ/mol The enthalpy of formation of anhydrous solid Al2Cl6 is :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q54 – JEE-MAIN 2026

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Home Chemistry Chemical ThermodynamicsQuestion #294

Consider the following data. (i)

2 A l (s) + 6 H C l (a q) \to A l_{2} C l_{6} (a q) + 3 H_{2} (g) + 1200 kJ/mol

(ii)

H_{2} (g) + C l_{2} (g) \to 2 H C l (g) + 164 kJ/mol

(iii)

H C l (g) + a q \to H C l (a q) + 83 kJ/mol

(iv)

A l_{2} C l_{6} (s) + a q \to A l_{2} C l_{6} (a q) + 663 kJ/mol

The enthalpy of formation of anhydrous solid

A l_{2} C l_{6}

is :

Clue 1 of 3: Hess's Law Application

The problem requires calculating the enthalpy of formation for a target reaction using a series of given reactions. Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway taken, as long as the initial and final conditions are the same.

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Step 1: Define the Target Reaction and Identify Required Components✦ Active

The target reaction for the enthalpy of formation of anhydrous solid $A l_{2} C l_{6}$ from its elements in their standard states is:

2 A l (s) + 3 C l_{2} (g) \to A l_{2} C l_{6} (s)

We will use Hess's Law to combine the given reactions to achieve this target.

Step 2: Manipulate Given Reactions and Their Enthalpies○ Expand

1. Keep reaction (i) as is to get $2 A l (s)$ on the reactant side:

2 A l (s) + 6 H C l (a q) \to A l_{2} C l_{6} (a q) + 3 H_{2} (g) Δ H_{1} = + 1200 kJ/mol

2. Multiply reaction (ii) by 3 to get $3 C l_{2} (g)$ on the reactant side and $6 H C l (g)$ on the product side:

3 \times [H_{2} (g) + C l_{2} (g) \to 2 H C l (g)] ⟹ 3 H_{2} (g) + 3 C l_{2} (g) \to 6 H C l (g) Δ H_{2^{'}} = 3 \times (+ 164) = + 492 kJ/mol

3. Multiply reaction (iii) by 6 to convert $6 H C l (g)$ to $6 H C l (a q)$ for cancellation:

6 \times [H C l (g) \to H C l (a q)] ⟹ 6 H C l (g) \to 6 H C l (a q) Δ H_{3^{'}} = 6 \times (+ 83) = + 498 kJ/mol

4. Reverse reaction (iv) to get $A l_{2} C l_{6} (s)$ on the product side and cancel $A l_{2} C l_{6} (a q)$ :

A l_{2} C l_{6} (a q) \to A l_{2} C l_{6} (s) Δ H_{4^{'}} = - 663 kJ/mol

💡 Teacher's Secret Hint

Ensure correct multiplication factors and sign changes for reversed reactions.

Step 3: Sum the Manipulated Reactions and Enthalpies○ Expand

Summing the manipulated reactions and canceling common species on both sides:

2 A l (s) + 6 H C l (a q) + 3 H_{2} (g) + 3 C l_{2} (g) + 6 H C l (g) + A l_{2} C l_{6} (a q) \to A l_{2} C l_{6} (a q) + 3 H_{2} (g) + 6 H C l (g) + 6 H C l (a q) + A l_{2} C l_{6} (s)

This simplifies to the target reaction: $2 A l (s) + 3 C l_{2} (g) \to A l_{2} C l_{6} (s)$ . Now, sum the enthalpy changes:

Δ H_{f o r m a t i o n} = Δ H_{1} + Δ H_{2^{'}} + Δ H_{3^{'}} + Δ H_{4^{'}}

Δ H_{f o r m a t i o n} = (+ 1200) + (+ 492) + (+ 498) + (- 663)

Δ H_{f o r m a t i o n} = 1200 + 492 + 498 - 663 = 2190 - 663 = 1527 kJ/mol

The enthalpy of formation of anhydrous solid $A l_{2} C l_{6}$ is $1527 kJ/mol$ .

💡 Teacher's Secret Hint

Carefully perform the arithmetic summation of the enthalpy values.

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Chemistry Question 54 – JEE-MAIN 2026

Clue 1 of 3: Hess's Law Application

🧠 Full Solution Path