JEE Main 2026 Apr 2 Shift 1, Chemistry Q4: Hess’s Law and Bond Energy

JEE Main2026Apr 2, Shift 1Chemistry

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 :

Solution

The target reaction is $2 A l (s) + 3 C l_{2} (g) \to A l_{2} C l_{6} (s)$ , $Δ_{f} H = ?$

The energy values given are released, so each $Δ_{r} H$ is negative. Add the following weighted reactions to reach the target:

Reaction (i) as written: $Δ H_{1} = - 1200$ kJ/mol

Reaction (ii) multiplied by $3$ : $Δ H_{2} = 3 \times (- 163) = - 489$ kJ/mol

Reaction (iii) multiplied by $6$ : $Δ H_{3} = 6 \times (- 83) = - 498$ kJ/mol

Reaction (iv) reversed: $Δ H_{4} = + 663$ kJ/mol

Summing:

$Δ_{f} H = - 1200 - 489 - 498 + 663 = - 1527$ kJ/mol

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.