JEE Main 2026 · 2 Apr · Shift 1 · Chemistry Q54

ChemistrySingle correct+4 / -1Question 54 of 75

Consider the following data.

(i) $\ce{2Al(s) + 6HCl(aq) -> Al2Cl6(aq) + 3H2(g)}$ + 1200 kJ/mol

(ii) $\ce{H2(g) + Cl2(g) -> 2HCl(g)}$ + 164 kJ/mol.

(iii) $\ce{HCl(g) + aq -> HCl(aq)}$ + 83 kJ/mol.

(iv) $\ce{Al2Cl6(s) + aq -> Al2Cl6(aq)}$ + 663 kJ/mol

The enthalpy of formation of anhydrous solid $\ce{Al2Cl6}$ is :

-648

kJ mol

^{-1}

-1350

kJ mol

^{-1}

-2002

kJ mol

^{-1}

-1527

kJ mol

^{-1}

Answer and worked solution

Official answer: D

Target. $\ce{2Al(s) + 3Cl2(g) -> Al2Cl6(s)}$ . The heats given are released, so each ΔH is negative.

Hess's law.
$(\text{i})+3(\text{ii})+6(\text{iii})-(\text{iv}):\quad\Delta H_f=-1200-3(164)-6(83)+663=-1527\text{ kJ mol}^{-1}.$

(The 6 HCl(g) formed in 3 × (ii) dissolve to give the 6 HCl(aq) used in (i); reversing (iv) turns $\ce{Al2Cl6(aq)}$ into the solid.)

(D)

Topics:Hess's law and enthalpies of reaction

JEE Main 2026 · 2 Apr · Shift 1, Chemistry question 54