JEE Main 2026 — Chemistry Physical Chemistry

The given reactions with their enthalpy changes are:

(i) $2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g)$2Al(s)+6HCl(aq)→Al2​Cl6​(aq)+3H2​(g), $\Delta H_1 = -1200$ΔH1​=−1200 kJ/mol

(ii) $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$H2​(g)+Cl2​(g)→2HCl(g), $\Delta H_2 = -164$ΔH2​=−164 kJ/mol

(iii) $HCl(g) + aq \rightarrow HCl(aq)$HCl(g)+aq→HCl(aq), $\Delta H_3 = -83$ΔH3​=−83 kJ/mol

(iv) $Al_2Cl_6(s) + aq \rightarrow Al_2Cl_6(aq)$Al2​Cl6​(s)+aq→Al2​Cl6​(aq), $\Delta H_4 = -663$ΔH4​=−663 kJ/mol

The required reaction for the enthalpy of formation of anhydrous solid $Al_2Cl_6$Al2​Cl6​ is:

$$2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s)$$2Al(s)+3Cl2​(g)→Al2​Cl6​(s)

This target equation can be obtained by applying the following algebraic operations on the given equations:

Equation (i) $+ 3 \times$+3× Equation (ii) $+ 6 \times$+6× Equation (iii) $-$− Equation (iv)

Let us verify by adding the modified equations:

$$2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g)$$2Al(s)+6HCl(aq)→Al2​Cl6​(aq)+3H2​(g)

$$3H_2(g) + 3Cl_2(g) \rightarrow 6HCl(g)$$3H2​(g)+3Cl2​(g)→6HCl(g)

$$6HCl(g) + aq \rightarrow 6HCl(aq)$$6HCl(g)+aq→6HCl(aq)

$$Al_2Cl_6(aq) \rightarrow Al_2Cl_6(s) + aq$$Al2​Cl6​(aq)→Al2​Cl6​(s)+aq

Summing the above reactions and canceling common terms on both sides, we get:

$$2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s)$$2Al(s)+3Cl2​(g)→Al2​Cl6​(s)

The enthalpy of formation is calculated as:

$$\Delta H_f = \Delta H_1 + 3\Delta H_2 + 6\Delta H_3 - \Delta H_4$$ΔHf​=ΔH1​+3ΔH2​+6ΔH3​−ΔH4​

$$\Delta H_f = -1200 + 3(-164) + 6(-83) - (-663)$$ΔHf​=−1200+3(−164)+6(−83)−(−663)

$$\Delta H_f = -1200 - 492 - 498 + 663$$ΔHf​=−1200−492−498+663

$\Delta H_f = -2190 + 663 = -1527$ΔHf​=−2190+663=−1527 kJ/mol

Answer: $-1527$−1527 kJ mol$^{-1}$−1