Question

Consider the following equations: (i) \[ 2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g) + 1200 \, kJ/mol \] (ii) \[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) + 164 \, kJ/mol \] (iii) \[ HCl(g) + aq \rightarrow HCl(aq) + 83 \, kJ/mol \] (iv) \[ Al_2Cl_6(s) + aq \rightarrow Al_2Cl_6(aq) + 663 \, kJ/mol \] The enthalpy of formation of anhydrous solid \(Al_2Cl_6\) is:

• A. \(-648\) kJ mol\(^{-1}\)
• B. \(-1350\) kJ mol\(^{-1}\)
• C. \(-2002\) kJ mol\(^{-1}\)
• D. \(-1527\) kJ mol\(^{-1}\)

Answer 1

The Correct Option is D

Concept: Use Hess's law. The enthalpy change for a reaction equals the sum of enthalpy changes of individual steps. We combine the given equations to obtain the formation reaction: \[ 2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s) \]
Step 1:Manipulate given reactions} Use reaction (ii) to generate \(HCl(g)\). Convert \(HCl(g)\) to \(HCl(aq)\) using reaction (iii). Use reaction (i) for formation of \(Al_2Cl_6(aq)\). Finally convert \(Al_2Cl_6(aq)\) to \(Al_2Cl_6(s)\) using reverse of reaction (iv).
Step 2:Sum enthalpies} Adding appropriate reactions and subtracting hydration enthalpy: \[ \Delta H_f = -1527 \, kJ\,mol^{-1} \] \[ \boxed{-1527 \, kJ\,mol^{-1}} \]