Consider the following data. (i) 2Al(s) + 6HCl(aq) Al_2Cl_6(aq) + 3H_2(g) + 1200 kJ/mol (ii) H_2(g) + Cl_2(g) 2HCl(g) + 164 kJ/mol (iii) HCl(g) + aq HCl(aq) + 83 kJ/mol (iv) Al_2Cl_6(s) + aq Al_2Cl_6(aq) + 663 kJ/mol The enthalpy of formation of anhydrous solid Al_2Cl_6 is :
D. -1527 kJ mol^ -1
The given reactions with their enthalpy changes are: (i) 2Al(s) + 6HCl(aq) Al_2Cl_6(aq) + 3H_2(g), H_1 = -1200 kJ/mol (ii) H_2(g) + Cl_2(g) 2HCl(g), H_2 = -164 kJ/mol (iii) HCl(g) + aq HCl(aq), H_3 = -83 kJ/mol (iv) Al_2Cl_6(s) + aq Al_2Cl_6(aq), H_4 = -663 kJ/mol The required reaction for the enthalpy of formation of anhydrous solid Al_2Cl_6 is: 2Al(s) + 3Cl_2(g) Al_2Cl_6(s) This target equation can be obtained by applying the following algebraic operations on the given equations: Equation (i) + 3 Equation (ii) + 6 Equation (iii) - Equation (iv) Let us verify by adding the modified equations: 2Al(s) + 6HCl(aq) Al_2Cl_6(aq) + 3H_2(g) 3H_2(g) + 3Cl_2(g) 6HCl(g) 6HCl(g) + aq 6HCl(aq) Al_2Cl_6(aq) Al_2Cl_6(s) + aq Summing the above reactions and canceling common terms on both sides, we get: 2Al(s) + 3Cl_2(g) Al_2Cl_6(s) The enthalpy of formation is calculated as: H_f = H_1 + 3 H_2 + 6 H_3 - H_4 H_f = -1200 + 3(-164) + 6(-83) - (-663) H_f = -1200 - 492 - 498 + 663 H_f = -2190 + 663 = -1527 kJ/mol Answer: -1527 kJ mol^ -1
Related: Chemistry — Thermodynamics (C) · All PYQ Banks