Consider the following data. (i) 2Al(s) + 6HCl(aq) Al2Cl6(aq) + 3H2(g) + 1200 kJ/mol (ii) H2(g) + Cl2(g) 2HCl(g) + 164 kJ/mol (iii) HCl(g) + aq HCl(aq) + 83 kJ/mol (iv) Al2Cl6(s) + aq Al2Cl6(aq) + 663 kJ/mol The enthalpy of formation of anhydrous solid Al2Cl6 is :
- -648 kJ mol-1
- -1350 kJ mol-1
- -2002 kJ mol-1
- -1527 kJ mol-1
Answer
(4) -1527 kJ mol-1
Solution
Heat written on the product side is heat evolved, so:
(i) H1 = -1200 kJ, (ii) H2 = -164 kJ, (iii) H3 = -83 kJ, (iv) H4 = -663 kJ
Target equation: 2Al(s) + 3Cl2(g) Al2Cl6(s)
Add (i) + 3 × (ii) + 6 × (iii):
2Al(s) + 3Cl2(g) + aq Al2Cl6(aq) (6HCl(aq), 3H2 and 6HCl(g) cancel)
Then subtract (iv), i.e. add Al2Cl6(aq) Al2Cl6(s) + aq ( H = +663 kJ):
⇒ f H = H1 + 3 H2 + 6 H3 - H4
⇒ f H = -1200 + 3(-164) + 6(-83) - (-663)
⇒ f H = -1200 - 492 - 498 + 663
∴ f H = -1527 kJ mol-1
Answer: (4)