Consider the following reactions: (i) H+(aq) + OH-(aq) H2O(l), H = -X1 kJ mol-1 (ii) H2(g) + 12O2(g) H2O(l), H = -X2 kJ mol-1 (iii) CO2(g) + H2(g) CO(g) + H2O(l), H = -X3 kJ mol-1 (iv) C2H2(g) + 52O2(g) 2CO2(g) + H2O(l), H = +X4 kJ mol-1 Enthalpy of formation of H2O(l) is
- +X3 kJ mol-1
- -X4 kJ mol-1
- +X1 kJ mol-1
- -X2 kJ mol-1
Answer
(4) -X2 kJ mol-1
Solution
As per NCERT, the standard enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (reference states).
⇒ Only reaction (ii) forms 1 mol of H2O(l) from its elements H2(g) and O2(g).
(i) is neutralisation, (iii) and (iv) have compounds as reactants.
∴ f H [H2O(l)] = -X2 kJ mol-1
Answer: (4)