Consider the reaction 2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g) The magnitude of enthalpy change for the reaction in kJ mol-1 is . (Nearest integer) Given : f H (H2S) = -20.1 kJ mol-1 f H (H2O) = -286.0 kJ mol-1 f H (SO2) = -297.0 kJ mol-1
Answer
1126
Solution
r H = f H (products) - f H (reactants) ( f H of O2 = 0)
= [2(-286.0) + 2(-297.0)] - [2(-20.1) + 0]
= -1166.0 + 40.2
= -1125.8 kJ mol-1
∴ Magnitude ≈ 1126
Answer: 1126