Change in enthalpy for reaction, 2H2O2(l) 2H2O(l) + O2(g) if heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively, is
- -196 kJ/mol
- +196 kJ/mol
- +948 kJ/mol
- -948 kJ/mol
Answer
(1) -196 kJ/mol
Solution
r H = f H(products) - f H(reactants)
∵ f H of O2(g) (element in standard state) = 0
⇒ r H = 2(-286) + 0 - 2(-188)
= -572 + 376
= -196 kJ/mol
Answer: (1)