Question

100 ml of 3 mol H.SO, reacts with 100 ml of 3 mol
NaOH. Enthalpy of neutralisation of reaction will be
(1) - 57.1 kJ/mol
(2) -2 * 57.1 kJ
(3) -0.3 * 57.1 kJ
(4) - 3* 57.1 kJ​

Answer 1

Answer:

-57.1 Kj.

Explanation:

Since, in the question it is given that there is equal amount of moles of acid of H2SO4 and base NaOH. That is 3 moles of the acid and base is  used for the neutralization reaction. Hence, the enthalpy will be same for both of the acid and the base and will be combined to give the neutralization enthalpy.

Since, the enthalpy of neutralization for equivalent amount of acid and base is the energy evolved as -57.1 Kj. So, for this reaction the enthalpy of neutralization is -57.1 Kj.