Heat of neutralization of strong acid and strong base under 1 atm and 25 o C is -13.7 Kcal/equivalent. If standard Gibb's energy change for dissociation of water to H + and O H - is -19.14 Kcal/mol, the change in standard entropy for dissociation of water in cal K - 1 m o l - 1 is:

Step-by-step solution

Δ H o for neutralization of strong acid and base is − 13.7 kcal/equivalent H + + O H − → H 2 O Hence, for dissociation, enthalpy change will be + 13.7 kcal/equivalent H 2 O → H + + O H − Δ G o for dissociation is given as: − 19.14 kcal Δ G o = Δ H o − T Δ S o ⇒ Δ S o = ( Δ H o − Δ G o ) T = 13.7 − ( − 19.14 ) 298 ⇒ Δ S o = ( 32.84 k c a l 298 K ) = 110.2 c a l K − 1 m o l − 1