Question

500 ml of 1 M HCN is neutralized completely by 500 ml of 1 M KOH
solution and 1.36 K.cal of heat is liberated. The A H of dissociation of HCN
1) 10.98 Kcal/mol
3) 12.305 K.cal/mol
2) -10.98 K.cal
4) 8.12 Kcal/mol​

Answer 1

Answer:10.98 k cal per mole

Explanation:n=(500×1)/1000 =1/2 mole = 0.5 mole Heat released for 0.5 mol is 1.36kcal Then for 1mole =1.36×2 = 2.72 kcal Heat change in dissociating HCN = 13.7kcal -2.72kcal= 10.98kcal