Given that for the reaction KOH (aq) + HCI(aq) → KCI...
Given that for the reaction KOH (aq) + HCI(aq) → KCI(aq) + H2O(L) H = - 54 Kj mol-1. What is the quantity of heat involved in the reaction?2NaOH(aq) + H2SO 4(aq) → Na2SO4(aq) + 2H 2O(l)?
- 108Kj
-54 kj
-27 kj
+27 kj
+54 kj
Correct answer is A
Given that KOH(aq)+HCl(aq)→KCl(aq)+H2O(l);ΔH=−54kJ/mol
The standard heat of neutralization is the amount of heat evolved when 1 mole of hydrogen ion from an acid reacts with 1 mole of hydroxide ion from the base to form one mole of water, under standard conditions.
∴
\therefore \Delta H = -108 kJmol^{-1}