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
The correct answer is: A
Explanation
Given that \(KOH_{(aq)} + HCl_{(aq)} \to KCl_{(aq)} + H_{2}O_{(l)} ; \Delta H = -54 kJ/mol\)
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.
\(\therefore 2NaOH_{(aq)} + H_{2}SO_{4(aq)} \to Na_{2}SO_{4(aq)} + 2H_{2}O_{(l)} ; \Delta H = -(2 \times 54 kjmol^{-1})\)
\(\therefore \Delta H = -108 kJmol^{-1}\)