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)?
-
A.
– 108Kj -
B.
-54 kj -
C.
-27 kj -
D.
+27 kj -
E.
+54 kj
Correct Answer: Option 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})