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Question

The enthalpy of neutralisation of strong acid by a strong base is 57.32 kJmol1. The enthalpy of formation of water is 285.84 kJmol1. The enthalpy of formation of hydroxyl ion is :

A
+228.52 kJmol1
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B
114.26 kJmol1
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C
228.52 kJmol1
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D
+114.2 kJmol1
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Solution

The correct option is C 228.52 kJmol1
Neutralisation of strong acid and strong base-
H+(aq)+OH(aq)H2O(l) ΔH=57.32 kJmol1
reversing the reaction we get-
H2O(l)H+(aq)+OH(aq) ΔH=57.32 kJmol1.....1
Formation of water-
H2(g)+12O2(g)H2O(l) ΔH=285.84 kJmol1.....2

add equation 1 & 2 to get the enthalpy of formation of OH.
therefore,

H2O(l)+H2(g)+12O2(g)H+(aq)+OH(aq)+H2O(l) ΔH=57.32285.84 kJmol1

H2(g)+12O2(g)H+(aq)+OH(aq) ΔH=228.52 kJmol1

hence, the enthalpy of formation of OH is 228.52 kJmol1

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