ThermodynamicsHard

Question

On the basis of the following thermochemical data : (ᐃGfoH+(aq) = 0)
H2O(l) → H+ (aq) + OH-(aq); ᐃH = 57.32 kj
H2(g) + O2(g) → H2O(l) ; ᐃH = - 286.20kJ
The value of enthalpy of formation of OH- ion at 25oC is :

Options

A.-22.88 kJ
B.-228.88 kJ
C.+228.88 kJ
D.-343.52 kJ

Solution

By adding the two given equations, we have
H2(g) + O2(g) → H+2(g) + OH-2(g) ; ᐃH = -228.88 Kj
Here ᐃHfo of H+(aq) = 0
∴     ᐃHfo of OH+ = -228.88 kJ

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