By adding the two given equations, we have H2( g)+21O2( g)→H(aq)++OH(aq)−;ΔH=−228.88Kj Here ΔHf∘ of H(aq)+=0 ∴ΔHf∘ of OH−=−228.88 kJ
On the basis of the following thermochemical data: (Δ fG° H_ (aq) ^+=0 ) … The value of enthalpy of formation of OH^-ion at 25° C is:
JEE Main 2009 — Chemistry Physical Chemistry
On the basis of the following thermochemical data: (ΔfG∘H(aq) +=0) H2O(ℓ)→H+(aq)+OH−(aq);ΔH=57.32 kJH2( g)+21O2( g)→H2O(ℓ);ΔH=−286.20 kJ The value of enthalpy of formation of OH−ion at 25∘C is:
Held on 30 Apr 2009 · Verified 6 Jul 2026.
Options
- A
−22.88 kJ
- B
−228.88 kJ
- C
+228.88 kJ
- D
−343.52 kJ
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