On the basis of the following thermochemical data: (Δ f GoH+(aq)=0) H2O(ℓ) arrow H+(aq)+OH-(aq); Δ H=57.32kj H2(g)+(1/2)O2(g) arrow H2O(ℓ); Δ H=-286.20kj The value of enthalpy of formation of OHâ ion at 25o C is :

Question

On the basis of the following thermochemical data: (Δ f GoH+(aq)=0) H2O(ℓ) arrow H+(aq)+OH-(aq); Δ H=57.32kj H2(g)+(1/2)O2(g) arrow H2O(ℓ); Δ H=-286.20kj The value of enthalpy of formation of OHâ ion at 25o C is : (A) -22.88 kJ (B) -228.88 kJ (C) +228.88 kJ (D) -343.52 kJ

Tardigrade Admin · Accepted Answer

By adding the two given equations, we have H2(g)+(1/2)O2(g) arrow H+(aq)+OH-(aq); Δ H=-228.88 Kj Here Δ Hof of H+(aq)=0 ∴ Δ Hof of OH-=-228.88 kj

Q. On the basis of the following thermochemical data $: (Δ f G^{o} H_{(a q)}^{+} = 0)$
$H_{2} O (ℓ) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 kj$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ); Δ H = - 286.20 kj$
The value of enthalpy of formation of $O H^{-}$ ion at $2 5^{o} C$ is :

$- 22.88 k J$

$- 228.88 k J$

$+ 228.88 k J$

$- 343.52 k J$

By adding the two given equations, we have
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{(a q)}^{+} + O H_{(a q)}^{-}; Δ H = - 228.88 K j$
Here $Δ H_{f}^{o} o f H_{(a q)}^{+} = 0$
$∴ Δ H_{f}^{o} o f O H^{-} = - 228.88 kj$

Q. On the basis of the following thermochemical data :(ΔfGoH(aq)+​=0) H2​O(ℓ)→H+(aq)+OH−(aq);ΔH=57.32kj H2​(g)+21​O2​(g)→H2​O(ℓ);ΔH=−286.20kj The value of enthalpy of formation of OH− ion at 25oC is :

−22.88kJ

−228.88kJ

+228.88kJ

−343.52kJ