On the basis of the following thermochemical data : ( \(\Delta \mathrm{fG}^{\circ} \mathrm{H}_{(\mathrm{aq})}^{+}=0\) ) \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; \Delta \mathrm{H}=57.32 \mathrm{~kJ}\) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286.20 \mathrm{~kJ}\) The value of enthalpy of formation of \(\mathrm{OH}^{-}\)ion at \(25^{\circ} \mathrm{C}\) is :

JEE Main · 2009 · 12 Apr 2009 · Offline · Chemical Thermodynamics

On the basis of the following thermochemical data : (

Δ fG^{\circ} H_{(aq)}^{+} = 0

)

H_{2} O (ℓ) \to H^{+} (aq) + OH^{-} (aq); 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

OH^{-}

ion at

2 5^{\circ} C

is :

(A)-22.88 kJ
(B)-228.88 kJ✓ Correct
(C)+228.88 kJ
(D)-343.52 kJ

Answer: (B)

Solution

Adding the two thermochemical equations cancels the liquid water and leaves

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H^{+} (aq) + O H^{-} (aq)

with

Δ H = - 286.20 + 57.32 = - 228.88 kJ

. Since the enthalpy of formation of the aqueous hydrogen ion is taken as zero by convention, this whole value belongs to the hydroxide ion, so its enthalpy of formation is

- 228.88 kJ mo l^{- 1}

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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JEE Main · 2009 · 12 Apr 2009 · Offline · Chemical Thermodynamics

Solution

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