Question

On the basis of the following thermochemical data : ($\Delta$ ƒGŗH+(aq) = 0)

H2O($\mathcal{l}$)$\longrightarrow$H+ (aq) + OH– (aq.) ; $\Delta$H = 57.32 kJ

H2(g) +$\frac{1}{2}$ O2 (g) $\longrightarrow$H2O ($\mathcal{l}$) ; $\Delta$H = –286.20 kJ

The value of enthalpy of formation of OH– ion at 25⁰C is :

• A. –228.88 kJ
• B. +228.88 kJ
• C. –343.52 kJ
• D. –22.88 kJ

Answer 1

Answer: (A)

–228.88 kJ

Answer 2

H2(g) + $\frac{1}{2}$ O2(g) $\longrightarrow$H2O ($\mathcal{l}$) $\Delta$H = –286.20 kJ

$\Delta$Hr = $\Delta$Hf (H2O, $\mathcal{l}$) –$\Delta$Hf (H2 , g) $- \frac{1}{2}\Delta$Hf (O2 , g)

–286.20 = DHf (H2O ($\mathcal{l}$))

So $\Delta$Hf (H2O, $\mathcal{l}$) = –286.20 KJ/mole

H2O ($\mathcal{l}$)$\longrightarrow$H+ (aq) + OH– (aq) $\Delta$H = 57.32 kJ

$\Delta$Hr = $\Delta$Hŗf (H+, aq) + $\Delta$Hŗf(OH–, aq) – $\Delta$Hŗf (H2O, $\mathcal{l}$)

57.32 = 0 + $\Delta$Hŗf (OH–, aq) – (–286.20)

$\Delta$Hŗf (OH–, aq) = 57.32 – 286.20 = –228.88 kJ.