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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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On the basis of the following thermo chemical data $ [ \Delta G_f^{\circ} (H^+ )(aq) = 0] $ \[\] $H_2O(l) \rightarrow H^+ (aq) + OH^-(aq);\: \: \: \: \: \Delta H = 57.32\: kJ $ \[\] $H_2(g) + \large\frac{1}{2} (g) \rightarrow H_2O(l); \: \: \: \: \: \: \Delta H = -286\: kJ$ \[\] The value of enthalpy of formation of $OH^-\: ion $ at $25^{\circ}C$ is

$\begin {array} {1 1} (A)\;-22.88\; kJ & \quad (B)\;-228.88\: kJ \\ (C)\;228.88 \: kJ & \quad (D)\;-343.52\: kJ \end {array}$

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For heat of formation of $ H_2O,$
$\Delta H_r = \Delta H_f (H_2O,l) - \Delta H_f\: (H_2,g) – \large\frac{1}{2} $$ \Delta H_f(O_2,g) – 286.2 = -286.2\: kJ$
For ionization of $ H_2O,$
$ \Delta H_r = \Delta H_f(H^+, aq) + \Delta H_f ( OH^-, aq) - \Delta H_f (H_2, O, l)$
$ \Rightarrow 57.32 = 0 + \Delta H_f (OH^- , aq) – (-286.2)$
Thus, $ \Delta H_f (OH^- , aq) = 57.32 – 286.2 = -228.88\: kJ$
Ans : (B)
answered Mar 17, 2014 by thanvigandhi_1
 

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