આપેલ:

$(I)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);$

$\Delta {H^o_{298\,K}} = - 285.9\,kJ\,mo{l^{ - 1}}$

$(II)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);$

$\Delta {H^o_{298\,K}} = - 241.8\,kJ\,mo{l^{ - 1}}$

તો પાણીની મોલર બાષ્પાયન એન્થાલ્પી .....$kJ\,mol^{-1}$

Question

આપેલ:

$(I)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);$

$\Delta {H^o_{298\,K}} = - 285.9\,kJ\,mo{l^{ - 1}}$

$(II)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);$

$\Delta {H^o_{298\,K}} = - 241.8\,kJ\,mo{l^{ - 1}}$

તો પાણીની મોલર બાષ્પાયન એન્થાલ્પી .....$kJ\,mol^{-1}$

JEE MAIN 2013, Diffcult

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Solution

c
Given

${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);$

$\Delta {H^o} = - 285.9\,kJ\,mo{l^{ - 1}}........(1)$

${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);$

$\Delta {H^o}= - 241.8\,kJ\,mo{l^{ - 1}}........(2)$

We have to calculate

${H_2}O(l) \to {H_2}O(g);\,\Delta {H^o} = ?$

On substracting eqn. $(2)$ from eqn. $(1)$ we get

${H_2}O(l) \to {H_2}O(g);$

$\Delta {H^o} = - 241.8 - ( - 285.9)$

$ = 44.1\,kJ\,mo{l^{ - 1}}$

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