H_{2(g)} + frac{1}{2} O_{2(g)} to H_2O_{(g)}B | vedclass

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(C) The heat of formation of liquid water is given by the reaction: $H_{2(g)} + \frac{1}{2} O_{2(g)} 	o H_2O_{(l)}$.Using bond energies,the enthalpy

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$x_1 + \frac{x_2}{2} - 2x_3 - x_4$

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(C) The heat of formation of liquid water is given by the reaction: $H_{2(g)} + \frac{1}{2} O_{2(g)} 	o H_2O_{(l)}$.Using bond energies,the enthalpy change for the formation of gaseous water is: $\Delta H_{g} = \sum (BE)_{	ext{reactants}} - \sum (BE)_{	ext{products}}$.$\Delta H_{g} = [x_1 + \frac{1}{2} x_2] - [2x_3]$.Since we need the heat of formation for liquid water,we subtract the heat of vaporization $(x_4)$ from the enthalpy of formation of gaseous water: $\Delta H_f = \Delta H_{g} - x_4$.Therefore,$\Delta H_f = x_1 + \frac{x_2}{2} - 2x_3 - x_4$.

$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:

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$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;$B.E. (O-H) = x_3$Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is: