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The heat of formation of liquid methyl alcohol in kilo joules per mole is found to be $- (260 + X) k J / m o l,$ using the following data. Heat of vaporization of liquid methyl alcohol = 38 kJ/mol. Heat of formation of gaseous atoms from the elements in their standard states : $H : 218 k J / m o l; C : 715 k J / m o l; O : 249 k J / m o l .$ Average bond energies :
$E_{C - H} = 415 k J / m o l$
$E_{C - O} = 356 k J / m o l$
$E_{O - H} = 463 k J / m o l$
Here ‘X’ is :

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answer is 6.

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Detailed Solution

$C (s) + 2 H_{2} (g) + \frac{1}{2} O_{2} (g) \to C H_{3} O H (l); Δ H = ?$

$Δ H$ = Bond energy data for formation + Bond energy data for dissociation + Energy released during liquefaction

$= - [3 \times e_{C - H} + e_{C - O} + e_{O - H}] + [C_{s \to g} + 2 \times e_{H - H} + \frac{1}{2} \times e_{O - O}] - [C H_{3} O H_{g \to l}]$

$= - [3 \times 415 + 356 + 463] + [715 + 2 \times 436 + 249] - 38 = - 266 k J m o l^{- 1}$

$- (260 + \underset{X}{6}) kJ / mol = - 266 kJ / mol$

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