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Compute the heat of formation of liquid methanol in kilojoules per mole, using the given data:  $\Delta H_{vap}$ of liquid methanol $= -38\: kJ/mol, \Delta H_f^{\circ}\: of\: H(g), C(g), O(g)$ respectively are $218, 715\: and\: 247\: kJ/mol$. Also, $\Delta H_{C-H}, \Delta H_{C-O}, \Delta H_{O-H}$ being $415, 356, 463 \: kJ/mol$ respectively.

$\begin {array} {1 1} (A)\;228\: kJ/mol & \quad (B)\;-228\: kJ/mol \\ (C)\;266 \: kJ/mol & \quad (D)\;-266 \: kJ/mol \end {array}$

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$C(graphite) + 2H_2(g) + \large\frac{1}{2}$$O_2(g) \rightarrow CH_3OH(g) \rightarrow CH_3OH(l) \Delta H = [ \Delta H_fC(g) + 2\Delta H_{H-H} + \large\frac{1}{2}$$ \Delta H_{O=O}] – [\Delta H_{C-O} + 3\Delta H_{C-H} + \Delta H_{O-H}] - \Delta H_{cond}(CH_3OH)$
$= [\Delta H_fC(g) + 4\Delta H_f(H) + \Delta H_f(O)] – [\Delta H_{C-O} 3\Delta H_{C-H} + \Delta H_{O-H}] - \Delta H_{cond}(CH_3OH)$
$= (715 + 4 \times 18 + 249) – (356 + 3 \times 415 + 463) – 38 = -266\: kJ/mol$
Ans : (D)
answered Mar 19, 2014

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