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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Heat of combustion of $ CH_4(g), \: C(graphite),\: H_2(g)$ are -20 kcal, -40 kcal and -10 kcal respectively. The heat of formation of methane is

$\begin {array} {1 1} (A)\;-40 \: kcal/mol & \quad (B)\;40\: kcal/mol \\ (C)\;-80\: kcal/mol & \quad (D)\;80\: kcal/mol \end {array}$

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$CH_4(g) + O_2(g) \rightarrow 2H_2O + CO_2; \Delta H = -20\: kcal$
$-20 = \Delta f (CO_2) + 2 \Delta H_f (H_2O) - \Delta H_f (CH_4)$
$ \Rightarrow \Delta H_f (CH_4) = -40 + 2(-10) – (-20) = -40\: kcal/mol$
Ans : (A)
answered Mar 17, 2014 by thanvigandhi_1
 

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