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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Calculate the bond energy of $ C-H$ bond in methane from the following data: \[\] $C(s) + 2H_2(g) \rightarrow CH_4(g);\: \: \: \Delta H = -74.8\: kJ$ \[\] $ H_2(g)\rightarrow 2H(g);\Delta H = 435.4\: kJ$ \[\] $ C(s) \rightarrow C(g);\Delta H = 718.4 \: kJ$

$\begin {array} {1 1} (A)\;1664\: kJ/mol & \quad (B)\;1514.4\: kJ/mol \\ (C)\;416\: kJ/mol & \quad (D)\;378.6\: kJ/mol \end {array}$

1 Answer

$ \Delta H_f\: of\: CH_4 = -74.8\: kJ/mol\: \Delta H_f\: of\: H = \large\frac{435.4}{2}$$ = 217.7 \: kJ/mol$
$\Delta H_f\: of \: C = 718.4\: kJ/mol$
Now, $ \Delta H = \sum \Delta H_f^{\circ}(products) - \sum \Delta H_f^{\circ}o(reactants)$
$ = (4\times217.7 + 718.4) – (-74.8) = 1664\: kJ$
$ \Delta H_{C-H} = \large\frac{Total\: energy \: required}{No.\: of\: C-H \: broken }$$ = \large\frac{1664}{4}$$ = 416\: kJ/mol$
Ans : (C)
answered Mar 19, 2014 by thanvigandhi_1
 

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