Consider the reactions:C_{(s)} + 2H_{2(g)} to | vedclass

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(D) The bond energy of a $C-H$ bond is defined as the energy required to break one mole of $C-H$ bonds in a gaseous molecule.From the second reaction:

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$x_1/4 \ kcal \ mol^{-1}$

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(D) The bond energy of a $C-H$ bond is defined as the energy required to break one mole of $C-H$ bonds in a gaseous molecule.From the second reaction: $C_{(g)} + 4H_{(g)} 	o CH_{4(g)}, \Delta H = -x_1 \ kcal$.This represents the formation of $4$ moles of $C-H$ bonds from gaseous atoms.Therefore,the energy released for $4$ moles of $C-H$ bonds is $x_1 \ kcal$.The bond energy of one $C-H$ bond is the average energy required to break one mole of $C-H$ bonds,which is $\frac{x_1}{4} \ kcal \ mol^{-1}$.Thus,the correct option is $D$.

Consider the reactions:
$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}, \Delta H = -x \ kcal$
$C_{(g)} + 4H_{(g)} \to CH_{4(g)}, \Delta H = -x_1 \ kcal$
$CH_{4(g)} \to CH_{3(g)} + H_{(g)}, \Delta H = +y \ kcal$
The bond energy of $C-H$ bond is:

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Consider the reactions:$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}, \Delta H = -x \ kcal$$C_{(g)} + 4H_{(g)} \to CH_{4(g)}, \Delta H = -x_1 \ kcal$$CH_{4(g)} \to CH_{3(g)} + H_{(g)}, \Delta H = +y \ kcal$The bond energy of $C-H$ bond is: