Consider the following data:Delta_{f}H^{Theta | vedclass

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(A) The thermochemical equation for the formation of methane is: $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$The enthalpy of formation is given by: $\D

Vedclass · Accepted Answer

$\frac{X+Y+2Z}{4}$

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(A) The thermochemical equation for the formation of methane is: $C_{(s)} + 2H_{2(g)} ightarrow CH_{4(g)}$The enthalpy of formation is given by: $\Delta_{f}H^{\Theta} = \sum 	ext{Bond Enthalpies of Reactants} - \sum 	ext{Bond Enthalpies of Products}$$-X = [\Delta H_{sub}(C) + 2 	imes B.E.(H-H)] - [4 	imes B.E.(C-H)]$Substituting the given values: $-X = Y + 2Z - 4 	imes B.E.(C-H)$Rearranging for $B.E.(C-H)$: $4 	imes B.E.(C-H) = X + Y + 2Z$$B.E.(C-H) = \frac{X+Y+2Z}{4}$

Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by:

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Consider the following data:$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$The bond enthalpy of $C-H$ bond is given by: