At 25^{circ} C and 1 atm pressure,the enthal | vedclass

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(C) The formation reaction of ethane is: $2C_{	ext{(graphite)}} + 3H_{2(g)} ightarrow C_{2}H_{6(g)}$The enthalpy of formation is given by: $\Delta_

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$-86.0 \ kJ \ mol^{-1}$

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(C) The formation reaction of ethane is: $2C_{	ext{(graphite)}} + 3H_{2(g)} ightarrow C_{2}H_{6(g)}$The enthalpy of formation is given by: $\Delta_{f}H^{\Theta}(C_{2}H_{6}) = [2 	imes \Delta_{c}H^{\Theta}(C) + 3 	imes \Delta_{c}H^{\Theta}(H_{2})] - \Delta_{c}H^{\Theta}(C_{2}H_{6})$Substituting the given values:$\Delta_{f}H^{\Theta}(C_{2}H_{6}) = [2 	imes (-394.0) + 3 	imes (-286.0)] - (-1560.0)$$\Delta_{f}H^{\Theta}(C_{2}H_{6}) = [-788.0 - 858.0] + 1560.0$$\Delta_{f}H^{\Theta}(C_{2}H_{6}) = -1646.0 + 1560.0 = -86.0 \ kJ \ mol^{-1}$

Substance	$H_{2(g)}$	$C_{\text{(graphite)}}$	$C_{2}H_{6(g)}$
$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$	$-286.0$	$-394.0$	$-1560.0$

At $25^{\circ} C$ and $1 \ atm$ pressure,the enthalpies of combustion are as given below:

Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$

$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$

The enthalpy of formation of ethane is ........ .

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At $25^{\circ} C$ and $1 \ atm$ pressure,the enthalpies of combustion are as given below: Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$ $\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$ The enthalpy of formation of ethane is ........ .