At 25$^\circ$C and 1 atm pressure, the enthalpies of combustion are as given below :
| Substance | ${H_2}$ | C (graphite) | ${C_2}{H_6}(g)$ |
|---|---|---|---|
| ${{{\Delta _c}{H^\Theta }} \over {kJ\,mo{l^{ - 1}}}}$ | $- 286.0$ | $- 394.0$ | $- 1560.0$ |
The enthalpy of formation of ethane is
Solution
$2 \mathrm{C}$ (graphite) $$+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})$$<br/><br/>
$\Delta \mathrm{H}_{\mathrm{r}}=+1560+2(-394)+3(-286)$<br/><br/>
$=-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$<br/><br/>
Enthalpy of formation of C<sub>2</sub>H<sub>6</sub>(g) = –86.0 kJ mol–1
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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