JEE MAIN - Chemistry (2022 - 24th June Evening Shift - No. 4)
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
+54.0 kJ mol$$-$$1
$$-$$68.0 kJ mol$$-$$1
$$-$$86.0 kJ mol$$-$$1
+97.0 kJ mol$$-$$1
Explanation
$$2 \mathrm{C}$$ (graphite) $$+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})$$
$$ \Delta \mathrm{H}_{\mathrm{r}}=+1560+2(-394)+3(-286) $$
$$=-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1} $$
Enthalpy of formation of C2H6(g) = –86.0 kJ mol–1
$$ \Delta \mathrm{H}_{\mathrm{r}}=+1560+2(-394)+3(-286) $$
$$=-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1} $$
Enthalpy of formation of C2H6(g) = –86.0 kJ mol–1
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