What is the value of $\Delta \mathrm{H}^{\circ}$ for the formation of ethanol from ethene gas and liquid…
What is the value of $\Delta \mathrm{H}^{\circ}$ for the formation of ethanol from ethene gas and liquid water from following data?
(i) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(1)}+3 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}^{\circ}=-1368 \mathrm{~kJ}$
(ii) $\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \Delta \mathrm{H}^{\circ}=-1410 \mathrm{~kJ}$
$-1326 \cdot 0 \mathrm{~kJ}$
$-4188 \cdot 0 \mathrm{~kJ}$
$-42 \cdot 0 \mathrm{~kJ}$
$-2778 \cdot 0 \mathrm{~kJ}$
Solution
The standard enthalpy of combustion of \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(I)}\) i.e. \(\Delta \mathrm{c} \mathrm{H}^{\circ} \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)}=-1368 \mathrm{~kJ}\)
The standard enthalpy of combustion of ethene i.e. \(\Delta \mathrm{c} \mathrm{H}^{\circ} \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}=-1410 \mathrm{~kJ}\)
To find: \(\Delta H^{\circ}\) for the enthalpy of formation of liquid \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
Calculation: Given equations are,
\(\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(\mathrm{l})}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} ; \Delta \mathrm{cH}^{\circ}=-1368 \mathrm{~kJ} \ldots(\mathrm{C} \\
& \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})}-2 \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} ; \Delta \mathrm{cH}=-1410 \mathrm{~kJ} \ldots(2)
\end{aligned}\)
The required equation is,
\(\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(\mathrm{l})}\)
To get required equation, reverse equation (1) and add to equation (2).
\(\begin{aligned}
& 2 \mathrm{CO}_{2(l)}+3 \mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)}+3 \mathrm{O}_{2(g)}=+1368 \mathrm{~kJ} \\
& \mathrm{C}_2 \mathrm{H}_{4(g)}+3 \mathrm{O}_{2(g)} \longrightarrow 2 \mathrm{CO}_{2(g)}+2 \mathrm{H}_2 \mathrm{O}_{(l)} \quad \Delta \mathrm{H}^{\circ}=-1410 \mathrm{~kJ} \\
& \mathrm{C}_2 \mathrm{H}_{4(g)}+\mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \quad \Delta \mathrm{H}^{\circ}=+1368 \mathrm{~kJ}-1410 \mathrm{~kJ} \\
& =-42 \mathrm{~kJ} \text {. }
\end{aligned}\)
The calculated \(\Delta H^{\circ}=-42 \mathrm{~kJ}\) is not the enthalpy of formation of liquid ethanol because the reaction does not involve the formation of liquid ethanol from its constituent elements.