Question

For complete combustion of ethanol:

$C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2{CO}_2\left(g\right)+3H_{2}O\left(l\right)$,

the amount of heat produced as measured in bomb calorimeter is $1364.47kJ{mol}^{-1}$ at ${25}^{o}C$.

Assuming ideality, the enthalpy of combustion, ${\Delta }_{C}H$ for the reaction will be:

$[R=8.314J{K}^{-1}{mol}^{-1}]$

• A. $-1460.50kJ{mol}^{-1}$
• B. $-1350.50kJ{mol}^{-1}$
• C. $-1366.95kJ{mol}^{-1}$
• D. $-1361.96kJ{mol}^{-1}$

Answer 1

Answer: (C)

$-1366.95kJ{mol}^{-1}$

$C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)⟶2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

enthalpy of combustion, ${\Delta }_{C}H$

${\Delta }_{C}H=\Delta U_C+(\Delta n{)}_{g}RT$

Where $\Delta U_C\to$ heat produced.

$(\Delta n{)}_g\to$ change in number of moles of gaseous molecules.

$T={25}^{o}C=273+25=298K$

$\therefore {\Delta }_{C}H=-1364.47KJ+(-1)\times 8.314\times {10}^{-3}\times 298K$

$=-(1364.47kJ+2.48kJ)$

$=-1366.95kJ/mol$

$\therefore$ Enthalpy of combustion for ethanol.

${\Delta }_{C}H=-1366.95KJ/mol$

Hence, the correct option is $\text{C}$