Question

For complete combustion of ethanol
$C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)⟶2{CO}_2\left(g\right)+3H_{2}O\left(l\right)$
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:
$(R=8.314kJ{mol}^{-1})$

• A. $-1366.95kJ{mol}^{-1}$
• B. $-1361.95kJ{mol}^{-1}$
• C. $-1460.95kJ{mol}^{-1}$
• D. $-1350.50kJ{mol}^{-1}$

Answer 1

The correct option is A $-1366.95kJ{mol}^{-1}$

Heat produced in bomb calorimeter $=-\Delta U=1364.47$ $kJ/mol$

$\therefore \Delta U=-1364.47$ $kJ/mol$

$\Delta H=\Delta U+\Delta n_{g}RT$

Where, $\Delta n_g=$ difference between gaseous meters of reactants and products.

Here, $\Delta n_g=2-3=-1$

$\therefore \Delta H=-1364.47-1\times 8.314\times 298\times {10}^{-3}$

$⟹\Delta H=(-1364.47-2.48)$ $kJ/mol$

$⟹\Delta H=-1366.95$ $kJ/mol$