Question
For complete combustion of ethanol,
$${C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right) \to $$ $$2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right),$$
the amount of heat produced as measured in bomb calorimeter, is $$1364.47\,kJ\,mo{l^{ - 1}}$$ at $${25^ \circ }C.$$ Assuming ideality the enthalpy of combustion, $${\Delta _c}H,$$ for the reaction will be :
$$\left( {R = 8.314\,kJ\,mo{l^{ - 1}}} \right)$$
A.
$$ - 1366.95\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 1361.95\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 1460.95\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 1350.50\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 1366.95\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right) \cr
& {\text{Bomb calorimeter gives}}\,\,\Delta U\,{\text{of}}\,{\text{the}}\,{\text{reaction}} \cr
& {\text{Given, }}\Delta U = {\text{ - }}1364.47\,kJ\,mo{l^{ - 1}} \cr
& \Delta {n_g} = - 1 \cr
& \Delta H = \Delta U + \Delta {n_g}RT \cr
& = - 1364.47 - \frac{{1 \times 8.314 \times 298}}{{1000}} \cr
& = - 1366.93\,kJ\,mo{l^{ - 1}} \cr} $$