JEE MAIN - Chemistry (2014 (Offline) - No. 6)
For complete combustion of ethanol, C2H5OH(l) + 3O2(g) $$\to$$ 2CO2(g) + 3H2O(l) the amount of heat
produced as measured in bomb calorimeter, is 1364.47 kJ mol–1 at 25oC. Assuming ideality the Enthalpy of combustion, $$\Delta _CH$$, for the reaction will be : (R = 8.314 kJ mol–1)
–1460.50 kJ mol–1
– 1350.50 kJ mol–1
– 1366.95 kJ mol–1
– 1361.95 kJ mol–1
Explanation
$${C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right)$$
$$ \to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right)$$
Bomb calorimeter gives $$\Delta U$$ of the reaction
Given, $$\Delta U = - 1364.47\,kJ\,mo{l^{ - 1}}$$
$$\Delta {n_g} = - 1$$
$$\Delta H = \Delta U + \Delta {n_g}RT$$
$$ = - 1364.47 - {{1 \times 8.314 \times 298} \over {1000}}$$
$$ = - 1366.93\,kJ\,mo{l^{ - 1}}$$
$$ \to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right)$$
Bomb calorimeter gives $$\Delta U$$ of the reaction
Given, $$\Delta U = - 1364.47\,kJ\,mo{l^{ - 1}}$$
$$\Delta {n_g} = - 1$$
$$\Delta H = \Delta U + \Delta {n_g}RT$$
$$ = - 1364.47 - {{1 \times 8.314 \times 298} \over {1000}}$$
$$ = - 1366.93\,kJ\,mo{l^{ - 1}}$$
Comments (0)
