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  4. The enthalpy of combustion of 2 moles of benzene at 27° C differs from the value determined in bomb calorimeter by

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Q. The enthalpy of combustion of $2$ moles of benzene at $27^{\circ} C$ differs from the value determined in bomb calorimeter by

BITSATBITSAT 2013

Solution:

By bomb calorimeter we get $\Delta E$.
$2 C _{6} H _{6}( l )+15 O _{2}( g ) \longrightarrow 12 CO _{2}( g )+6 H _{2} O (l)$
$\Delta H -\Delta E =\Delta nRT$
$=(12-15) \times 8.314 \times 300=-7.483\, kJ$