Question

The heat liberated on complete combustion of \[7.8g\] benzene in\[327KJ\]. This heat has been measured at constant volume and at \[{27^ \circ }C\]. Calculate the heat combustion of benzene at constant pressure.
(A) \[ - 323.7KJmo{l^{ - 1}}\]
(B) \[ - 3273.7KJmo{l^{ - 1}}\]
(C) \[273.7KJmo{l^{ - 1}}\]
(D) \[3273.7KJmo{l^{ - 1}}\]

Answer 1

Hint: Heat of combustion is represented as delta U. First find the heat liberated by using proper reaction and quantities then by putting values in the equation of first law of thermodynamics, we can get the correct answer.

Complete step-by-step solution:The heating value is the amount of heat released during the combustion of a specified amount of it.
Molecular weight of \[{C_6}{H_6} = 78g\]
Heat liberated on combustion of \[7.8g\] of benzene = \[327KJ\]
Heat liberated on combustion of \[78g\] benzene
\[\dfrac{{327}}{{7.8}} \times 78 = 3270KJ\]
\[\therefore \Delta E = - 3270KJ\]
The complete reaction is as follow:-
\[{C_6}{H_6} + 7{\dfrac{1}{2}}{O_2} \to 6C{O_2} + 3{H_2}O\]
From above reaction,
\[\Delta {n_g} = \Delta {n_P} - \Delta {n_R}\]
$\Rightarrow \Delta {n_g}= 6-\dfrac{15}{2}= \dfrac{-3}{2}$
Temperature (T) \[ = {27^ \circ }C = (25 + 273)K = 298K\]
On further solving,
From first law of thermodynamics-
\[\Delta H = \Delta E + \Delta {n_g}RT\]
\[\Delta H = \Delta E + (\dfrac{{ - 3}}{2}) \times 8.314 \times {10^{ - 3}} \times 300\]
\[\Delta H = - 3270 - 3.74 = - 3273.74KJ/mol \approx - 3273.7KJmo{l^{ - 1}}\]

Hence, the correct answer is option (B) that is \[ - 3273.7KJmo{l^{ - 1}}\]

Note:Heat of combustion is also known as calorific value or energy value. Combustion is always exothermic because heat is released during combustion, the enthalpy change for the reaction is negative, \[\Delta H\] has a negative sign which indicates that energy is liberated. In the endothermic heat is absorbed by the system which is just opposite to the exothermic reaction.