Question

The heat of combustion of benzene at ${\text{2}}{{\text{7}}^{\text{0}}}{\text{C}}$ found by bomb calorimeter i.e. for the reaction ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}} + 7\dfrac{1}{2}{{\text{O}}_2} \to 6{\text{C}}{{\text{O}}_2} + 3{{\text{H}}_{\text{2}}}{\text{O}}$ is $780{\text{K}}{\text{.Cal mo}}{{\text{l}}^{{\text{ - 1}}}}$. The heat evolved on burning $39{\text{g}}$ of benzene in an open vessel will be
A.$390$ ${\text{K}}{\text{.Cal}}$
B.$780.9$ ${\text{K}}{\text{.Cal}}$
C.$390.45$ ${\text{K}}{\text{.Cal}}$
D.$780$ ${\text{K}}{\text{.Cal}}$

Answer 1

The heat of combustion can be defined as the amount of heat released for the complete combustion of a compound in its standard state to form standard products in their stable states.
Formula Used: ${{\Delta H = \Delta U + \Delta nRT}}$
H is enthalpy, U is internal energy, n is the moles, R is a constant, T is the temperature.

Complete step by step answer:
The equation for the combustion of benzene is:
${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}} + 7\dfrac{1}{2}{{\text{O}}_2} \to 6{\text{C}}{{\text{O}}_2} + 3{{\text{H}}_{\text{2}}}{\text{O}}$
Here the change in the number of moles for the reaction = no. of moles of the product - no. of moles of the reactant = $6-7.5$= $- 1.5$
According to the question, the heat of combustion for benzene = $- 780{\text{K}}{\text{.Cal mo}}{{\text{l}}^{{\text{ - 1}}}}$. This is the change in the internal energy of the reaction.
Now, the mathematical formula for the heat of combustion, $\Delta H = \Delta U + \Delta \left( {PV} \right)$
From the ideal gas law equation, ${\text{P V = nRT}}$
Hence, $\Delta H = \Delta U + \Delta nRT$
Substituting the values:
$\Delta H = - 780 + \left( {\dfrac{{ - 1.5 \times 2 \times 300}}{{1000}}} \right)$
$\Delta H = - 780 - 0.9 = - 780.9$ ${\text{K}}{\text{.Cal}}$
Since the molecular weight of benzene is \left[ {\left( {6 \times 12} \right) + \left( {6 \times 1} \right)} \right] = 76$$${\text{g/mol}}$$and this value of heat is liberated for one mole of benzene which is equal to its molecular weight 76$${\text{g/mol}}$$. Therefore, the amount of heat liberated for 39{\text{g}}$of benzene will be$ - \dfrac{{780.9}}{2}$=$ - 390.45$${\text{K}}{\text{.Cal}}$.

Hence, the correct answer is option C.
Note:
The calorific value of a substance is equal to the heat released when a substance undergoes complete combustion with oxygen under standard conditions. There are different units of expressing the calorific value: the energy/mole of the fuel, the energy/ mass of the fuel, the energy/ volume of the fuel. The values are generally measured using the Bomb Calorimeter.