
the combustion of benzene(l) gives $C{O_2}$(g) and ${H_2}O$ (l). Given that heat of combustion of benzene at constant volume is $-3263.9kJmo{l^{ - 1}}$ at ${25^\circ}C$ ; heat of combustion(in $kJmo{l^{ - 1}}$) of benzene constant pressure will be:
($R= 8.314J{K^{ - 1}}mo{l^{ - 1}}$)
A. 415.6
B. -452.46
C. 3260.2
D. -3265.9
Answer
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Hint: The heat of combustion is defined as the amount of heat which is liberated when a given amount of substance undergoes the process of combustion. It is also known as energy value or the calorific value. If the calorific value is high then the value of heating would also be high. The heat of combustion can be calculated in joules and kilojoules.
Complete answer:
The reaction which takes place benzene and oxygen during combustion is given below:
${C_6}{H_6}(l) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l)$
So we see that the carbon dioxide and water is formed on combustion. Now after balancing the above equation we get,
${C_6}{H_6}(l) + {\dfrac{15}{2}} {O_2}(g) \to 6C{O_2}(g) + 3{H_2}O(l)$
So the formula for the heat of combustion is the following:
$\Delta H = \Delta U + \Delta ngRT$
The $\Delta ng$ is the subtraction of the moles of gases of the product and reactants
$\Delta U$ is the heat of combustion of the benzene in this question
R is $8.314J{K^{ - 1}} mo{l^{ - 1}}$ and
T is ${25^\circ}$C, but we need to convert it into kelvin so
T = ${25^\circ}$C + 273= 298K
So now we will calculate $\Delta ng$ so
$\Delta ng$ = 6 - $\dfrac{{15}}{2}$ = -$\dfrac{3}{2}$
So now substituting the values in the formula we get,
$\Delta H = - 3263.9 - \dfrac{3}{2} \times 8.314 \times 298 = - 3265.9KJmo{l^{ - 1}}$
So the correct answer is option D.
Note: The calorific value reflects the amount of energy which has been emitted when the fuel undergoes the complete combustion process. The high calorific value is a term which denotes the amount of the heat or the amount of energy which has been emitted from the fuel burning when its by products are concentrated in form to recover some of the energy generated from burning. Whereas the low calorific value is referred to the amount of energy which is emitted from the fuels combustion when its by product are able to escape freely from the reaction.
Complete answer:
The reaction which takes place benzene and oxygen during combustion is given below:
${C_6}{H_6}(l) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l)$
So we see that the carbon dioxide and water is formed on combustion. Now after balancing the above equation we get,
${C_6}{H_6}(l) + {\dfrac{15}{2}} {O_2}(g) \to 6C{O_2}(g) + 3{H_2}O(l)$
So the formula for the heat of combustion is the following:
$\Delta H = \Delta U + \Delta ngRT$
The $\Delta ng$ is the subtraction of the moles of gases of the product and reactants
$\Delta U$ is the heat of combustion of the benzene in this question
R is $8.314J{K^{ - 1}} mo{l^{ - 1}}$ and
T is ${25^\circ}$C, but we need to convert it into kelvin so
T = ${25^\circ}$C + 273= 298K
So now we will calculate $\Delta ng$ so
$\Delta ng$ = 6 - $\dfrac{{15}}{2}$ = -$\dfrac{3}{2}$
So now substituting the values in the formula we get,
$\Delta H = - 3263.9 - \dfrac{3}{2} \times 8.314 \times 298 = - 3265.9KJmo{l^{ - 1}}$
So the correct answer is option D.
Note: The calorific value reflects the amount of energy which has been emitted when the fuel undergoes the complete combustion process. The high calorific value is a term which denotes the amount of the heat or the amount of energy which has been emitted from the fuel burning when its by products are concentrated in form to recover some of the energy generated from burning. Whereas the low calorific value is referred to the amount of energy which is emitted from the fuels combustion when its by product are able to escape freely from the reaction.
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