
The combustion of benzene (I) gives \[C{{O}_{2}}\](g) and \[{{H}_{2}}O\](l). given that heat of combustion of benzene at constant volume is – 3263.9$kJmo{l^{ - 1}}$ at${25^o}C$; heat of combustion (in$kJmo{l^{ - 1}}$) of benzene at constant pressure will be : (R= 8.314$J{K^{ - 1}}mo{l^{ - 1}}$)
A)415.6
B)-452.46
C)3260.2
D)-3267.6
Answer
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Hint: The heat of combustion is defined as the amount of the heat which has been liberated when a certain amount of the substance undergoes the process of the combustion. The other name of the heat of combustion is known as the calorific value and energy value. In the typical combustion reaction we observe that the carbon containing compound with oxygen tends to form carbon dioxide and water as that of the products.
Complete step-by-step answer: When benzene is being reacted with oxygen and the process of combustion takes place the formation of carbon dioxide and water takes place. So the reaction of this combustion is the following:
${C_6}{H_6}(l) + \dfrac{{15}}{2}{O_2}(g) \to 6C{O_2}(g) + 3{H_2}O(l)$
So let us find the change in the gaseous moles that is $\Delta {n_g}$:
$\Delta {n_g}$=6-7.5= -1.5
The heat of combustion of benzene or $\Delta U$or$\Delta E$is equal to – 3263.9$kJmo{l^{ - 1}}$
The formula which we will be using here is the following:
$\Delta H = \Delta U + \Delta {n_g}RT$
So the values given to us are as follows:
$\Delta {n_g}$=-1.5
$\Delta U$= – 3263.9$kJmo{l^{ - 1}}$
R= 8.314$J{K^{ - 1}}mo{l^{ - 1}}$
T= 298K
So on substituting the values in formula we get,
$\Delta H = - 3263.9 + ( - 1.5) \times 8.314 \times {10^{ - 3}} \times 298 = - 3267.6kJ$
So the value of heat of combustion of benzene at the constant pressure is equal to the $ - 3267.6kJ$ ,so the correct option is options d.
Note: The process of combustion or burning at the high temperature is known as the exothermic redox chemical reaction. The combustion generally takes place between the fuel which is the reductant and the oxidant is the oxygen. This tends to produce the gaseous product in a mixture known as the smoke.
Complete step-by-step answer: When benzene is being reacted with oxygen and the process of combustion takes place the formation of carbon dioxide and water takes place. So the reaction of this combustion is the following:
${C_6}{H_6}(l) + \dfrac{{15}}{2}{O_2}(g) \to 6C{O_2}(g) + 3{H_2}O(l)$
So let us find the change in the gaseous moles that is $\Delta {n_g}$:
$\Delta {n_g}$=6-7.5= -1.5
The heat of combustion of benzene or $\Delta U$or$\Delta E$is equal to – 3263.9$kJmo{l^{ - 1}}$
The formula which we will be using here is the following:
$\Delta H = \Delta U + \Delta {n_g}RT$
So the values given to us are as follows:
$\Delta {n_g}$=-1.5
$\Delta U$= – 3263.9$kJmo{l^{ - 1}}$
R= 8.314$J{K^{ - 1}}mo{l^{ - 1}}$
T= 298K
So on substituting the values in formula we get,
$\Delta H = - 3263.9 + ( - 1.5) \times 8.314 \times {10^{ - 3}} \times 298 = - 3267.6kJ$
So the value of heat of combustion of benzene at the constant pressure is equal to the $ - 3267.6kJ$ ,so the correct option is options d.
Note: The process of combustion or burning at the high temperature is known as the exothermic redox chemical reaction. The combustion generally takes place between the fuel which is the reductant and the oxidant is the oxygen. This tends to produce the gaseous product in a mixture known as the smoke.
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