Question

The difference between heat of reaction at constant pressure and constant volume for the reaction:
\[{\rm{2}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}\left[ {\rm{l}} \right] + {\rm{15}}{{\rm{O}}_{\rm{2}}}\left[ {\rm{g}} \right] \to {\rm{12C}}{{\rm{O}}_{\rm{2}}}\left[ {\rm{g}} \right] + {\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left[ {\rm{l}} \right]\] in kJ is:
A. \[ - 7.43\]
B. \[ + 3.72\]
C. \[ - 3.72\]
D. \[ + 7.43\]

Answer 1

Hint: At constant pressure, the heat of reaction is equal to the change enthalpy of the system \[{\rm{\Delta H}}\]. At constant volume, the heat of reaction is equal to the change in internal energy of the system \[{\rm{\Delta U}}\]. These two quantities are related to each other by an equation [given below]. We calculate the difference between heat of reaction at constant pressure and constant volume for the given reaction by substituting the appropriate values in the equation.
Formula used:
 \[{\rm{\Delta H}} = {\rm{\Delta U}} + {\rm{[\Delta }}{{\rm{n}}_{\rm{g}}}{\rm{]RT}}\] [Eq. 1]
Where \[{\rm{\Delta H}}\] is change in enthalpy of the system, \[{\rm{\Delta U}}\] is change in internal energy of the system, \[{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}\] is the change in number of gaseous moles in the reaction, R is a constant \[\left[ {{\rm{8}}{\rm{.314 kJ mo}}{{\rm{l}}^{{\rm{ - 1}}}}} \right]\] and T is the temperature of the system.

Complete step by step answer:
The heat of reaction at constant pressure is simply the change in enthalpy of the system \[{\rm{\Delta H}}\] and the heat of reaction at constant volume is the change in internal energy of the system \[{\rm{\Delta U}}\] . The Eq. 1 given can be rewritten as:
\[{\rm{\Delta H}} - {\rm{\Delta U = }}\left[ {{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}} \right]{\rm{RT}}\]
The left hand side of the equation is the difference in heat of reaction at constant pressure and at constant volume. Since, in the above reaction, the only species in gaseous form are \[{{\rm{O}}_{\rm{2}}}\] and \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] .
\[ \Rightarrow {\rm{\Delta }}{{\rm{n}}_{\rm{g}}}{\rm{ = }}{{\rm{n}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} - {{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}\]
\[ \Rightarrow {\rm{\Delta }}{{\rm{n}}_{\rm{g}}} = 12 - 15 = - 3\]
As the temperature is not mentioned in the given question, we shall assume it to be normal room temperature, i.e. \[{\rm{T}} = 298.15{\rm{ K}}\]
Substituting these values in the equation, we get:
\[{\rm{\Delta H}} - {\rm{\Delta U = }}\left[ { - 3} \right] \times 8.314 \times 298.15{\rm{ J}}\]
Solving this, we get:
\[{\rm{\Delta H}} - {\rm{\Delta U = }} - 7436.4{\rm{ J}}\]
\[ \Rightarrow {\rm{\Delta H}} - {\rm{\Delta U = }} - 7.43{\rm{ kJ}}\]
So, the difference between heat of reaction in kJ at constant pressure and constant volume for the reaction is \[ - 7.43\] .

\[\therefore\] The correct option is option A, i.e. \[ - 7.43\].

Note:
In the question, when temperature is not given, we assume it to be \[298.15{\rm{K}}\]. Sometimes, it is given that the reaction is occurring at STP, in this case the temperature will be \[273.15{\rm{K}}\]. We must be careful not to get confused between these two values. Also, \[{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}\] is always the difference between the moles of product and the moles of reactant, the positive and negative sign of \[{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}\] cannot be ignored as it is used in the calculation.