Solveeit Logo
Login

Question

Question: The difference between heat of reaction at constant pressure and constant volume for the reaction: ...

The difference between heat of reaction at constant pressure and constant volume for the reaction:
2C6H6[l]+15O2[g]→12CO2[g]+6H2O[l]{\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 - 7.43
B. +3.72 + 3.72
C. −3.72 - 3.72
D. +7.43 + 7.43

Explanation

Solution

At constant pressure, the heat of reaction is equal to the change enthalpy of the system ΔH{\rm{\Delta H}}. At constant volume, the heat of reaction is equal to the change in internal energy of the system ΔU{\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:
ΔH=ΔU+[Δng]RT{\rm{\Delta H}} = {\rm{\Delta U}} + {\rm{[\Delta }}{{\rm{n}}_{\rm{g}}}{\rm{]RT}} [Eq. 1]
Where ΔH{\rm{\Delta H}} is change in enthalpy of the system, ΔU{\rm{\Delta U}} is change in internal energy of the system, Δng{\rm{\Delta }}{{\rm{n}}_{\rm{g}}} is the change in number of gaseous moles in the reaction, R is a constant [8.314kJmol−1]\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 ΔH{\rm{\Delta H}} and the heat of reaction at constant volume is the change in internal energy of the system ΔU{\rm{\Delta U}} . The Eq. 1 given can be rewritten as:
ΔH−ΔU=[Δng]RT{\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 O2{{\rm{O}}_{\rm{2}}} and CO2{\rm{C}}{{\rm{O}}_{\rm{2}}} .
⇒Δng=nCO2−nO2\Rightarrow {\rm{\Delta }}{{\rm{n}}_{\rm{g}}}{\rm{ = }}{{\rm{n}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} - {{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}
⇒Δng=12−15=−3\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. T=298.15K{\rm{T}} = 298.15{\rm{ K}}
Substituting these values in the equation, we get:
ΔH−ΔU=[−3]×8.314×298.15J{\rm{\Delta H}} - {\rm{\Delta U = }}\left[ { - 3} \right] \times 8.314 \times 298.15{\rm{ J}}
Solving this, we get:
ΔH−ΔU=−7436.4J{\rm{\Delta H}} - {\rm{\Delta U = }} - 7436.4{\rm{ J}}
⇒ΔH−ΔU=−7.43kJ\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 - 7.43 .

∴\therefore The correct option is option A, i.e. −7.43 - 7.43.

Note:
In the question, when temperature is not given, we assume it to be 298.15K298.15{\rm{K}}. Sometimes, it is given that the reaction is occurring at STP, in this case the temperature will be 273.15K273.15{\rm{K}}. We must be careful not to get confused between these two values. Also, Δng{\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 Δng{\rm{\Delta }}{{\rm{n}}_{\rm{g}}} cannot be ignored as it is used in the calculation.