Question

The difference between ∆H and ∆E on a molar basis for the combustion of n-octane at $ {25^ \circ }C $ would be:
A. $ - 13.6kJ $
B. $ - 1.14kJ $
C. $ - 11.15kJ $
D. $ + 11.15kJ $

Answer 1

Hint: We have to first write the balanced equation of the reaction of combustion of n-octane. Then, the number of moles is calculated from the difference of moles in product side and reactant side. The difference between $ \Delta H $ and $ \Delta E $ is calculated by the product of this change in number of moles, universal gas constant and temperature.

Complete Step by step answer:
Octane is a hydrocarbon and an alkane with the chemical formula $ {C_8}{H_{18}} $. It is a colourless liquid with a smell of gasoline. Combustion of n-octane is given in this question i.e., reaction of octane with oxygen. So, first we’ll write this reaction in an equation form.
 $ {C_8}{H_{18}}\left( l \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right) + {H_2}O\left( g \right) $ where l represents liquid state and g represents gaseous state.

Now, we’ll balance this equation,
 $ {C_8}{H_{18}}\left( l \right) + \dfrac{{25}}{2}{O_2}\left( g \right) \to 8C{O_2}\left( g \right) + 9{H_2}O\left( g \right) $ as 8-carbons, 18-hydrogens are present initially in the reactant side. So, we have multiplied 8 with $ C{O_2} $ and 9 with $ {H_2}O $ to get carbon and hydrogen balanced. Now, there are 25 atoms of oxygen in the product side. So, we’ll multiply $ {O_2} $ with $ \dfrac{{25}}{2} $ to get 25 as both the two’s would get reduced.
 $ \Delta H $ and $ \Delta E $ represents the change in enthalpy and internal energy of the system. Enthalpy means the sum of the internal energy of a system and the product of the system’s pressure $ \left( P \right) $ and volume $ \left( V \right) $ i.e., $ \Delta H = \Delta E + PV $. From the ideal gas law, we know that $ PV = nRT $ where n is the number of moles, R is universal gas constant and T is temperature. So, we’ll substitute the value of PV in the first equation.
 $ \Delta H = \Delta E + \Delta nRT $ where $ \Delta n $ represents the change in number of moles.
 $ \Rightarrow \Delta H - \Delta E = \Delta nRT $
Now, we’ll calculate $ \Delta n $,
 $ \Delta n $ = \[number{\text{ }}of{\text{ }}moles{\text{ }}in{\text{ }}product{\text{ }}side{\text{ - }}number{\text{ }}of{\text{ }}moles{\text{ }}in{\text{ }}reactant{\text{ }}side\]
 $ \Rightarrow \Delta n $ = $ \left( {8 + 9} \right) - \dfrac{{25}}{2} $ [number of moles of octane is not taken as it is in a liquid state]
 $ \Delta n $ = $ 17 - 12.5 $
 $ \Rightarrow \Delta n $ = $ 4.5 $
Now, we’ll put this value in the above equation,
 $ \Delta H - \Delta E = 4.5 \times 8.314 \times 298 $ [Temperature $ \left( T \right) = 25 + 273i.e.,298K $ ]
 $ \Rightarrow \Delta H - \Delta E = 11149.074J $
 $ \Rightarrow \Delta H - \Delta E = 11.149kJ $ as $ 1kJ = 0.001J $
Or $ \Delta H - \Delta E = + 11.15kJ $ (rounded-off)

Therefore, option D is correct.

Note: Kindly remember that the number of moles of n-octane is not considered in the calculation as it is present in the liquid state and only moles of the compounds which are present in gaseous state are considered. The temperature in the formula of ideal gas law is taken in Kelvin so we have to change the temperature given in $ ^ \circ C $ to K.