Solveeit Logo
Login

Question

Question: Standard free energies of formation \(\left( {{\rm{in}}\;{\rm{kJ/mol}}} \right)\) at \({\rm{298}}\;{...

Standard free energies of formation (in  kJ/mol)\left( {{\rm{in}}\;{\rm{kJ/mol}}} \right) at 298  K{\rm{298}}\;{\rm{K}} are −237.2,−394.4 - 237.2, - 394.4 and −8.2 - 8.2 for H2O(l),CO2(g){{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right){\rm{,C}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) and pentane (g) respectively. The value of Ecell0{\rm{E}}_{{\rm{cell}}}^{\rm{0}} for the pentane oxygen fuel cell is:
A. 1.0968  V1.0968\;{\rm{V}}
B. 0.0968  V0.0968\;{\rm{V}}
C. 1.968  V1.968\;{\rm{V}}
D. 2.0968  V2.0968\;{\rm{V}}

Explanation

Solution

We know that Gibbs free energy is that quantity which can generally be utilized to measure the extreme amount of work done in the thermodynamic system and the temperature and pressure kept constant during the whole process.

Complete step by step solution
The reaction of pentane oxygen fuel cell, which is also the combustion reaction of pentane is shown below.
C5H12+8O2→5CO2+6H2O{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}} + {\rm{8}}{{\rm{O}}_{\rm{2}}} \to {\rm{5C}}{{\rm{O}}_{\rm{2}}} + {\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
The formula for standard free energies formation is shown below.
ΔG=n×G(Products)−n×G(Reactants){\rm{\Delta G}} = {\rm{n}} \times {\rm{G}}\left( {{\rm{Products}}} \right) - {\rm{n}} \times {\rm{G}}\left( {{\rm{Reactants}}} \right)
Where, ΔG{\rm{\Delta G}} is the standard free energy, G(Products){\rm{G}}\left( {{\rm{Products}}} \right) is the free energy of products, and G(Reactants){\rm{G}}\left( {{\rm{Reactants}}} \right) is the free energy of reactants.

Substitute the respective values in the above equation.

ΔG=5×(−394.4)+6×(−237.2)−1×(−8.2)+8×0 =−1972−1423.2+8.2  kJ/mol =−3387  kJ/mol{\rm{\Delta G}} = 5 \times \left( { - 394.4} \right) + 6 \times \left( { - 237.2} \right) - 1 \times \left( { - 8.2} \right) + 8 \times 0\\\ = - 1972 - 1423.2 + 8.2\;{\rm{kJ/mol}}\\\ = - 3387\;{\rm{kJ/mol}}

The value of Ecell0{\rm{E}}_{{\rm{cell}}}^{\rm{0}} can be calculated by using the formula shown below.
ΔG=−nFEcell0{\rm{\Delta G}} = - {\rm{nFE}}_{{\rm{cell}}}^{\rm{0}}

Where, Ecell0{\rm{E}}_{{\rm{cell}}}^{\rm{0}} is the standard cell potential, n is the number of moles of electrons involved, F is the faraday constant, and ΔG{\rm{\Delta G}} is the standard free energy.
As oxygen is moving from oxidation state 0  to  −20\;{\rm{to}}\; - 2.
The number of moles of oxygen involved Is 3232.
The value of faraday in coulomb is 1  F=96500  coulombs{\rm{1}}\;{\rm{F}} = 96500\;{\rm{coulombs}}
Substitute the respective values in the above equation.

⇒−3387  kJ/mol=−32×96500×Ecell0 ⇒−3387×103  J/mol=−32×96500×Ecell0 ⇒Ecell0=−3387×103  J/mol−32×96500 ⇒Ecell0=1.0968  V\Rightarrow - 3387\;{\rm{kJ/mol}} = - {\rm{32}} \times {\rm{96500}} \times {\rm{E}}_{{\rm{cell}}}^{\rm{0}}\\\ \Rightarrow - 3387 \times {10^3}\;{\rm{J/mol}} = - {\rm{32}} \times {\rm{96500}} \times {\rm{E}}_{{\rm{cell}}}^{\rm{0}}\\\ \Rightarrow {\rm{E}}_{{\rm{cell}}}^{\rm{0}} = \dfrac{{ - 3387 \times {{10}^3}\;{\rm{J/mol}}}}{{ - {\rm{32}} \times {\rm{96500}}}}\\\ \Rightarrow {\rm{E}}_{{\rm{cell}}}^{\rm{0}} = 1.0968\;{\rm{V}}

**Hence, the correct choice for this question is A that is 1.0968  V1.0968\;{\rm{V}}.

Note: **
The combustion reaction mainly occurs in the presence of atmospheric oxygen. The product of that combustion reaction is always carbon dioxide and water vapour.