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Question: Calculate \(\Delta {{G}^{\circ }}\) for the conversion of oxygen to ozone; \(\dfrac{3}{2}{{O}_{2(g)}...

Calculate ΔG∘\Delta {{G}^{\circ }} for the conversion of oxygen to ozone; 32O2(g)⇌O3(g)\dfrac{3}{2}{{O}_{2(g)}}\rightleftharpoons {{O}_{3(g)}} at 298K, if KP{{K}_{P}} ​ for this conversion is 2.47×10−292.47\times {{10}^{-29}} .

Explanation

Solution

Gibbs Free Energy is a quantity used to measure the amount of work in a thermodynamic system when temperature (T) and pressure (P) are kept constant or it can be defined as maximum amount of work extracted from a closed system.

Formula used: We will use the following formula:-
ΔG∘=−RTln⁡(KP)\Delta {{G}^{\circ }}=-RT\ln ({{K}_{P}})

Complete step-by-step answer: -As we know that the free energy change of the reaction in any state, ΔG\Delta {{G}} (when equilibrium has not been attained) is related to the standard free energy change of the reaction, ΔG∘\Delta {{G}^{\circ }} (that is equal to the difference in the free energies of formation of the products and reactants, both being their standard states) is as follows:-
ΔG=ΔG∘+RTln⁡Q\Delta G=\Delta {{G}^{\circ }}+RT\ln Q
where Q = reaction quotient.
-At equilibrium,
ΔG\Delta {{G}}= 0 and Q (reaction quotient) becomes equal to the equilibrium constant. Hence the equation will be as follows:-
ΔG∘=−RTln⁡(KP)\Delta {{G}^{\circ }}=-RT\ln ({{K}_{P}})
As we know that ln⁡x=2.303log⁡KP\ln x=2.303\log {{K}_{P}}
Therefore, the relation can be written as:-
ΔG∘=−2.303RTlog⁡(KP)\Delta {{G}^{\circ }}=-2.303RT\log ({{K}_{P}})
Where,
R = universal gas constant = 8.314Jmol⋅K8.314\dfrac{J}{mol\cdot K}
T= temperature (on Kelvin scale)
KP{{K}_{P}}= equilibrium constant calculated from the partial pressures of a reaction.
-The values are given as follows:-
T= 298K
KP{{K}_{P}}= 2.47×10−292.47\times {{10}^{-29}}
On substituting all these values in the formula, we get:-
⇒ΔG∘=−2.303RTlog⁡(KP) ⇒ΔG∘=−2.303×8.314Jmol⋅K×298Klog⁡(2.47×10−29) ⇒ΔG∘=−2.303×8.314Jmol⋅K×298K×(−28.607) ⇒ΔG∘=−163228.93J/mol ⇒ΔG∘=−163.22893kJ/mol \begin{aligned} & \Rightarrow \Delta {{G}^{\circ }}=-2.303RT\log ({{K}_{P}}) \\\ & \Rightarrow \Delta {{G}^{\circ }}=-2.303\times 8.314\dfrac{J}{mol\cdot K}\times 298K\log (2.47\times {{10}^{-29}}) \\\ & \Rightarrow \Delta {{G}^{\circ }}=-2.303\times 8.314\dfrac{J}{mol\cdot K}\times 298K\times (-28.607) \\\ & \Rightarrow \Delta {{G}^{\circ }}=-163228.93J/mol \\\ & \Rightarrow \Delta {{G}^{\circ }}=-163.22893kJ/mol \\\ \end{aligned}
-Therefore, ΔG∘\Delta {{G}^{\circ }} for the conversion of oxygen to ozone is -163.229kJ/mol.

Note: Always remember to put values in formula according to the universal gas constant and similarly do the conversion as well of the given values. Other gas constant values which can be preferred are as follows:-
-8.314J⋅K−1⋅mol−1J\cdot {{K}^{-1}}\cdot mo{{l}^{-1}} -8.314m3⋅Pa⋅K−1⋅mol−1{{m}^{3}}\cdot Pa\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}
-2 cal⋅K−1⋅mol−1cal\cdot {{K}^{-1}}\cdot mo{{l}^{-1}} -8.205m3⋅atm⋅K−1⋅mol−1{{m}^{3}}\cdot atm\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}
-0.082L⋅atm⋅K−1⋅mol−1L\cdot atm\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}
-Also in a reversible reaction, the free energy of the reaction mixture is lower than the free energy of reactants and products.