Question:
Calculate $\Delta G^{o}$ for conversion of oxygen?
To ozone $3 / 2 O _{2}(g) \rightarrow O _{3}(g)$ at $298 \,K$, if $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ :
Calculate $\Delta G^{o}$ for conversion of oxygen?
To ozone $3 / 2 O _{2}(g) \rightarrow O _{3}(g)$ at $298 \,K$, if $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ :
Updated On: June 02, 2025
- $163\, kJ \,mol ^{-1}$
- $2.4 \times 10^{2} \,k J\, mol ^{-1}$
- $1.63\,kJ\, mol ^{-1}$
- $2.38 \times 10^{6}\, kJ \,mol ^{-1}$
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The Correct Option is A
Solution and Explanation
Use the following formula to find $\Delta G^{\circ}$.
$\Delta G^{\circ}=-2.303 R T \log K_{p}$
Given, $K_{p}=2.47 \times 10^{-29} $
$\therefore \Delta G^{\circ}=2.303 \times 8.314 \times 298 \times \log 2.47 \times 10^{-29} $
$=163000\, J\,mol ^{-1}$
$=163\, kJ\,mol ^{-1}$
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NEET Notification
Concepts Used:
Gibbs Free Energy
The energy associated with a chemical reaction that can be used to do work.It is the sum of its enthalpy plus the product of the temperature and the entropy (S) of the system.
The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system. In completely reversible process maximum enthalpy can be obtained.
ΔG=ΔH−TΔS
The Conditions of Equilibrium
If both it’s intensive properties and extensive properties are constant then thermodynamic system is in equilibrium. Extensive properties imply the U, G, A.
