Question:

The following reaction is performed at $298\, K ?$ $2 NO ( g )+ O _{2}( g ) \rightleftharpoons 2 NO _{2}( g )$ The standard free energy of formation of $NO ( g )$ is $86.6 \,kJ /\, mol$ at $298 \,K .$ What is the standard free energy of formation of $NO _{2}( g )$ at $298 \,K ?\left( K _{n}=1.6 \times 10^{12}\right)$

Updated On: Aug 21, 2024
  • $R (298) \, in \, (1.6 \times 10^{12}) - 86600$
  • $86600 + R(298) \, in \, (1.6 \times 10^{12})$
  • $86600 - \frac{In \, (1.6 \times 10^{12})}{R(298)}$
  • $0.5[2 \times 86600 - R(298) \, In \, (1.6 \times 10^{12})]$
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The Correct Option is D

Solution and Explanation

$-\frac{ R \times 298 \,\ell\, n \,1.6 \times 10^{12}}{2}$ $=\Delta G _{ r }^{ O }=2 \Delta G _{ NO _{2}}^{0}-2 \Delta G _{ NO }^{0}$ $\Delta G _{ NO _{2}}^{0}=86.6 \times 10^{3}-\frac{298\, K \,\ell\, n 1.6 \times 10^{12}}{2}$
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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.