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 $298 \,K ?\left( K _{n}=1.6 \times 10^{12}\right)$

• A. $R (298) \, in \, (1.6 \times 10^{12}) - 86600$
• B. $86600 + R(298) \, in \, (1.6 \times 10^{12})$
• C. $86600 - \frac{In \, (1.6 \times 10^{12})}{R(298)}$
• D. $0.5[2 \times 86600 - R(298) \, In \, (1.6 \times 10^{12})]$

Answer 1

Answer: (D)

The relationship between the standard Gibbs free energy change for reaction and the equilibrium constant is as shown below.
$\Delta G _{ rxn }^{0}=- RT \ln K _{ p } \ldots \ldots(1)$
The standard Gibbs free energy change for the reaction is given by the expression
$\Delta G _{ rxn }^{0}=2 \Delta G _{ f }^{0}\left( NO _{2}\right)-2 \Delta G _{ f }^{0}( NO ) \ldots \ldots . .(2)$
From equations (1) and (2).
$2 \Delta G _{ f }^{0}\left( NO _{2}\right)-2 \Delta G _{ f }^{0}( NO )=- RT \ln K _{ p }$
$\Delta G _{ f }^{0}\left( NO _{2}\right)=0.5 \times\left[2 \Delta G _{ f }^{0}( NO )- RT \ln K _{ p }\right]$
Substitute values in the above expression.
$\Delta G _{ f }^{0}\left( NO _{2}\right)=0.5 \times\left[2 \times 86600- R (298) \ln 1.6 \times 10^{12}\right] \,J / mol$