Question

The following reaction is performed at $298$ K.
$2NO\left(g\right)+O_2\left(g\right)\leftrightharpoons 2N{O}_2\left(g\right)$

The standard free energy of formation of $NO\left(g\right)$ is $86.6$ kJ/mol at $298$ K. What is the standard free energy of formation of $N{O}_2\left(g\right)$ at 298 K? $(K_p=1.6\times {10}^{12})$

• A. $R\left(298\right)ln(1.6\times {10}^{12})$ J/mol
• B. $86600+R\left(298\right)ln(1.6\times {10}^{12})$ J/mol
• C. $8660-$ $\frac{(1.6\times {10}^{12})}{R\left(298\right)}$ J/mol
• D. $0.5[2\times 86,600-R(298\left)ln\right(1.6\times {10}^{12}\left)\right]$ J/mol

Answer 1

The correct option is D $0.5[2\times 86,600-R(298\left)ln\right(1.6\times {10}^{12}\left)\right]$ J/mol

The relationship between the standard Gibbs free energy change for reaction and the equilibrium constant is as shown below.

$\Delta G_{rxn}^0=-RTln{K}_p$......(1)

The standard Gibbs free energy change for the reaction is given by the expression

$\Delta G_{rxn}^0=2\Delta G_f^0\left(N{O}_2\right)-2\Delta G_f^0\left(NO\right)$......(2).

From equations (1) and (2).

$2\Delta G_f^0\left(N{O}_2\right)-2\Delta G_f^0\left(NO\right)=-RTln{K}_p$

$\Delta G_f^0\left(N{O}_2\right)=0.5\times \left[2\Delta G_f^0\right(NO)-RTln{K}_p]$

Substitute values in the above expression.

$\Delta G_f^0\left(N{O}_2\right)=0.5\times [2\times 86600-R(298)ln1.6\times {10}^{12}]J/mol$