The standard Gibbs free energy change $\Delta G^o$ is related to the equilibrium constant $K_p$ as:

At $227^{\circ} C$,dinitrogen tetraoxide is $60 \%$ dissociated. What is the standard free energy change at this temperature and at $1 \ atm$ pressure?

For a system at equilibrium,$\Delta G = 0$ under which of the following conditions?

Assuming ideal behaviour,the magnitude of $\log \, K$ for the following reaction at $25^{\circ} C$ is $x \times 10^{-1}$. The value of $x$ is $......$. (Integer answer)
$3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
[Given: $\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5 \, J \, mol^{-1}$
$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5 \, J \, mol^{-1}; R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

Calculate $\Delta_{r} G^{\ominus}$ for the conversion of oxygen to ozone,$\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$ at $298 \, K$,if $K_{p}$ for this conversion is $2.47 \times 10^{-29}$.

$K_p$ for the conversion of oxygen to ozone at $400 \ K$ is $1.0 \times 10^{-30}$,its standard Gibbs energy change in $kJ \ mol^{-1}$ is approximately