The $INCORRECT$ match in the following is

Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

Consider the reaction $X \rightleftharpoons Y$ at $300 \text{ K}$. If $\Delta H^\circ$ and $K$ are $28.40 \text{ kJ mol}^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^\circ$ for the reaction in $\text{J K}^{-1} \text{ mol}^{-1}$ is . . . . . . . (Nearest integer) (Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $\ln 10 = 2.3$, $\log 3 = 0.48$, $\log 2 = 0.30$)

At $320 \ K,$ a gas $A_2$ is $20 \%$ dissociated to $A_{(g)}.$ The standard free energy change at $320 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is approximately $(R = 8.314 \ J \ K^{-1} \ mol^{-1}; \ \ln \ 2 = 0.693; \ \ln \ 3 = 1.098).$

In an equilibrium reaction for which $\Delta G^o = 0$,the equilibrium constant $K = $