At 298 K , if the standard Gibbs energy change $\Delta_r G^{\circ}$ of a reaction is -115 kJ , the value of $\log _{10} K_p$ will be $\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$

One mole of an ideal gas at 300 K and 20 atm expands to 2 atm under isothermal and reversible conditions. The work done by the gas is $-x \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $x$ is $\left(R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

At 298 K , the enthalpy change ( in kJ ) for the reaction given below is

$$ \mathrm{CH}_4(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(l) $$

$$ \begin{aligned} Given:\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) & \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\ominus}=-286 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g) ; \Delta H^{\ominus}=-394 \mathrm{~kJ} \\ \mathrm{CH}_4(g)+2 \mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \Delta H^{\ominus}=-890 \mathrm{~kJ}\end{aligned} $$