The temperature in $K$ at which $\Delta G = 0$, for a given reaction with $\Delta H = -20.5 \ kJ \ mol^{-1}$ and $\Delta S = -50.0 \ J \ K^{-1} \ mol^{-1}$ is:

The condition of spontaneity of a process is

For a certain reaction,the enthalpy change and entropy change are $40.63 \ kJ \ mol^{-1}$ and $100 \ J \ K^{-1} \ mol^{-1}$ respectively. What will be the value of $\Delta G$ at $27 \ ^oC$?

In the conversion of limestone to lime,$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$,the values of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are $+179.1 \ kJ \ mol^{-1}$ and $160.2 \ J \ K^{-1} \ mol^{-1}$ respectively at $298 \ K$ and $1 \ bar$. Assuming that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not change with temperature,the temperature above which the conversion of limestone to lime will be spontaneous is ........... $K$.

For the vaporization of water at $1 \ atm$ pressure,the values of $\Delta H$ and $\Delta S$ are $40.63 \ kJ/mol$ and $108.8 \ J/K \cdot mol$ respectively. At what temperature $(K)$ will the change in Gibbs energy $(\Delta G)$ be zero (in $.4$)?

$A$ process has $\Delta H = 200 \ J \ mol^{-1}$ and $\Delta S = 40 \ J \ K^{-1} \ mol^{-1}$. Choose the minimum temperature above which the process will be spontaneous.