What is the nature of the reaction at $298 \ K$,if the entropy change and enthalpy change for a chemical reaction are $7.4 \ cal \ K^{-1}$ and $-2.5 \times 10^3 \ cal$,respectively?

When enthalpy and entropy change for a chemical reaction are $-2.5 \times 10^3 \ cal$ and $7.4 \ cal \ K^{-1}$ respectively,predict if the reaction at $298 \ K$ is:

The relation between $\Delta G$ and $\Delta H$ is

The values of $\Delta H$ and $\Delta S$ of a certain reaction are $-400 \text{ kJ mol}^{-1}$ and $-20 \text{ kJ mol}^{-1} \text{ K}^{-1}$ respectively. The temperature below which the reaction is spontaneous, is

The densities of graphite and diamond at $298 \, K$ are $2.25 \, g \, cm^{-3}$ and $3.31 \, g \, cm^{-3}$ respectively. If the standard free energy difference $(\Delta G^o)$ is $1895 \, J \, mol^{-1}$, the pressure at which graphite will be transformed into diamond at $298 \, K$ is:

For the reaction $C_{(graphite)} + CO_{2(g)} \rightarrow 2CO_{(g)}$,$\Delta H = 170 \ kJ$ and $\Delta S = 170 \ J/K$. At what temperature $(K)$ will the reaction become spontaneous?