The free energy for a reaction having $\Delta H = 31400 \, cal$ and $\Delta S = 32 \, cal \, K^{-1} \, mol^{-1}$ at $1000 \, ^oC$ is ....... $cal$.

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 $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$ at $298 \ K$ and $1 \ atm$ pressure,the values of $\Delta H^o$ and $\Delta S^o$ are $+179.1 \ kJ \ mol^{-1}$ and $+160.2 \ J \ K^{-1} \ mol^{-1}$ respectively. Assuming that $\Delta H^o$ and $\Delta S^o$ do not change with temperature,at what temperature (in $K$) will the process become spontaneous?

Calculate the work done if $1 \ mole$ of an ideal gas is compressed isothermally and reversibly from $12 \ dm^3$ to $6 \ dm^3$ at $300 \ K$. $\left[R = 8.314 \ J \ K^{-1} \ mol^{-1}\right]$ (in $kJ$)

The entropy and enthalpy changes for the reaction $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ at $300 \ K$ and $1 \ atm$ are respectively $-42.4 \ J \ K^{-1}$ and $-41.2 \ kJ$. The temperature at which the reaction will go in the reverse direction is (in $K$)

For the change $H_2O_{(l)} \to H_2O_{(g)}$ at $P = 1 \ atm$ and $T = 373 \ K$,the free energy change $\Delta G = 0$. This indicates that: