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?

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:

Calculate the work done in the following reaction at $27^{\circ} C$.
$4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$
$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)

Assertion: Many endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature.
Reason: Entropy of the system increases with increase in temperature.

If for a certain reaction $\Delta_{r} H$ is $30 \; kJ \; mol^{-1}$ at $450 \; K$,the value of $\Delta_{r} S$ (in $J \; K^{-1} \; mol^{-1}$) for which the same reaction will be spontaneous at the same temperature is

For the given reaction: $H_{2(g)} + S_{(s)} \to H_2S_{(g)}$; $\Delta H_r = 100\, kJ/mol$ and $\Delta S_r = 400\, J/mol\cdot K$. The temperature at which the above reaction occurs reversibly is (Assuming $\Delta H_r$ and $\Delta S_r$ are independent of temperature) ..... $K$.