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

A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If $$\Delta H$$ is the enthalpy change and $$\Delta E$$ is the change in internal energy, then

The value of enthalpy change $$(\Delta H)$$ for the reaction $$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+ 3 \mathrm{H}_2 \mathrm{O}(l)$$, at $$27^{\circ} \mathrm{C}$$ is $$-1366 \cdot 5 \mathrm{~kJ} \mathrm{~mol}^{-1}$$. The value of internal energy change for the above reaction at this temperature will be

Standard entropy of X₂, Y₂ and XY₃ are 60, 40 and 50 JK^$$-$$1mol^$$-$$1, respectively. For the reaction, $$\frac{1}{2}$$X₂ + $$\frac{3}{2}$$Y₂ $$\to$$ XY₃, $$\Delta$$H = $$-$$30 kJ, to be at equilibrium, the temperature will be