For the reaction : 2NO(g) + O_(2)(g) rarr 2NO_(2)(g), the enthalpy and entropy changes are - 113.0 kJ mol^(-1) and -145 JK^(-1) mol^(-1) respectively. Find the temperature below which this reaction is spontaneous.

For the reaction : 2NO(g) + O_(2)(g) rarr 2NO_(2)(g), the enthalpy and

1.	For the reaction : 2NO(g) + O_(2)(g) rarr 2NO_(2)(g), the enthalpy and entropy changes are - 113.0 kJ mol^(-1) and -145 JK^(-1) mol^(-1) respectively. Find the temperature below which this reaction is spontaneous.
Answer» Solution :`DELTAG = DELTAH - T DeltaS`. When reaction is in equilibrium , `DeltaG =0.``:. T = (DeltaH)/(DeltaS) = ( -113000J mol^(-1))/( - 145 JK^(-1)mol^(-1)) = 779.3K`. For reaction to be SPONTANEOUS , `Delta G ` should be -ve which can be so if temperature is below ` 779.3K`