For the reaction N2(g) + O2(g)2NO(g) H° = 181 kJ and S° = 24.9 J/K G° would be negative at temperatures (above, below) K. Enter above or below in the first box and enter the temperature in the second box. Assume that H° and S° are constant.

ΔG deg will be negative above 7.27e+3 K.

Explanation:

The ΔG deg with the temperature can be found using the formula and the formula is given belowΔG deg = ΔH deg - T ΔS degGiven data, ΔH deg = 181kJ and ΔSdeg=24.9J/K-T ΔS deg will be always negative and ΔG deg = ΔH deg  will  be  positive and ΔG deg will be negative at relatively high temperatures and positive at relatively low temperatures solving the equation and substitute  ΔGdeg=0ΔGdeg = ΔHdeg - T ΔSdegT= ΔHdeg/ΔSdegT=181 kJ / 2.49e-2 kJK-1 By simplification we getT=7.27 × 10^3 K.Therefore, Go will be negative above   7.27 × 10^3 KSince ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.ΔG deg will be negative above 7.27e+3 K.