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.
The ΔG deg with the temperature can be found using the formula and the formula is given below
ΔG deg = ΔH deg - T ΔS deg
Given 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=0
ΔGdeg = ΔHdeg - T ΔSdeg
T= ΔHdeg/ΔSdeg
T=181 kJ / 2.49e-2 kJK-1
By simplification we get
T=7.27 × 10^3 K.
Therefore, Go will be negative above 7.27 × 10^3 K
Since ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.
