J.R. S. answered 07/13/22
Ph.D. University Professor with 10+ years Tutoring Experience
ln (K2/K1) = -∆H/R (1/T2 - 1/T1)
K1 = 6.22x10-4
K2 = ?
T1 = 2000C + 273 = 2273K
T2 = 1000C + 273 = 1273K
∆H = 180.6 kJ
R = 8.314 J/Kmol = 0.008314 kJ/Kmol
ln(K2/6.22x10-4) = -180.6/0.008314 (1/1273 - 1/2273)
ln(K2/6.22x10-4) = -21,722 (7.86x10-4 - 4.40x10-4)
ln(K2/6.22x10-4) = -21,722 * 3.46x10-4
ln(K2/6.22x10-4) = -7.516
(K2/6.22x10-4) = 5.44x10-4
K2 = 3.39x10-7

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