The free energy of formation of NO is 78 kJ mol -1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K? 1 2 N 2 g + 1 2 O 2 g → NO g (Anti log 0.9263 = 8.459)

Step-by-step solution

K e q = a n t i l o g - Δ G o 2.303 R T Δ G o = 78 k J / m o l = 78000 J / m o l R = 8.314 J K - 1 m o l - 1 , T = 1000 K Putting these values in eqn. (i) we get, K e q = a n t i l o g - 78000 2.303 × 8.314 × 1000 K e q = a n t i l o g ( - 4.0737 ) or K e q = 8.4 × 1 0 - 5