Question:
For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

NO (g) + O₃ (g) ↔ NO₂ (g) + O₂(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?

Answer:

It is given that K_cfor the forward reaction is 6.3 × 10¹⁴

Then, K_c for the reverse reaction will be,

K^’_c = 1 / K_c

= 1 / 6.3 × 10¹⁴

= 1.59 x 10^-15

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Question:
For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

NO (g) + O₃ (g) ↔ NO₂ (g) + O₂(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?

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Question: For the following equilibrium, Kc = 6.3 × 1014 at 1000 K NO (g) + O3 (g) ↔ NO2 (g) + O2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

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Question:
For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

NO (g) + O₃ (g) ↔ NO₂ (g) + O₂(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?