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skad [1K]
3 years ago
10

Today, ammonia is synthesized through a series of reactions (called the Haber-Bosch process) that take place between methane, ai

r (which is four parts N2, one part O2), and potassium carbonate:
7 CH4(g) + 8 N2(g) +2 O2(g) + 17 H2O(g) + 7 K2CO3(s) ?? 16 NH3(g) + 14 KHCO3(s)
What is the equilibrium expression for the Harber-Bosch process?
Chemistry
1 answer:
djyliett [7]3 years ago
8 0

Answer:

K = [NH₃]¹⁶ [KHCO₃]¹⁴ / [CH₄]⁷ [N₂]⁸ [O₂]² [H₂O]¹⁷ [K₂CO₃]⁷

Explanation:

The equation for the reaction is given below:

7CH₄(g) + 8N₂(g) +2O₂(g) + 17H₂O(g) + 7K₂CO₃(s) —> 16NH₃(g) + 14KHCO₃(s)

Equilibrium constant (K) =..?

Equilibrium constant, K for a reaction is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, the equilibrium expression for the Harber-Bosch process for the synthesis of ammonia can be written as shown below:

K = [NH₃]¹⁶ [KHCO₃]¹⁴ / [CH₄]⁷ [N₂]⁸ [O₂]² [H₂O]¹⁷ [K₂CO₃]⁷

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1)Given data:

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