If the equilibrium constant K for a reaction at a given temperature is much greater than 1, what does this imply about the position of equilibrium?

Prepare for the NANTeL Chemistry Certification and Engineering Fundamentals Test with multiple choice questions, detailed explanations, and key insights to boost your understanding and confidence. Ace your exam!

Multiple Choice

If the equilibrium constant K for a reaction at a given temperature is much greater than 1, what does this imply about the position of equilibrium?

Explanation:
When a reaction reaches equilibrium, the equilibrium constant expresses the ratio of product activities to reactant activities at that temperature. If this constant is much greater than 1, the numerator dominates, so the equilibrium mixture has far more products than reactants. That means the position of equilibrium lies toward the product side, with products being favored. If the constant were much less than 1, reactants would be favored. The idea that the system is “necessarily at equilibrium with K > 1” is a misunderstanding—K describes the ratio at equilibrium, not a statement about the current state of a system unless it has actually reached equilibrium. And calling the reaction not feasible isn’t accurate here; a large K indicates the forward process is thermodynamically favorable (though kinetics could still affect how fast it proceeds).

When a reaction reaches equilibrium, the equilibrium constant expresses the ratio of product activities to reactant activities at that temperature. If this constant is much greater than 1, the numerator dominates, so the equilibrium mixture has far more products than reactants. That means the position of equilibrium lies toward the product side, with products being favored.

If the constant were much less than 1, reactants would be favored. The idea that the system is “necessarily at equilibrium with K > 1” is a misunderstanding—K describes the ratio at equilibrium, not a statement about the current state of a system unless it has actually reached equilibrium. And calling the reaction not feasible isn’t accurate here; a large K indicates the forward process is thermodynamically favorable (though kinetics could still affect how fast it proceeds).

Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy