The equilibrium: 2 NO2 (g) / Longleftrightarrow⇔ N2O4 (g) has Kc = 4.7 at 100ºC. What is true about the rates of the forward (ratefor) and reverse (raterev) reactions initially and at equilibrium if an empty container is filled with just NO2?

Initial: forward rate reverse rate

Initial: forward rate > reverse rate Equilibrium: forward rate > reverse rate

Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate

Initial: forward rate = reverse rate Equilibrium: forward rate = reverse rate

Question

Sebastian Kline

Chemistry

14 December, 06:49

The equilibrium: 2 NO2 (g) / Longleftrightarrow⇔ N2O4 (g) has Kc = 4.7 at 100ºC. What is true about the rates of the forward (ratefor) and reverse (raterev) reactions initially and at equilibrium if an empty container is filled with just NO2?

Initial: forward rate reverse rate

Initial: forward rate > reverse rate Equilibrium: forward rate > reverse rate

Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate

Initial: forward rate = reverse rate Equilibrium: forward rate = reverse rate

+1

Answers (1)

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Hassan Wiley · Answer 1

Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate

Explanation:

2NO₂ (g) → N₂O₄ (g) Kc=4.7

The definition of equilibrium is when the forward rate and the reverse rate are equal.

Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.