If K_c equals 0.110 at 25°C for the reaction: N₂O₄(g)⇌ 2 NO₂(g),what is K_c for the reaction: 6NO₂(g)⇌ 3 N₂O₄(g)?

Question

Quizplus · Accepted Answer

For the reverse reaction, the equilibrium constant $K_{c}$ is the reciprocal of the original reaction. Given the stoichiometry of the reverse reaction is multiplied by 3 (from 2 N2O(g) to 6 N2O(g)), the new $K_{c}$ is the reciprocal of the original $K_{c}$ raised to the power of 3. Therefore, $K_{c} = (1/0.110)^3 = 1/(0.110)^3 = 751$.

If Kc Equals 0

If K_c Equals 0