at-500-k-the-equilibrium-constant-kc-for-the-following-reaction-is-x-n2-g-2-o2-g

Question

At 500 K, the equilibrium constant (K_{c}) for the following reaction is 1.68.N_{2}(g) + 2 O_{2}(g) → $$N_{2}O$$_{4}(g)If the reaction is performed in a 1.00 L vessel at 500 K, how much $$N_{2}O$$_{4} must be placed in the vessel to contain 0.178 mol of N_{2} at equilibrium?

Accepted Answer

0.216 M

Answer

0.127 M

Answer

0.356 M

Answer

0.178 M

At 500 K, the equilibrium constant (Kc) for the following reaction is 1.68.N2(g) + 2 O2(g) → N2O4(g)If the reaction is performed in a 1.00 L vessel at 500 K, how much N2O4 must be placed in the vessel to contain 0.178 mol of N2 at equilibrium?

At 500 K, the equilibrium constant (K_c) for the following reaction is 1.68.
N₂(g) + 2 O₂(g) → N₂O₄(g)If the reaction is performed in a 1.00 L vessel at 500 K, how much N₂O₄ must be placed in the vessel to contain 0.178 mol of N₂ at equilibrium?