Consider the reaction N₂(g)+ O₂(g) 2NO(g),for which K_c = 0.10 at 2000ºC.Starting with initial concentrations of 0.040 M for N₂ and 0.040 M for O₂ ,what is the equilibrium concentration of NO?

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From the balanced equation, we see that 1 mole of N₂(g) reacts with 1 mole of O₂(g) to produce 2 moles of NO(g). Let x be the equilibrium concentration of NO(g). Then, at equilibrium, the equilibrium concentrations of N₂(g) and O₂(g) will both be (0.040 - x) M. Using the equilibrium constant expression, we have: K_c = [NO]2/[(N₂)(O₂)] 0.10 = (2x101.075)/[(0.040 - x)2] 0.10(0.040 - x)2 = 2x101.075 0.008(0.0016 - 0.00008x + x2) = 2x101.075 1.6x10^-5 - 8x10^-7 x + 8x10^-9 x^2 = 2x10^-1 8x10^-9 x^2 - 8x10^-7 x + (1.6x10^-5 - 2x10^-1) = 0 4x10^-7 x^2 - 4x10^-5 x + (8x10^-4 - 1) = 0 Using the quadratic formula, we find: x = 0.011 M or x = 0.037 M Since x represents the equilibrium concentration of NO, it cannot be higher than the initial concentration of N₂ or O₂, which is 0.040 M. Therefore, the only valid solution is x = 0.011 M. Thus, the equilibrium concentration of NO is 0.011 M.

Consider the Reaction N2(g)+ O2(g) 2NO(g),for Which Kc =

Consider the Reaction N₂(g)+ O₂(g) 2NO(g),for Which K_c =