General Chemistry I: How do you figure out a simple proportional/ratio equilibrium problem?

Calculate the value of the equilibrium constant for the reaction 2A + 3B %26lt;%26gt; C + D if 2.00 moles of A and 3.00 moles of B are introduced into a 2.00 liter reaction vessel and allowed to come to equilibrium, at which point 0.400 mole of A remain.General Chemistry I: How do you figure out a simple proportional/ratio equilibrium problem?
2A + 3B %26lt;----%26gt; C + D...





you started with 2.00 moles A


you have 0.400 moles A remaining..


so 1.60 moles A have been consumed...





from the coefficients of the balanced equation


2 moles A reacts with 3 moles B


2 moles A ---%26gt; 1 mole C


2 moles A ---%26gt; 1 mole D





so moles B consumed = 1.6 A x (3 B / 2 A) = 2.4


moles B remaining = 3.0 - 2.4 = 0.6


moles C produced = 1.6 A x (1 C / 2 A) = 0.8


moles D produced = 1.6 A x (1 D / 2 A) = 0.8





at equilibrium...


[A] = 0.4 moles / 2 L = 0.2 moles / L


[B] = 0.6 moles / 2 L = 0.3 moles / L


[C] = 0.8 moles / 2 L = 0.4 moles / L


[D] = 0.8 moles / 2 L = 0.4 moles / L





Keq for the reaction aA + bB %26lt;---%26gt; cC + dD = [C]^c x[D]^d / ([A]^a x [B^b])





so Keq = (0.4 x 0.4) / (0.2虏 x 0.3鲁) = 148General Chemistry I: How do you figure out a simple proportional/ratio equilibrium problem?
no problem..





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Set up an ICE chart to show what happens when equilibrium is established.





Initial molarity A = 2.00 moles / 2.00 L = 1.00 M


Initial molarity B = 3.00 moles / 2.00 L = 1.50 M





Molarity . . . . .2A .+ .3B .==%26gt; C .+ ,D


Initial . . . . . .1.00 . . 1.50 . . . 0 . . . 0


Change . . . . .-2x . . .-3x . . . ..x . . . x


Final . . . . .1.00-2x .1.50-3x . .x . . . x





When equilibrium is reached, x molar each of C and D form. Then, according to the balanced equation, A will be reduced by 2x molar and B will be reduxed by 3x molar.





We are told that the final [A] = 0.400 moles / 2.00 L = 0.500 M = 2x


Then x = 0.250 M





The final concentrations are





[A] = 0.500 M


[B] = 1.50 - 3x = 1.50 - (3)(0.250) = 0.750 M


[C] = x = 0.250 M


[D] = x = 0.250 M





Kc = ([C][D]) / ([A]^2 [B]^3) = ((0.250)(0.250) / ((0.500)^2 (0.750)^3) =


0.595