Presentation is loading. Please wait. # Chemical Equilibrium. Rate of forward Rx = Rate of reverse Rx As a system approaches equilibrium, both the forward and reverse reactions are occurring.

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Chemical Equilibrium

Rate of forward Rx = Rate of reverse Rx As a system approaches equilibrium, both the forward and reverse reactions are occurring.

A System at Equilibrium Once equilibrium is achieved… the amount of each reactant and product remains constant

Equilibrium In a system at equilibrium both the forward and reverse reactions are being carried out equation written with a double arrow N 2 O 4 (g) 2 NO 2 (g)

Equilibrium Which arrow is larger if….. 1) [Reactants] > [Products] Would the theoretical yield by high or low? 2) [Reactants] < [Products] Would the theoretical yield by high or low?

Law of Mass Action Rate of any Chemical Reaction is proportional to: 1) Product of the masses of reacting substances AND 2) Mass of each substance raised to the power of the coefficient in a chemical equation

Mathematical Relationships in Equilibrium Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ]

Mathematical Relationships in Equilibrium Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2

Mathematical Relationships in Equilibrium At Equilibrium: Rate f = Rate r therefore k f [N 2 O 4 ] = k r [NO 2 ] 2 kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

Equilibrium Equation Ratio of the rate constants is a constant at that temperature expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

Equilibrium Expression and Constant For any reaction: aA + bBcC + dD K c = [C] c [D] d [A] a [B] b Equilibrium expression for this reaction Equilibrium Constant

What Does the Value of K Mean? If K >> 1 reaction is product-favored; product predominates at equilibrium. If K << 1 reaction is reactant-favored; reactant predominates at equilibrium.

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