I. the temperature of the system II. the nature of the reactants and products III. the concentration of the reactants IV. the concentration of the products.

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Presentation transcript:

I. the temperature of the system II. the nature of the reactants and products III. the concentration of the reactants IV. the concentration of the products The value of the equilibrium constant, K, is dependent on:

I.the temperature of the system II. the nature of the reactants and products

At a given temperature, K = for the equilibrium: PCl 5 (g) PCl 3 (g) + Cl 2 (g) What is K for: Cl 2 (g) + PCl 3 (g) PCl 5 (g)?

ANSWER 59

CONSIDER THE CHEMICAL SYSTEM : CO + CL 2 COCL 2 ; K = 4.6 X 10 9 How do the equilibrium concentrations of the reactants compare to the equilibrium concentration of the product? A)They are much smaller. B)They are much bigger. C)They are about the same. D)They have to be exactly equal. E)You can't tell from the information given.

ANSWER A: They are much smaller

If the concentration of the product were to halved, what would happen to the equilibrium constant? A) B) C) D) E

ANSWER D: It would not change its value

Find the value of the equilibrium constant (K) (at 500 K) for N 2 (g) + 3H 2 (g) 2NH 3 (g). The value for K p at 500 K is 1.5 x 10 –5

ANSWER 0.025

Consider the following reaction: 2HF(g) H 2 (g) + F 2 (g) (K = 1.00 x 10 –2 ) Given 1.00 mole of HF(g), mole of H 2 (g), and mole of F 2 (g) are mixed in a 5.00 L flask, determine in which direction this reaction will shift to reach equilibrium. Support your work with Q.

ANSWER Q=0.272 Q>K therefore will shift towards reactants

Consider the reaction H 2 + I 2 2HI for which K = 44.8 at a high temperature. If an equimolar mixture of reactants gives the concentration of the product to be 0.50 M at equilibrium, determine the equilibrium concentration of the hydrogen. A)A) B)B) C)C) D)D) E)E)

ANSWER 7.5 x M

GIVEN THE EQUATION 2A(G) 2B(G) + C(G). AT A PARTICULAR TEMPERATURE, K = 1.6 X If you mixed 5.0 mol B, 0.10 mol C, and mol A in a one-liter container, which direction would the reaction initially proceed? A) To the left. B) To the right. C) The above mixture is the equilibrium mixture. D) Cannot tell from the information given. E) None of these (A-D).

ANSWER A: To the left

GIVEN THE EQUATION 2A(G) 2B(G) + C(G). AT A PARTICULAR TEMPERATURE, K = 1.6 X Addition of chemical B to an equilibrium mixture of the above will A) cause [A] to increase B) cause [C] to increase C) have no effect D) cannot be determined E) none of the above

ANSWER A: Cause [A] to increase

GIVEN THE EQUATION 2A(G) 2B(G) + C(G). AT A PARTICULAR TEMPERATURE, K = 1.6 X At a higher temperature, K = 1.8 x 10 –5. Placing the equilibrium mixture in an ice bath (thus lowering the temperature) will A) cause [A] to increase B) cause [B] to increase C) have no effect D) cannot be determined E) none of the above

ANSWER B: Cause [B] to increase

GIVEN THE EQUATION 2A(G) 2B(G) + C(G). AT A PARTICULAR TEMPERATURE, K = 1.6 X Raising the pressure by lowering the volume of the container will A) cause [A] to increase B) cause [B] to increase C) have no effect D) cannot be determined E) none of the above

ANSWER A: cause [A] to increase