1. a) Calculate Kc for the following reaction if at equilibrium the [A] = .482 M

Question

1. a) Calculate Kc for the following reaction if at equilibrium the [A] = .482 M , [B] = .482 M and the [C] = 1.08 M.
                                      A + B   =  2C           Kc =  ________










b) When the above reaction is at equilibrium someone removes some of product, C. Initially the resultant concentrations are [A] = .482 M , [B] = .482 M and the [C] = .100 M. Calculate Q for this.
                                                            A + B  =   2C           Q=  ________









c) which direction will the reaction shift (left or right) to return to equilibrium:
_________________


d) Calculate the new equilibrium concentrations of all components, [A], [B] and [C], for this same reaction when equilibrium is restored.
                                                A + B  =   2C

Explanation / Answer

Kc = 1.08^2 / 0.482 x 0.482 = 5.02

Q = 0.1^2 / 0.482^2 = 0.043

since Q< Kc

The reaction will shift in forward direction

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