The reaction described by the equation ch 3 cl + naoh → ch 3 oh + nacl follows the second-order rate law, rate = k [ ch 3 cl ] [ naoh ] . when this reaction is carried out with starting concentrations [ ch 3 cl ] = 0.2 m and [ naoh ] = 1.0 m , the measured rate is 1 × 10 − 4 mol l − 1 s − 1 . what is the rate after one-half of the ch 3 cl has been consumed? (caution: the initial concentrations of the starting materials are not identical in this experiment. hint: determine how much of the naoh has been consumed at this point and what its new concentration is, compared with its initial concentration.)

The rate is

Explanation:

Stoichiometry

Kinetics

The rate constant K can be calculated by replacing with the initial data

Taking as a base of calculus 1L, when half of the is consumed the mixture is composed by

(half is consumed)

(by stoicheometry)

Then, the rate is

The reaction rate decreases because there’s a smaller concentration of reactives.

Absorbed

Released

Released

Explanation:

The formation of a cation is an endothermic process because energy must be absorbed in order to remove an electron from an atom.

Similarly, energy is evolved when an electron is added to an atom to form a negative ion.

The formation of an ionic compound is an exothermic process. Since ionic compounds are more stable than the individual ions separated by a distance, the excess energy of the isolated ions is evolved when the ionic compound is formed.