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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.)

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mairarmol

Answer:

The rate is [tex] 4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]

Explanation:

Stoichiometry

[tex]CH_{3}Cl+NaOH \rightarrow CH_{3}OH+NaCl [/tex]

Kinetics

[tex]-r_{A}=k \times [CH_{3}Cl] \times [NaOH] [/tex]

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

[tex] 1 \times 10^{-4}\frac{mole}{Ls}=k \times [0,2M] \times [1,0M]  =5 \times 10^{-4}\frac{L}{mole s}[/tex]

Taking as a base of calculus 1L, when half of the [tex] CH_{3}Cl [/tex] is consumed the mixture is composed by

[tex] 0,1 mole CH_{3}Cl [/tex] (half is consumed)

[tex] 0,9 mole NaOH [/tex] (by stoicheometry)

[tex] 0,1 mole CH_{3}OH [/tex]  

[tex] 0,1 mole NaCl [/tex]

Then, the rate is

[tex]-r_{A}=5 \times 10^{-4} \frac{L}{mole s}\times 0,1M \times 0,9 M=4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]

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