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Consider the reaction of ethyl bromide with sodium hydroxide: [tex]CH_3CH_2Br(aq) + NaOH(aq) \rightleftharpoons CH_3CH_2OH(aq) + NaBr(aq)[/tex]The reaction is first order in NaOH and second order overall. What is the rate law?

Respuesta :

Answer:

rate law = K [CH3CH2BR] [NaOH]

Explanation:

What is applied here is the principle of rate of reaction and the rate law.

where the rate is directly proportional to concentration and K is the rate constant.

The explanation is as attached below

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