The following ionic substitution reaction has a rate constant r=k[CH 3 CH 2 Br]

Question

The following ionic substitution reaction has a rate constant r=k[CH3CH2Br]1[NaOCH3]1.

a) How much will the overall rate increase or decrease if the concentration of bromoethane doubles?

b) How much will the overall rate increase or decrease if the concentration of NaOCH3 doubles?

c) How much will the overall rate increase or decrease if you use the same amount of each reactant, but you double the amount of solvent that you use?

I'm having a really hard time understanding the Rate Laws and reaction rates. My chemistry book doesn't go into much detail explaining the rate laws and reaction rates. I also did some online research with respect to the subject but I'm still not grasping it. Can someone please answer the above questions and thoroughly explain how you arrived at the answer(s). Please, include any equations or formulas used to arrive at the answer(s). Thanks!

Explanation / Answer

a)

consider the rate law

r = k [CH3CH2Br] [NaOCH3]

now

concentration of CH3CH2Br is doubled

so

new rate = k [2 x CH3CH2Br] [ NaOCH3]

new rate = 2 x k [CH3CH2Br] [NaOCH3]

new rate = 2 r

so

the rate is increased by 2 times if the concentration of CH3CH2Br is doubled

b)

similarly

the rate is increased by 2 times if the concentration of NaOCH3 is doubled

c)

we know that

concentration = moles / volume

given

the amount remains same but the volume of solvent is doubled

so

new concentration = moles / 2 x volume

new concentration = 0.5 x moles / volume

so

the concentration of each reactant is halved

now

new rate = k [0.5 x CH3CH2Br] [ 0.5 x NaOCH3]

new rate = 0.25 x k [CH3CH2Br] [NaOCH3]

new rate = 0.25 r = r/4

so

the overall rate is decreased by 4 times , if you double the amount of solvent

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.