Respuesta :
Answer:
Explanation:
CH₃Br+NaOH⟶CH₃OH+NaBr
It is a single step bimolecular reaction so  order of reaction is 2 , one for CH₃Br and one for NaOH .
rate of reaction = k x [CH₃Br] [ NaOH]
.008 = k x .12 x .12
k = .55555
when concentration of CH₃Br is doubled
rate of reaction = .555555 x [.24] [ .12 ]
= .016 M/s
when concentration of NaOH Â is halved
rate of reaction = .555555 x [.12] [ .06 ]
= .004 Â M/s
when concentration of both CH₃Br and Na OH is made 5 times
rate of reaction = .555555 x .6 x .6
= 0.2 M/s
if the concentration of the CH3Br in the reaction is doubled; then the rate of the reaction is also doubled and it becomes 0.016 M/s. If the concentration of NaOH is being halved, the rate of reaction also gets halved. If both concentrations for CH3Br and NaOH are increased by a factor of 5, the rate of reaction will increase by 25 times.
The single-step, bimolecular reaction given shows the chemical combination between two different molecular substances in the given reaction.
CH₃Br + NaOH ⟶ CH₃OH + NaBr
The rate of the reaction (r) = K[CH₃Br][NaOH].
where;
[tex]\mathbf{rate = -\dfrac{ d[CH_3Br ] }{dt} = -\dfrac{ d[NaOH] }{dt} }[/tex]
From the given information; if the concentration of the CH3Br in the reaction is doubled;
 Â
[tex]\mathbf{rate = -\dfrac{1}{2}\dfrac{d[CH_3Br]}{dt}}[/tex]
[tex]\mathbf{2 \times rate = -\dfrac{d[CH_3Br]}{dt}}[/tex]
[tex]\mathbf{2 \times 0.008 \ M/s = -\dfrac{d[CH_3Br]}{dt}}[/tex]
[tex]\mathbf{ -\dfrac{d[CH_3Br]}{dt} =0.016 \ M/s }[/tex]
If the concentration of NaOH is being halved, then, we have:
[tex]\mathbf{rate = -\dfrac{d[NaOH]}{dt}}[/tex]
[tex]\mathbf{rate = -2 \times \dfrac{d[NaOH]}{dt}}[/tex]
[tex]\mathbf{\dfrac{rate}{2} = - \times \dfrac{d[NaOH]}{dt}}[/tex]
where;
- rate = 0.0080 M/s
[tex]\mathbf{\dfrac{0.0080 \ m/s}{2} = - \dfrac{d[NaOH]}{dt}}[/tex]
[tex]\mathbf{- \dfrac{d[NaOH]}{dt} = 0.004 \ M/s}[/tex]
If both concentrations for CH3Br and NaOH are increased by a factor of 5, then:
[tex]\mathbf{rate = - \dfrac{1}{5}\dfrac{d[CH_3Br]}{dt} = -\dfrac{1}{5} \dfrac{d[NaOH]}{dt}}[/tex]
[tex]\mathbf{5 \times 5 \times rate = -\dfrac{d[CH_3Br]}{dt} = -\dfrac{d[NaOH]}{dt} }[/tex]
[tex]\mathbf{5 \times 5 \times 0.0080 \ M/s = -\dfrac{d[CH_3Br]}{dt} = -\dfrac{d[NaOH]}{dt} }[/tex]
[tex]\mathbf{0.2 \ M/s = -\dfrac{d[CH_3Br]}{dt} = -\dfrac{d[NaOH]}{dt} }[/tex]
Thus, the rate of the reaction is increased by 25 times faster.
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