Question

Consider the single‑step, bimolecular reaction.

CH3Br + NaOH --> CH3OH + NaBr

When the concentrations of CH3Br and NaOH are both 0.100 M, the rate of the reaction is 0.0080 M/s.

What is the rate of the reaction if the concentration of CH3Br is doubled?

What is the rate of the reaction if the concentration of NaOH is halved?

What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of 6?

Answer 1

Single step, Bimolecular reaction

CH,Br+ NaOH → CH3OH + NaBr

r=K[CH3 Br][NaOH)

And

d[CH3 Br] rate = - d[NaOH] dt

Now, If the concentration of [CH₃Br] is doubled then

1 d[CH3 Br] rate= 2 dt

[CH3 Br] 2 x rate = -

2 x0.008 M. [CH3Br] at

[-9°H]]p À 9100

Rate of reaction becomes double (0.016 M/s).

If the concentration of [NaOH] is halved, then

rate= d[NaOH at

d[NaOH] rate = -2 x dt

rate _ _d[NaOH) dt

0.009 M d[NaOH] dt

0.004 M S d[NaOH] dt

Rate of reaction (0.004 M/s) get halved also.

If the concentration of [CH₃Br] and [NaOH] both increased by a factor of 6, the

1 d[CH3Br] 6 dt rate= 12[NaOH) 6 dt

d[CH3 Br] 6 X 6 X rate = - dt d[NaOH) dt

d[CH3 Br] d[NaOH] 6 x 6 x 0.008 dt

0.288M - d[CH; Br] _ a[NaOH] dt at

Rate of reaction will increased by 36 times.