Sapling Learning The following reaction is a single-step, bimolecular reaction:

Question

Sapling Learning The following reaction is a single-step, bimolecular reaction: CH,Br+ NaOHCHOH+ NaBr When the concentrations of CHaBr and NaOH are both 0.140 M, the rate of the reaction is 0.0030 M/s. (a) What is the rate of the reaction if the concentration of CHsBr is doubled? Number M/ s (b) What is the rate of the reaction if the concentration of NaOH is halved? Number Mis (c) What is the rate of the reaction if the concentrations of CH Br and NaOH are both increased by a factor of four? Number M/ s nSwEr

Explanation / Answer

(a) Ans : 0.0060 M/s

(b) Ans: 0.0015 M/s

(c) Ans: 0.0480 M/s

Solution :

The given reaction is

CH3Br + NaOH -------------) CH3OH + NaBr

Now, Rate of reaction (rA) = k(CH3Br)1(NaOH)1

k=(rA) /(CH3Br)1 *(NaOH)1

And given that Rate of reaction = 0.0030 M/s

Concentration of CH3Br = 0.140 M

Concentration of NaOH = 0.140 M

Now, k= (0.0030)/(0.140)*(0.140)

=0.153061224 M-1s-1

(a) solution :

If the concentration of CH3Br is doubled that is

(CH3Br) = 2*0.140 = 0.280 M

(NaOH) =0.140 M

Now, Rate of reaction is

(ra) = (0.153061224)*(0.280)*(0.140)= 0.006 M/s

(b) solution :

If the concentration of NaOH is halved that is

(NaOH) = (0.140)/2 =0.070 M

(CH3Br) = 0.140 M

Now, Rate of reaction is

(rb) = (0.153061224)*(0.140)*(0.070) = 0.0015 M/s

(c) solution :

If the concentrations of both CH3Br, NaOH are increased by a factor of four, then

(CH3Br) =4*(0.140) = 0.560 M

(NaOH) =4*(0.140)= 0.560 M

Now, Rate of reaction is

(rc) = (0.153061224)*(0.560)*(0.560) = 0.048 M/s

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