A student mixed the following amounts of KIOi_3 and H_2 SO_3, in order to determ

Question

A student mixed the following amounts of KIOi_3 and H_2 SO_3, in order to determine the rate of the following reaction: If the stock solutions for KIO_3, and H_2 SO_3 were 0.200 and 0.100 M respectively, use the volumes provided below to calculate the concentration of IO_3^- and SO_3^-2 in every dilution. Use the time provided to calculate the rate and fill all of your numbers in the table. Use your data to determine the rate for each reactant and write the rate law for this reaction. What is the overall order for this reaction? Calculate the rate constant for this reaction.

Explanation / Answer

2 KIO3 + 5 H2SO3 I2 + H2O + 2 KHSO4 +4H2SO4

a) A) As molarity*volume=moles

[IO3-1]=0.2M*15ml=0.2 mol/L*0.015 L=0.003 moles

Or, [IO3-1]=0.003/total volume=0.003moles/100ml=0.003mol/0.1L=0.03 M

[SO32-]=0.1 M*5.5ml=0.1 mol/L*0.0055 L=0.00055 M

B

[IO3-1]= 0.2M*15ml=0.2 mol/L*0.015 L=0.003 moles

[IO3-1]=0.003/total volume=0.003moles/100ml=0.003mol/0.1L=0.03 M

[SO32-]=0.1 M*11ml=0.1 mol/L*0.011 L=0.0011 M

C

[IO3-1]= 0.2M*30ml/total volume=0.2 mol/L*0.030L/0.1L=0.06M

[SO32-]=0.1 M*5.5ml=0.1 mol/L*0.0055 L=0.00055 M

b)Rate=C/t

solution A

, [IO3-1]=0.03 M

[SO32-]=0.00055 M

Here SO32- is the limiting reagent as it is lower in concentration. As they react in the molar ratio of 6:16 or 3:8

So amount of IO3- that will react with given SO3(2-) (per litre of solution )is 2/5 *0.00055M=0.0002 moles.

Or, Change in concentration=0.0002 mol/L

Rate=0.0002 mol/L/117s=1.71*10^-6 M/s

Solution B

[IO3-1]=0.03 M

[SO32-]=0.0011 M

Amount of IO3- that will react with given SO3(2-) (per litre of solution )=2/5*0.0011 moles=0.0004M

Rate= Change in concentration/time=0.0004 M/58 s=7.0*10^-6 M/s

Solution C

[IO3-1]=0.06M

[SO32-]=0.00055 M

Here SO32- is the limiting reagent as it is lower in concentration. As they react in the molar ratio of 2:5

So amount of IO3- that will react with given SO3(2-) (per litre of solution )is 2/5*0.00055M=0.0002moles.

Or, Change in concentration=0.0002 mol/L

Rate=0.000206 mol/L/117s=1.71*10^-6 M/s

Rate =k*[IO3-1]^m[SO32-]^n [Rate law]

K=rate constant

m,n=order of rxn with respect to IO3- and SO32- respectively

c) Rate =k*[IO3-1]^m[SO32-]^n

solution A

[IO3-1]=0.03 M

[SO32-]=0.00055 M

1.71*10^-6 M/s=k(0.03M)^m (0.00055 M)^n………………(1)

Solution B

[IO3-1]=0.03 M

[SO32-]=0.0011 M

7.0*10^-6 M/s=k(0.03M)^m (0.0011M)^n……………..(2)

Solution C

[IO3-1]=0.06M

[SO32-]=0.00055 M

1.71*10^-6 M/s=k(0.06M)^m (0.00055 M)^n………………….(3)

Eqn 1/eqn 2

1.7*10^-6/7.0*10^-6=(0.00055/0.0011)^n

0.24=(0.5)^n

Log 0.24=n log 0.5

N=2

Also divide 2 by 3

4.0=(0.03/0.06)^m

4.0=(0.5)^m

Taking log on both sides

Log 4.0=mlog0.5

M=log4.0/log0.5=2.0

So overall order=m+n=2+2=4

d) put value of m,n in eqn 1

1.71*10^-6 M/s=k(0.03M)^2 (0.00055 M)^2………………(1)

K=1.71*10^-6/0.03M)^2 (0.00055 M)^2=6464.64

K=6280.99 s^-1

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