8/17/2019 Problem Nomer 3 Dan 9 Fix

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Problem

3. the catalysed decomposition of H2O2 in aqueous solution was followed by

titrating samples with KMnO4 at arious time interals to determine

undecomposed H2O2.

!ime"min# $ %& 2& 3& $&

'olume of

KMnO4

"cm3#

3()% 2*)+ %*), %2)3 $)&

-how graphically that the reaction first order and determine the rate constant.

Jawab :

!ime"min# $ %& 2& 3& $&

'olume of

KMnO4

"cm3#

3()% 2*)+ %*), %2)3 $)&

KMnO4/

"mol dm03#

&)&2( &)&33 &)&,& &)&+% &)2

1og"a0#

"mol dm

03

#

0%)$,* 0%)4+% 0%)222 0%)&*% 0&),**

0 10 20 30 40 50 60

-1.8

-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

time (min)

log (a-x) (mol/ dm 3)

rafi orde pertama dari decomposition of H2O2

-lope 5 &)&%*2$*

8/17/2019 Problem Nomer 3 Dan 9 Fix

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slope= −k t 2,303

0,019259= −k t 

2,303

k t =−0,019259×2,303=−0,044 /min

(. the neutralisation reaction of nitroethane in aqueous solution proceeds

according to the rate equation

OH 

¿

−¿¿−d [OH ]

dt   =

−d [C 2 H 5 NO2 ]dt 

  =k [C 2 H 5 NO2 ]¿

6periment at &   ℃  with initial concentration of both reactant equal to &)&%

mol dm03 gie a alue of %$&s for the reaction half0life. 7alculate the

corresponding rate constant at &   ℃ .

8awab 9

t 0,5=

  1

k t a

(a

2)

( a2 )=

  1

k t  a

:ietahui

t 0,5=

150s

a=0,01moldm−3

8adi)t 0,5=

  1

k t a

150s=  1

k t 0,01moldm−3

8/17/2019 Problem Nomer 3 Dan 9 Fix

3/5

k t 0,01moldm−3=

  1

150s

k t 0,01moldm−3=1,667×10−3 s−1

k t =

1,667×10−3s−1

0,01moldm−3  =0,667dm

3mol

−1s−1

k t =0,667 dm3mol

−1s−1

*. in the second order reaction between isobutyl bromide and sodium ethoidein ethanol at *$   ℃ ) the initial concentration of bromide was &)&$&$ mol dm03 

and ethoide was &)&(,2 mol dm03. !he decrease in the concentration of both

reactants was measured as follows

.

%&3"mol

dm3#

& $)* %&)( %,), 23)& 2()( 33)$

!ime.min & $ %& %( 3& 4& ,&  7alculate the rate constant for the reaction.

Jawab

if the reaction second order.

1

a− x =kt 

 1

a

a=0,0762moldm−3

.

%&3"mol

dm3#

& $)* %&)( %,), 23)& 2()( 33)$

%;"a0# %3)%2

3

%4)22

$

%$)2,

(

%,)((

+

%+)(*

(

2&)4*

%

24)$(

&

slope

!ime.mi

n

& $ %& %( 3& 4& ,& &)%+,&4

,

8/17/2019 Problem Nomer 3 Dan 9 Fix

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0 10 20 30 40 50 60 70

0

5

10

15

20

25

30

Time (min)

1/(a-x)

-lope 5 t 5 &)%+,&4, mol;dm03

*. 8awaban

Hasil data beriut ini diperoleh dari pengurangan dari onsentrasi reatan

%&3;mol dm3 & $)* %&)( %,), 23)& 2()( 33)$

!ime;min & $ %& %( 3& 4& ,&

-ehingga diperoleh data untu reasi orde 2) dengan persamaan 2.%% maa

dm3 mol0%;

"a0#

3+)*&4,

%

3+)+**+% 3+)+*3*% 3+)++($% 3+)++2+% 3+)+((&%

!ime;min $ %& %( 3& 4& ,&

Maa diperoleh grai

8/17/2019 Problem Nomer 3 Dan 9 Fix

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0 2 4 6 8 10 12

0

2

4

6

8

10

12

f(x) =

Linear ()

:iperoleh garis lurus sehingga reasi orde 2) maa

-lope 5  Kr (a−b)2,303

0&)&&&$ 5  Kr (0,0505−0,0763)

2,303

0&.&&&$ 5 Kr

 −0,02572,303

&)&&&$ 5 Kr &.&%%%

Kr 5 4)$ %&$ min0%