uas 15b 1201507 anelin osirikna

8/11/2019 Uas 15b 1201507 Anelin Osirikna

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FINAL EXAMINATION

PHYSICALCHEMISTRY 2

By:

NAME : ANELIN OSIRIKNA

NIM/TM : 1201480/2012

PRODI : PENDIDIKAN KIMIA

NO. SOAL : 15 B

DEPARTMENT OF CHEMISTRY

FACULTY OF MATHEMATICS AND SCIENCE

STATE UNIVERSITY OF PADANG

2014


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1. A 0.05 molal KCl splutionis electrolyzed in a Hittorf cell at 250. A silver coulometer in series

coulometer connected in series with the cell deposited 0.4068 g of silver. The anode solution of the

Hittorf cell weighed 123.9 g and was found to contain 186 mg of potassium. What is the trasference

numer of potassium ion. Atomic weights:

K = 39.1, Cl = 35.45, Ag = 107.868

Answer:

186 mg = 0,186 g

Molar mass of KCl= 74,55g

The anode solution of the Hittorf cell weighed 123.9 g and was found to contain 186 mg of

potassium.

Equivalent t =186 10-3 g Mr KCl/MrK

= 186 10-3 g 74.55/39.1

= 0.354 g

The weights of water in the solution: 123.9 g - 0.354 g

= 132,546 g

Before electrolysis: 0.05 molal

= 0.05molal KCl/1000 of the salt

132,546 g =

= 0.494 g KCl

loss of weights of KCl in anode compartment is equal to:

= 0.494 g0.354 g

= 0.14 g

Equivalent to =

= 1.878 10-3g

The total current passed trought the cell is given by = 0.4068 g of silver

Equal to = 0.4068 g/Ar Ag

= 0.4068 g / 107.868

= 0.00377 g

Actual weighed = 0.001878 g

The differences = 0.00377 g - 0.001878 g = 0.001892 g

t+ = = 0.51,

t-= (1- 0.51) = 0.49


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2. The emf at 298 K of the concentrationcell with tranference, H2(1atm) HCl (a= 0.0100)HCl (a =

0.0100)H2(1 atm) is 0.0190 V. What is the average transference number of the H+ion?

Answer:

Based on the literature we know that the reaction at the negatife electrode will be:

H2(g,1atm)=H+ + e

That at positive:

H+ + e = H2(g,1atm)

When we subtitute to the othe equation,the equation will be:

= (t_)(RT/F) ln (1/2)

0.0190 V = (t_ )0.0821/96500) ln 0.01/0.01

t_ = Cl-

t+=1- H+

From the literature we have known that 1


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10 ppm= 1 mg/L=10 x10-3g/L

Equiv /L (C) = (10 x10-3g/L)(58g/mol)

=0.17 x10-3 equiv /L

L= (AC/1000)A/d

= (1 cm2(0.17 x10-3)/1000) (1 cm2/0,2cm)

= 0.05 x 10-3ohm-cm-

L= 1/R

R=1/L

=1/0.05 x 10-3ohm-cm-

= 20. 10+3ohm

100 ppm= 1 mg/L=100 x10-3g/L

Equiv /L (C) = (100 x10-3g/L)(58g/mol)

=1,7 x10-3 equiv /L

L= (AC/1000)A/d

= (1 cm2(1,7 x10-3)/1000) (1 cm2/0,2cm)

= 0.5 x 10-3ohm-cm-

L= 1/R

R=1/L

=1/0.5 10-3ohm-cm-

= 2 x 10+3ohm


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4. The reaction of CO2(aq) + H2O H++ HCO3

-catalyzed by the enzymebovine carbonic anhydrase was

studied in a stopped-flow apparatus at pH 7.1 and temperature 0.5 0C. For an initial enzyme

concentration of 2.8 10-9mol/dm3, initial rates as a fuction of (CO2)0are (where c0= 1mol/dm3)

10 [CO ]0/c 1.25 2.50 5.00 20.0

10 r0 /(c s- ) 0.28 0.48 0.80 1.55

Find the k2 and Km from the plot!

Answer:

From the curve we see thatthe slope is 17,8 s and the intercept of y- axis is 3.73 x 104

Vmax = 1/(y-intercept)

= 1/(3.73 x 104)Lmol-s

= 0,268 x 10-4Lmol-s

Km = Vmax x slope

=(0,268 x 10-4Lmol-s )x (17,8 s)

=4.77 x 10-4Lmol-

K2 = Vmax/[E]0

= (0,268 x 10-4Lmol-s ) / (2.8 10-9mol/L)

=0.0957 x 105s-

-3

-2

-1

0

1

2

3

4

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

1/vx10-4

1/S0 x 10-3

a lineweaver-burk plot for enzyme-catalysed

proceed by michels menten

slope=Km/

Vmax

-1/Km

1/Vmax


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5.

For N2adsorbed in a certain sample of charcoal at -770C, adsorbed volume (recalculated 00C and 1 atm)

per gram charcoal vs. N2pressure are:

P/atm 3.5 10.0 16.7 25.7 33.5 39.2

V/(cm /g) 101 136 153 162 165 166

Answer:

a.)Fit the data with the Langmuir isoterm and give the values of vmonand b,

P/atm 3,5 10 16,7 25,7 33,5 39,2

p/v 0,03465 0,0735 0,109 0,1586 0,203 0,236

0.03465

0.0735

0.109

0.1586

0.203

0.236

0

0.05

0.1

0.15

0.2

0.25

0 5 10 15 20 25 30 35 40 45

P/v(atm/(cm3/g))

P(atm)

langmuir grafic for N2


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From the graph we can make the linear equation where:

y= ax +B

P1/V1= P1a+B

P1/V1= P1a+B

P2/V2= P2a+B -

()

()

interseptPP/v


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a.

Fit the data with the Freundlich isoterm and give the values of kand a

b.)Log V = log k +log p

Log V1 = log k +log p1

Log V2= log k + log p2 -

2,00 = log k +0,54

-

Soa= 0,28

Log v = log k +log p

2 = log k + (0,28 0,54)

Log k = 20,15

=1,85

So K = 70,79

1.95

2

2.05

2.1

2.15

2.2

2.25

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

logv

log p

freundlich graph of N2

log v

log p 0,544068 1 1,222716 1,409933 1,525045 1,593286

log v 2,004321 2,133539 2,184691 2,209515 2,217484 2,220108


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c.)Calculate v at 7.0 atm using both the Langmuir and Freundlich isoterm!

Langmuir isoterm

= ap + B where B = 0,0346535(0,0059 3,5)7 atm/v= 0,0059 (7) + 0,014 B = 0,014

V = 7/0,0553

= 132 cm3/g

Freundlich isoterm

Log V = log k +

log p

Log v = 1,85 + 0,28 (0,845)

Log v = 185 +0,23

Log v = 2,08

V = 120,2 cm3/g

uas 15b 1201507 anelin osirikna

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