Model Answer

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Model Answer

Revision Questions

12/27/2012 202committee KAU

1

202 committee at KAU/ Model answers of the revision questions (1) Actual mass = 24.986 g/mol

25Mg12

Protons = 12 , neutrons = 25-12 = 13

Calculated mass = (protons x mass of proton + neutrons x mass of neutrons) = [ 12 x 1.0078 + 13x 1.0086]

= 25.2054 a.m.u

Converted mass = calculated mass – actual mass = 25.2054 – 24. 986 = 0.2194 a.m.u.

B.E. = m c² = [0.2194 a.m.u. x 1.6605 x 10 ^-27 Kg ] x(3x 10⁸)² = 3.27 x 10^-11 joules

B.E./nucleon = 3.27 x 10^-11 joules/25 (mass number) = 1.3 x10 ^-12 joules/nucleon

(2) a) G^orxn= G^o products - G^o reactants H^orxn = H^o products - H^o reactants G^orxn = -301.25 kJ/mol – (-334.34 kJ/mol) = 33.09 kJ/mol

H^orxn = -313.8kJ/mol – (-394.34)kJ/mol = 80.54 kJ/mol

b) Non spontaneous because G^orxn has a positive value (more than zero) c) G^orxn = -RT ln kp

33.09 x1000 = - 8.314 x 298 ln kp Kp = e 33.09 x1000/ - 8.314 x 298

= 1.58 x 10^-6 (3) يطعملا لازتخلإا دهج ميق يلا رظنلاب دوثاكلا نم دونلأا ديدحت مت

دونلأا : لل رايتخلإا نوكي يلاتلابو ةبجوم ةميق لقأ وأ ةبلاس ةميق يلعأ هل نوكي

Fe³⁺

يريبعتلا ططخملا ةباتكل Cell notation دوثاكلا مث ةيحلملا ةرطنقلا يطخ مث دونلأاب ءدبلا متي a) Fe²⁺/ Fe³⁺ // Ag⁺/Ag

b) Fe²⁺ + Ag⁺ Fe³⁺ + Ag c) E^ocell = E^ocathode – E^o anode

= +0.799 – 0.770 = 0.029 V d) G^o = -n F E^ocell

= - 1 mol of electrons x 96500 C x 0.029 V = -2798.5 kJ/mol (spontaneous)

(4) Q = m x s x T

= (165x 1000 g) x 0.489 J/g.^oC x (95.7 – 15.7 ^oC) = 6454800 Joules (5) (i) First order

Rate = k [A]

0.016 mol L^-1 s^-1 = k x 0.4 mol. L^-1 K = 0.04 s^-1

(ii) Zero Order Rate = k = 0.016 mol L^-1 s^-1

2

202 committee at KAU/ Model answers of the revision questions (7) Mass percent = [mass of solute / mass of solute + mass of solvent] x100

= 25 g urea/ 25 g urea + 75 g water

Number of moles of solute = 25 g/ [12 + 16 + 28 +4] g.mol^-1 = 0.416 moles

Nuber of moles of solvent = 75 g/18 g mol^-1 = 4.16 moles (i) Mole fraction = n solute/ n solute + n solvent

= 0.416 moles/ 0.416 moles + 4.16 moles = 0.09

(ii) Molality = n solute / Mass of solvent (kg) = 0.416 moles / 0.075 kg = 5.546 m (8) Zn + 2Ag⁺ Zn²⁺ + 2Ag

E cell = E^ocell - 0.0257/n ln Q

1.37 V = 1.56 V – 0.0257/2 ln [Zn²⁺]/[ Ag⁺]² ln [Zn²⁺]/[ Ag⁺]² = 14.786

[Zn²⁺]/[ Ag⁺]² = e ^14.786 [0.01]/ [ Ag⁺]² = 26.39 x 10⁵

[ Ag⁺]² = 0.01 / 26.39 x 10⁵ = 3.78 x 10^-9 [ Ag⁺] = 6.15 x 10^-5 M

(9) Molality = n solute/ mass of solvent (kg) m = ? /0.25 Kg

Tf = Kf m

m = Tf/Kf = (33 - 22.8) ^oC/ 29.8 ^oC/m= 0.352 m 0.352 = n solute / 0.25 Kg

n solute = 0.352x 0.25 = 0.088 moles

Molar mass of unknown = mass (g) / n = 13.2 g / 0.088 moles =150.11 g/mol (10) Ln k1/k2 = Ea/R (T1-T2)/(T1xT2)

Ln [

0.286 L mol

^-1

.s

^-1

/ 0.5 L mol

^-1

.s

^-1

] =

Ea/ 8.314 x 10^-3 [ 1373-1473/1373 x 1473]

Ea = Ln [

0.286 L mol

^-1

.s

^-1

/ 0.5 L mol

^-1

.s

^-1

] x 8.314 x 10

^-3

/

[ 1373-1473/1373 x 1473]

= -0.558 x 0.008314/-4.94 x 10^-5 = 39.82 kJ/mol

(11) Al

2

O

3

= 2Al

³⁺

+ 3O

^2-

For two moles of Al = 6 e

^-

Then one mole of Al = 3 e

^-

96500 C x 3 = 1 mole of Al (27 g) 6500 C = x mole ( x g)

Mass (g) = 6500 C x 27 g/289500 C = 0.60 gram ii) 289500 C = 27 g

x = 10 g

3

202 committee at KAU/ Model answers of the revision questions

Q = 10 x 289500 /27 = 107222.22 C

I = 30 A

t = Q /I = 107222.22 /30 = 3574 seconds = 3574/60 = 59.56 min.

(12)

 G^o

= -RT ln K

p

H^o - TS^o = -RT ln Kp

ln Kp = [(-98.2 – 298 x0.0701) / (-8.314 x298 x 10^-3)] = 48.067 Kp = e ^answer = 7.504 x 10²⁰

(13) نفلك / لوجلا ةدحوب ةساقم ميقلا عيمج نأ وه حيحصتلاو هابتنلإا ءاجرب ةلأسملا يف يبورتنلأا ةدحو يف أطخ كانه لوم

Hº = 33.85 – [90.4 + 163.4] = -219.95 kJ/mol

Sº = [240.5 + 205] – [110.9 +210.6]= -124 J/K mol

 G^o

= Hº - T Sº = -219.95 – [298 x -124 x10

^-3

] = -182.99 kJ/mol

(14) (i) % mass = [ mass of KNO

3

/ mass of KNO

3

+ mass of water ]x 100 = [45 /45 + 295 ] x 100 = 13.23 %

(ii) Molality = n KNO

3

/ Mass of H

2

O (kg) = [45/ (39.098 + 14+ 3x16)] / 0.295 Kg = 1.51m

(15) G

^{o f}

NH

³

(g) = -16.6 kJ/mol

We have two moles of ammonia; therefore the  G

^o

reaction is equal to 2 x -16.6

= -33.2 kJ/mol

G

^o

= -RT ln Kp

-33.2 x 1000 = -8.314 x 298 ln Kp Kp = e

^13.40

= 6.6 x 10

⁵

(16) n

A

= mass of benzene / molar mass of benzene = 6.5 / (12x6 +6x1) = 0.0833 moles n

B

= mass of toluene / molar mass of toluene = 23 / (12x7 + 8) = 0.25 moles

X

A

= n

A

/n

A

+n

B

= 0.0833/0.0833 + 0.25 = 0.25 X

B

= n

B

/n

A

+n

B

= 0.25/0.0833+0.25 = 0.75 P

total

= X

A

P

A

+ X

B

P

B

= 0.25 molx 0.51 atm + 0.75 mol x 0.18 atm = 0.263 mol. atm

(17)

E

^ocell

= E

^ocathode

– E

^o anode

E

^o_cell

= 1.4 V – (-2.866 V)

= 4.266 V

4

202 committee at KAU/ Model answers of the revision questions

(18) Total time / half life time = 175/29 =6.03 = 6 periods

1 ½ ¼ 1/8 1/16 1/32 1/64 Remaining mass = 0.015 of the original mass

(19)

 = MRT

= n/v RT

n = 13.7/60 = 0.228

 = 0.228 mol/.5 L x 0.082 L.atm.K

^-1

mol

^-1

x 300 K = 5.6088 atm.

(20) ΔE˚ = ΔH˚-  n RT = - 200.0 kJ - 1 x 8.314 x10

^-3

x298 = -197.52