202 commettie presents the model answer of the ervision questions 2012-2013
Model Answer
Revision Questions
12/27/2012
202committee
KAU
202committee | Error! No text of specified style in document.
0
202 committee at KAU / Model answers of the revision questions
(1) Actual mass = 24.986 g/mol
25
Mg12
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) Go rxn= Go products - Go reactants
Horxn = Ho products - Ho reactants
Go rxn = -301.25 kJ/mol – (-334.34 kJ/mol)
= 33.09 kJ/mol
o
H rxn = -313.8kJ/mol – (-394.34)kJ/mol
= 80.54 kJ/mol
b) Non spontaneous because Go rxn has a positive value (more than zero)
c) Go rxn = -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) ‫تم تحديد األنود من الكاثود بالنظر الي قيم جهد اإلختزال المعطي‬
‫ يكون له أعلي قيمة سالبة أو أقل قيمة موجبة وبالتالي يكون اإلختيار لل‬:‫األنود‬
Fe3+
‫ لكتابة المخطط التعبيري‬Cell notation
‫يتم البدء باألنود ثم خطي القنطرة الملحية ثم الكاثود‬
a) Fe2+/ Fe3+ // Ag+/Ag
b) Fe2+ + Ag+
Fe3+ + Ag
c) Eo cell = Eo cathode – Eo anode
= +0.799 – 0.770 = 0.029 V
o
d) G = -n F Eo cell
= - 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
1
202 committee at KAU/ Model answers of the revision questions
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 = Eo cell - 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) Al2O3 = 2Al3+ + 3O2For two moles of Al = 6 eThen one mole of Al = 3 e96500 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
2
202 committee at KAU/ Model answers of the revision questions
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.
 Go = -RT ln Kp
(12)
Ho - TSo = -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
‫كلفن‬/‫هناك خطأ في وحدة األنتروبي في المسألة برجاء اإلنتباه والتصحيح هو أن جميع القيم مقاسة بوحدة الجول‬
(13)
‫مول‬
Hº = 33.85 – [90.4 + 163.4] = -219.95 kJ/mol
Sº = [240.5 + 205] – [110.9 +210.6]= -124 J/K mol
-3
 Go = Hº - T Sº = -219.95 – [298 x -124 x10 ] = -182.99 kJ/mol
(14)
(15)
(i) % mass = [ mass of KNO3/ mass of KNO3 + mass of water ]x 100
= [45 /45 + 295 ] x 100 = 13.23 %
(ii) Molality = n KNO3 / Mass of H2O (kg) = [45/ (39.098 + 14+ 3x16)] / 0.295 Kg
= 1.51m
Go f NH3(g) = -16.6 kJ/mol
We have two moles of ammonia; therefore the Go reaction is equal to 2 x -16.6
= -33.2 kJ/mol
Go = -RT ln Kp
-33.2 x 1000 = -8.314 x 298 ln Kp
Kp = e 13.40 = 6.6 x 105
(16) nA = mass of benzene / molar mass of benzene = 6.5 / (12x6 +6x1) = 0.0833 moles
nB = mass of toluene / molar mass of toluene = 23 / (12x7 + 8) = 0.25 moles
XA = nA/nA+nB = 0.0833/0.0833 + 0.25 = 0.25
XB = nB/nA+nB = 0.25/0.0833+0.25 = 0.75
P total = XA PA + XBPB
= 0.25 molx 0.51 atm + 0.75 mol x 0.18 atm
= 0.263 mol. atm
Eo cell = Eo cathode – Eo anode
(17)
Eo cell = 1.4 V – (-2.866 V)
= 4.266 V
3
202 committee at KAU/ Model answers of the revision questions
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 x10-3x298 = -197.52
4
202 committee at KAU/ Model answers of the revision questions