Download CHM_101_ASSIGNMENT_COPY_1_2

NAME: AKINPELUMI SAMUEL .O.
DEPARTMENT: ANATOMY
COLLEGE: MEDICAL AND HEALTH SCIENCE
CHEMISTRY 101
15/MHS03/004
1. Calculate the change in PH obtained on the addition of 0.03 mole of
solid NaOH to a buffer solution that consists of 0.15M sodium
acetate and 0.15M acetic acid solution, if we assume that there is
no change in volume (Ka = 1.8 x 10-5).
2. (a) The rate constant of a first order reaction is 2.5 ×10 -6/s and the
initial concentration is 0.1moldm-3, what is the initial rate in
moldm-3s-1.
(b) The initial rate of a second order reaction is 5.0×10-7moldm-3s-1, and
the initial concentrations of the two reacting substances are each
0.2moldm-3.What is the rate constant in dm3mol-1s-1?
(c) The first order rate constant for the decomposition of a certain
antibiotic in water at 250C is 1.30year-1. (i) If a 5.0×10-3moldm-3 solution
of this antibiotic is stored at 250C for 30 days, what is the concentration
of the antibiotic? (ii) How long will it take for the concentration of the
antibiotic to reach 1.0×10-3 moldm-3?
3. A reaction mixture at 800K consists of 0.65 M H2, 0.52M, N2 and 0.24
M NH3. Calculate the concentration reaction quotient QC at this
temperature for the formation of Ammonia.
4. At 500K 1 mole of NOCl was found to be 9.0% dissociated in a 1.00L
vessel.
2NOCl (g)
2NO (g) +Cl2 (g)
Calculate the pressure equilibrium constant Kp for the reaction at this
temperature. What is the partial pressure of chlorine in the vessel?
5. Write the expressions for the concentration equilibrium constant Kc
and pressure equilibrium constant Kp for the following reactions:
a)
NH4Cl(s)
<------> NH3 (g) + HCl (g)
b)
CO2(g) + 2NH3
<------->
CO(NH2)2 + H2O(l)
6. The half-life of 38Cl is 37.2 mins; determine (i) the decay constant
of 38Cl and (ii) the number of atoms of radioactive 38Cl in a sample that
has an activity of 10 µc
7. Differentiate between ionic and covalent compounds.
8. (a) 3.0g of 2-hydroxybenzoic acid is reflux with 6.5g of ethanoic
anhydride to produce aspirin and ethanoic acid.
(I) Which one of the two reagents is the limiting reagent?
(II) How much of excess reagent (in grams) is left at the end of the
reaction?
(b) 130g of ethyne was combusted in the presence of excess oxygen.
Calculate the mass of CO2 produced in the process.
(c) Discuss the three factors affecting ionization energy in the periodic
table.
1. CH3COONa= 0.15(sodium acetate)
CH3COOH= 0.15(acetic acid)
Ka= 1.8 x 10-5
Mole added=0.03
pH= pKa + log [Conjugate base]
[Acid]
= 4.74 + log [0.15]
[0.15]
= 4.74
NaOH + CH3COOH-> CH3COONa + H2O
Before
0.03
0.15
0.15
reaction
Change -0.03
-0.03
+0.03
due to
reaction
After
0
0.12
0.18
reaction
CH3COONa= 0.18M
CH3COOH= 0.12M
pH= pKa+ log [Conjugate base]
[Acid]
= 4.74 + log [0.18]
[0.12]
= 4.74 + 0.18
= 4.92
Change in pH= 4.92 - 4.74= 0.18
2. a) First Order reaction
R = K [A] 1
R = 2.5 × 10-6 × 0.1
R = 2.5 × 10-7mol/dm3s
b) R = K [A] 1[B] 1
Second Order Reaction
K= R ÷ ([A] [B])
K = 5.0 × 10-7 ÷ (0.2 × 0.2)
K = 5.0 × 10-7 ÷ (0.04)
K = 1.25 × 10-5
c)
K= 1 ln a
t (a-x)
4.12 × 10-8 = (2,592,000)-1 × (log a ÷ (a-x))
(ln a ÷ (a-x)) = 4.12 × 10-8 × 2,592,000
(ln 0.005 ÷ (0.005-x) = 0.1068
(-2.9957 ÷ (0.005-x) = 0.1068
0.005 – X = -2.9957 ÷ 0.1068
X = -28.0496 - 0.005
X = -28.0546 ÷ -1
X= 28.0546mol/dm3
3. N2 + H2 NH3
[N2]= O.52M
[H2] = 0.65M
[NH3] = 0.24M
QC = [NH3]
[H2] [N2]
QC = [0.24]
[0.65] [0.52]
QC = [0.24]
[0.338]
QC = 0.71
4. 2NOCl (g) 2NO+(g) + Cl2-(g)
Recall:
n = CV
C=n÷v
C=1÷1
C=1
But only 9% of NOCl dissociated which is equal to
9% of 1 = 0.09
2NOCl(g) <-> 2NO+(g) + Cl2-(g)
1M
<-> 2(0.09)
0.09
KC = [NO+]2 [Cl2-]
[NOCl]2
KC = (0.18)2(0.09) ÷ 12
KC = 0.002916 ÷ 1
KC = 0.002916
Recall:
Where Kc= 0.002916, R= 8.314, T= 800, where (c+d) (a+b) are the order of reaction
KP = KC RT (c+d) – (a+b)
KP = 0.002916 × 8.314 (800) (2+1) – (2+0)
KP = 0.002916 × 8.314×800
KP = 19.395
b) pCl = [Cl]RT(C+D)-(a+b)
pCl = 0.09 × 8.314 × 800
pCl = 598.608 Pa
5. NH4Cl(s) <-> NH3 (g) + HCl (g)
Kc = [NH3] [HCL]
[NH4Cl]
KP = PNH3 × PHCl
PNH4Cl
CO2 (g) + 2NH3 (g) <-> CO (NH2)2(g) + H2O (l)
Kc = [CO (NH2)2] [H2O]
[CO2] [NH3]2
Kp = PH2O* PCO (NH2)2
PCO2 × P (NH3)2
6. Data:
= 2,232
Recall:
K = 3.105 × 10-4
b)
Recall
= -KN
N = ÷ -K
N = -0.0322
7.
Electrovalent Compounds
Covalent compounds
Involves transfer of electrons from metal atom to a
Involves sharing of electrons either between
non-metal atom.
atoms of same or different element.
Formed between metals
Formed between non-metals
High boiling and melting point
Low melting point
Soluble in polar solvents like water
Soluble in non-polar solvents
They conduct electricity in molten state
Most covalent compounds do not conduct
electricity.
8. a)
(i) C7H6O3 + C4H6O3 C9H8O4 + CH3COOH
nC7H6O3 = = 0.02172mol
nC4H6O3 = = 0.06367mol
Mole ratio= 1
From the values given,
= 2.931
Since 2.9 is greater than 1, nC4H6O3 is in excess of nC7H6O3
Hence, C7H6O3 is the limiting reagent
(ii)
Excess nC4H6O3 = (2.931- 0.02172) mole
= 2.9mole
Excess C4H6O3 = 2.9 × 102.09 = 296.061g
b)
2C2H2 +5O2 4CO2 + 2H2O
nC2H2 = 5
Molar ratio of C2H2 to CO2 is 2: 4
Therefore
nCO2 = 10
But mass = molar mass × number of moles of CO2
Mass of CO2 = 44 × 10 = 440g
c) Factors affecting Ionization Energy
1. Distance of outermost electron from the nucleus: Across the period, as atomic
number increases, atomic radius decreases. As the distance decreases, the
attraction of the positive nucleus for the electron will increase. More energy is
needed to remove the outermost electron, thus the ionization energy increases.
2. Size of the positive nuclear charge: As the nuclear charge increases, its
attraction for the outermost electron increases, and so more energy is required
to remove the outermost electron. Hence, the ionization energy increases.
3. Shielding and Screening effect of the inner electrons: Down a group, the
shielding of outer electrons by inner electrons overcomes the influence on the
increasing nuclear charge, thus the outer electron is screened or shielded from
the nucleus by the repelling effect of the inner electrons. Across the group, the
reverse is the case; the increasing nuclear charge has greater effect. In general,
the screening effect by inner electrons is more effective, the closer they are to
the nucleus.