Download CHM_101_TUTORIAL_QUESTIONS_1

NAME: Bitrus Romanus Kura
DEPARTMENT: Chemical Engineering
COLLEGE: ENGINEERING
COURSE CODE: CHM 101
1. Required equation;
CH3COO- + Na+
CH3COONa
0.15M
0.15M
0.15M
CH3COO- + H3O
CH3COOH + H2O
0.15M
0.15M
From Henderson-Hasselbatch Equation
PH =pKa +log [Conjugate base]
[ Acid]
Where:
[Conjugate base]= [CH3COO-] = 0.15M
[ Acid] = [CH3COOH] = 0.15M
PH1= 4.744 + log (0.15 ÷ 0.15)
PH1 = 4.744 + log 1
PH1= 4.744 + O
PH1 = 4.74
On addition of 0.03 mole of NaOH the acid absorbs the hydroxide ion
CH3COOH + OH0.15
CH3COO- + H2O
0
- 0.O3
0.15
+0.03
0.12
0.18
Therefore;
PH2= pKa + log (0.18 ÷ 0.12)
PH2 = 4.74 + log (1.5)
PH2 = 4.74 + 0.176
PH2 = 4.916
Therefore
Change in PH = PH2 – PH1
Change in PH = 4.916 – 4.74
Change in pH = 0.176
a) For a first order reaction
R = K[A]x
R = 2.5 × 10-6 × 0.1
R = 2.5 × 10 -6 mol/dm3s
B) R = K[A]x[B]y
FOR A 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 -5dm3 mol -1s -1
C) K= t-1
ln a
(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
OR
i)K=1.30y-¹= 1.30÷(365x24x360)
K=4.12x10-8sx259200
Initial concentration=5.0x10-3
Time t=30days ;30x24x360=2592000s
K=1÷t·ln(a÷a-x)
In(a÷a-x)=kt=4.12x10-8x259200s
In(a÷a-x)=0.1068
a÷(a-x)=e0.1068=1.113
a=5.0x10-3mol/dm3
5.0x10-3÷5.0x10-3–x=1.113
5.0x10-3=1.113(5.0x10-3–x)
5.0x10-3=5.656x10-4-1.113x
x=0.565x10-4÷–1.113=5.08x10-4
Conc of antibiotic=5.0x10-3–5.08x10-4
Conc of antibiotic=4.49x10-8mol/dm3
ii)K=1÷t . ln (a÷a-x)
t=1÷k.ln (a÷a-x)
t=1÷4.12x10-8·ln5.0x10-3÷1.0x10-3
t=39,064.027s
t=39,064.027s
3) Required Equation:
N2 + 3H2
2NH3
[N2]= O.52M
[H2] = 0.65M
[NH3] = 0.24M
QC = [NH3]²
[H2]³ [N2]
QC = 0.24² ÷ (O.65)³ × (0.52)
QC = 0.403
4) 2NOCl(g)
Recall: n = CV
C=n÷v
C=1÷1
2NO(g)+ Cl2-(g)
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:
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 = 598.608pa
5a
NH4Cl(S)
KC = [NH3] [HCl]
KP = PNH3 × PHCl
NH3(g) + HCl(g)
CO2(g) + 2NH3(g) CO(NH2)2(g) + H2O(l)
Kc = [CO(NH2)2]
[CO2] [NH3]2
Kp = PCO(NH2)2
PCO2 × P(NH3)2
6)
Data:
= 2,232
Recall:
=
K=
K=
K = 3.105 × 10-4
b)
Recall
= -KN
N = ÷ -K
N=
N = -0.0322
7)
Covalent compounds
Ionic compounds
Formed through sharing of electrons
Formed through transferring of electrons
Are formed between atoms of nonmetals
Formed between atoms of metals and nommetals
Has low melting and boiling point
Has high melting and boiling point
Does not conduct electricity in molten or aqueous
form
They conduct electricity in molten or aqueous
form
8)
C7H6O3 + C4H6O3 C9H8O4 + CH3COOH
nC7H6O3 = = 0.02172mol
nC4H6O3 = = 0.06367mol
From the equation the mole ratio
= =1
But from the values given,
= = 2.931
Since 2.9 is > 1, nC4H6O3 is in excess of nC7H6O3
Therefore C7H6O3 is the limiting reagent
b) Excess nC4H6O3 = (2.931- 0.02172) mole
= 2.9mole
Excess C4H6O3 = 2.9 × 102.09 = 296.061g
b)
2C2H2 +5O2 4CO2 + 2H2O
Recall:
n=
nC2H2 = = 5
From the equation the molar ratio of C2H2 to CO2 is 2: 4
Therefore
nCO2 = 10
But mass = molar mass × nCO2
Mass of CO2 = 44 × 10 = 440g
c)
Factors affecting the ionization energy.
1. Atomic size: In small atoms electrons remains closer to nucleus and they feel more nuclear
attraction. So more ionization energy is required to remove electron from small atoms. While in big
sized atoms, valence electrons are away from nucleus , so they experience less nuclear attraction. Hence
it requires less ionization energy to remove electrons from big atoms.Therefore, Ionization energy is
inversely proportional to Atomic size.
2. Nuclear charge: By increasing the nuclear charge electrons feel more nuclear attraction.
Hence more ionization energy is required.Therefore, Ionization Energy is directly proportional to Nuclear
charge.
3. Penetration Power: Tendency of becoming nearer to the nucleus is called penetration
power.The order of penetration power of different sub-shells - s > p > d > f.Therefore, Ionization energy
is Directly proportional to Penetration Power.
4. Stability: In stable configuration we require more energy to release the electron as compared
to non stable configuration.Therefore, Ionization energy is directly proportional to Stability.Ionization
Energy is more of full-filled shell as compared to half-filled shell.
5. Screening & Shielding effect: Presence of other orbits between nucleus and last orbit
decreases the nuclear attraction. This effect is called screening effect but electron-electron repulsion is
called shielding effect which also decreases the nuclear attraction. Due to presence of these effects
ionisation energy decreases.
Therefore Screening and shielding effects are directly proportional to ionization energy.IONIC BOND
COVALENT BOND
It involves complete transfer
of electrons between the elements
It involves sharing of electrons among the elements combining
They conduct electricity in the molten state or in aqueous solution when their oppositely charged ions
are mobile
They do not conduct electricity in molten state because they do not contain ions
They dissolve readily in water (polar solvent) but not in non-polar solvent
They do not dissolve in polar solvents like water but usually dissolve in non-polar solvents
They are generally solids of high melting points because their oppositely charged ions are strongly held
together by electrostatic forces
They are generally gases or volatile liquids or low melting points solids because their molecules are held
together by weak intermolecular forces