Download chemisty_ass_2

NAME: AMODU OLUWATOYIN ABIGAIL.
COLLEGE: MEDICINE AND HEALTH SCIENCES.
DEPARTMENT: MEDICINE AND SURGERY.
COURSE CODE: CHEMISTRY 101.
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=1.8×10-5).
SOLUTION:
NaOH + CH3COOH → CH3COONa +H2O
0.03
0.15
0.15 (before reaction)
- 0.03 0.030.03+ (change due to reaction)
0
0.12
0.18
0.15M of CH3COONa
0.15M of CH3COOH
0.03 mole of NaOH
PH = pKa + log [conjugate base]/[Acid]
PKa = -log Ka
= - log 1.8×10-5
= 4.74.
PH =4.74 + log (0.18)/(0.12)
= 4.92
ΔH = 4.92-4.74
= 0.18.
2(a). the rate constant of a first order reaction is 2.5×10-6/s and the initial concentration is 0.1
moldm-3, what is the initial rate in moldm-3s-1.
K =2.5×10-6/sec
Concentration = 0.1moldm-3
Initial rate(R) =?
R=K[]
2.5×10-6 × 0.1
= 2.5× 10-7moldm-3s-1.
2(b). The initial rate of a second order reaction is 5× 10-7moldm-3s-1, and the initial
concentration of the two reacting substances are each 0.2moldm-3. What is the rate constant in
dm3mol-1s-1?
R2 = 5.0 × 10-7moldm-3s-1
C1 =0.2 moldm-3
C2 = 0.2 moldm-3
R=K[][]
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 × 10-5moldm-3s-1.
2c. ki= 1.30/year
a=5x10^-3moldm-3
t= 30 days= 0.082years.
R=k1(a-x)
in a/a-x=k1t
2.303loga/a-x=k1t
loga/a-x= 1.30x0.082/2.303
loga/a-x=4.63x10^-2
a/a-x= antilog of 4.63x10^-2
a/a-x= 1.112
5x10^-3/a-x =1.112
a-x = 4.5x10^-3moldm^-3
ii. a=5x10^\3
a-x=1x10^-3
t=?
2.303log a/a-x =k1t
2.303x log 5x10^-3/1x10^-3=130t
2.303log5=130t
1.61= 1.30t
t= 1.61/1.30
t=1.23year-1
3.
5.0 × 10-7 = K × 0.2 × 0.2
K= 1.25× 10-5moldm-3sec-1.
3H2 + N2 → 2NH3
Kc =[NH3]/[H2]^3 [N2]
= [0.24]^2/[0.65]^3 [0.52]
Qc = 0.4033
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: kp=kcRT^(c+d)-(atb)
Kp = 0.002916×8.314×(500)^2+1-2
Kp =12.12
4b. pCl = [Cl2]RT^(c+d)-(a+b)
= 0.09×8.314×500
= 374.13pa.
5a. NH4Cl → NH3 +HCl
KC = [NH3] [HCl]/[NH4Cl]
Kp=p[MH3] p[HCl]/P[NH4Cl]
B. CO2 +2NH3 → CO(NH2)2 +H2O
KC = [CO(NH2)2] [H2O]/ [CO2] [NH3]^2.
6a. t1/2 37.2mins (37.2 × 60)
= 2232 secs.
Decay constant(K) ?
t1/2 = 0.693/k
k = 0.693/t1/2
k = 0.693/ 2232
=3.1 × 10^-4s-1.
6b. k = 10µc
= 10 × 10^-6
No of atoms(n) =?
Avogadro’s mole constant = 6.02 ×10^23.
N=k× Avogadro’s constant
N =10× 10^-6 ×6.02 ×10^23
= 6.02 ×10^18.
7. DIFFERENCE BETWEEN IONIC AND COVALENT BONDING
IONIC BONDING
They are formed by sharing of electrons.
Bonds are formed usually between non-metals
They are insoluble in polar solvent
They do not dissociate into ions when dissolved in
water
They have low melting and boiling point
COVALENT BONDING
They are formed by transfer of electrons
Bonds are formed between metals and non-metals
They are soluble in polar solvent
They dissociate into ions when dissolved in water
They are of high melting and boiling point.
8. 2hydroxybenzoic acid + Ethanoic anhydride → aspirin + ethanoic acid
3.0g of hydroxybenzoic acid → 6.5g of anhydride
Before reaction 3.0g → 6.5g of anhydride
After reaction 3.0- 3.0 → 6.5- 3.0
0g
3.5g
A. the limiting agent is hydroxybenzoic acid.
B. 3.5g of the reagent was left behind after the reaction.
b. 2C2H2 + 5O2 → 2H20 +4CO2.
2 mole of C2H2 = 4 mole of CO2.
Number of moles of C2H2 combusted = 130/52
= 2.5
Numbers of moles of CO2 produced =
Mass of C02 produced/mass of C02 =2.5
Mass of CO2 produced = 176×2.5
=440g.
8c.(i). 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 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.
ii. 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.
iii. 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.