Posted by Kyo on November 03, 2007 at 17:09:31:
In Reply to: (04 Nov) Qns posted by Kyo on November 03, 2007 at 08:44:37:
Ans to Q33
The combustion equations are :
CH4 + 2O2 --> CO2 + 2H2O
and
2C2H6 + 7O2 --> 4CO2 + 6H2O
From info given in qn and stoichiometry, we have
Note : PartialPressure = P( )
P(CH4) + P(C2H6) = 294 mmHg (considering methane and ethane gases ~ ~ ~ equation 1)
1xP(CH4) + 2xP(C2H6) = 356 mmHg (considering cabon dioxide ~ ~ ~ equation 2)
Solving simultaneous equations, we find
P(C2H6) = 62 mmHg
P(CH4) = 232 mmHg
Since partial pressure is directly proportional to no. of mol (provided all other variables in PV = nRT remain constant, as is so in this question), hence
% of CH4 = (232/(232+62)) = 78.9%
% of C2H6 = (62/(232+62)) = 21.1%
Ans to Q34
A is para-(propan-1-one)toluene or 3-(para-toluene)-propan-1-one
B is para-(propan-1-ol)toluene or 3-(para-toluene)-propan-1-ol
C is para-(1-bromopropane)toluene or 3-(para-toluene)-1-bromopropane
D is para-(butan-1-nitrile)toluene or 3-(para-toluene)-butan-1-nitrile
E is para-(butanoic acid)toluene or 4-(para-toluene)-butan-1-oic acid
F is para-(butanoyl chloride)toluene or 4-(para-toluene)-butanoyl chloride
Mechanism from F to G is "intramolecular electrophilic aromatic substitution" (see Q38 for mechanism details).
Ans to Q35
a)
At pH 10, no. of mol of OH- = 1 x 10^-6
At pH 9, no. of mol of OH- = 1 x 10^-7
Hence, no of mol of OH- to be removed = no of mol of H+ to be added = 0.5 x no of mol of H2SO4 to be added.
Mass of H2SO4 = 4.41 x 10^-5 g
b)
CH3COO- + H+ <--> CH3COOH
Initial (mol) | 1.0 | 0.1 | 1.0
Change (mol) | -0.1 | -0.1 | +0.1
Final (mol) | 0.9 | 0 | 1.1
Above details neutralization process (between ethanoate base and hydrochloric acid) which occurs first.
Below details dissociation of protons from weak acid ethanoic acid, which occurs next.
CH3COOH <--> H+ + CH3COO-
Initial (molarity) | 1.1 | 0 | 0.9
Change (molarity) | -x | +x | +x
Equilibrium (molarity) | 1.1-x | x | 0.9+x
Ka = 1.8 x 10^-5 = ([H+][CH3COO-])/[CH3COOH) = [(x)(0.9+x)]/(1.1-x)
Approximating 0.9+x = 0.9 and 1.1-x = 1.1, we solve for x.
x = [H+] = 2.2 x 10^-5 mol/dm3
pH = -log(2.2 x 10^-5) = 4.66
c) (i)
No. of mol of NaOH = 1 x 10^-3 mol
No. of mol of CH3COOH = 2.5 x 10^-3 mol
CH3COOH + NaOH --> CH3COONa + H2O
Initial (mol) | 2.5 x 10^-3 | 1 x 10^-3 | 0 | (heck care)
Change (mol) | -1 x 10^-3 | -1 x 10^-3 | +1 x 10^-3 | (heck care)
Final (mol) | 1.5 x 10^-3 | 0 | 1 x 10^-3 | (heck care)
Ka = 1.8 x 10^-5 = [H+][CH3COO-] / [CH3COOH] = [H+](1.5 x 10^-3) / (1 x 10^-3)
Solve for [H+], we find [H+] = 2.7 x 10^-5 mol/dm3
Hence, pH = -log(2.7 x 10^-5) = 4.57
c) (ii)
No. of mol of NaOH = 2.5 x 10^-3 mol
No. of mol of CH3COOH = 2.5 x 10^-3 mol
Hence, at equivalence point, molarities of both acid and alkali is 0.0 mol/dm3. No. of mol of CH3COONa salt is 2.5 x 10^-3 mol. Molarity of CH3COONa salt is (2.5 x 10^-3)/((25+25)/1000) = 0.05 mol/dm3.
Since we know want to investigate the hydrolysis of the CH3COO- ions, that is to say, the uptake of H+ protons from water by the ethanoate ions, we now have to use Kb instead of Ka.
CH3COO- <--> OH- + CH3COOH
Initial (molarity) | 0.05 | 0 | 0
Change (molarity) | -x | +x | +x
Equilibrium (molarity) | 0.05-x | +x | +x
Kb = 5.6 x 10^-10 = [OH-][CH3COOH] / [CH3COO-] = x^2 / (0.05 - x)
Approximating 0.05-x = x, we solve for x which is [OH-], and find that [OH-] = 5.3 x 10^-6 mol/dm3. Hence pH = 8.72
c) (iii)
No. of mol of NaOH = 3.5 x 10^-3 mol
No. of mol of CH3COOH = 2.5 x 10^-3 mol
CH3COOH + NaOH --> CH3COONa + H2O
Initial (mol) | 2.5 x 10^-3 | 3.5 x 10^-3 | 0 | (heck care)
Change (mol) | -2.5 x 10^-3 | -2.5 x 10^-3 | +2.5 x 10^-3 | (heck care)
Final (mol) | 0 | 1 x 10^-3 | 2.5 x 10^-3 | (heck care)
Notice that two species are present that are responsible for making the solution alkaline : OH- and CH3COO-. However, because OH- is a much more powerful base (and as predicted by Le Chatelier's principle, the presence of excess OH- taking up protons will suppress the capacity of CH3COO- to take up protons, ie. suppress the the hydrolysis of ethanoate ion), hence we need only consider the molarity of OH- ions when calculating pH.
[OH-] = (1 x 10^-3)/((25+35)/1000) = 0.0167 mol/dm3
pOH = -log[OH-] = 1.78
pH = 14 - 1.75 = 12.22
Ans to Q36
a)(i)
Because AgNO3 is a soluble strong electrolyte, it dissociates complete to form 6.5 x 10^-3 mol/dm3 of Ag+ (participating) ions and 6.5 x 10^-3 mol/dm3 of NO3- (spectator) ions.
Let x be the molar solubility of AgCl in AgNO3 solution.
AgCl (s) <--> Ag+ (aq) + Cl- (aq)
Initial (molarity) | (heck care) | 6.5 x 10^-3 | 0
Change (molarity) | -x | +x | +x
Equilibrium (molarity) | (heck care) | 6.5 x 10^-3 + x | x
Ksp = 1.6 x 10^-10 = [Ag+][Cl-] = (6.5 x 10^-3 + x)(x)
Because AgCl is mostly insoluble and the presence of Ag+ ions from AgNO3 further lowers the solubility of AgCl (as predicted by Le Chatelier's principle), x is very small compared with 6.5 x 10^-3. Hence we apply the approximation (6.5 x 10^-3 + x) = 6.5 x 10^-3.
So we have 1.6 x 10^-10 = (6.5 x 10^-3)(x)
x = 2.5 x 10^-8 mol/dm3
Hence, solubility (g/dm3) = molar solubility x molar mass = (2.5 x 10^-8) x (143.5) = 3.59 x 10^-6 g/dm3.
a) (ii)
Ammonia reacts with water to produce OH- ions, which in sufficient molarity, combines with Fe2+(aq) to form Fe(OH)2 (s) precipitate.
First step is to ascertain the molarity of OH- needed to precipitate Fe(OH)2 (s).
Ksp = 1.6 x 10^-14 = [Fe2+]x[OH]^2 = (0.003)[OH-]^2
Hence [OH-] = 2.3 x 10^-6 mol/dm3
Second step is to calculate the molarity of NH3(aq) required to produce 2.3 x 10^-6 mol/dm3 of OH- ions. We use an ICE table :
NH3 (aq) + H2O(l) <--> NH4+ (aq) + OH- (aq)
Initial (molarity) | x | (heck care) | 0 | 0
Change (molarity) | - (2.3 x 10^-6) | (heck care) | +2.3 x 10^-6 | +2.3 x 10^-6
Equilibrium (molarity) | x - (2.3 x 10^-6) | (heck care) | 2.3 x 10^-6 | 2.3 x 10^-6
Here, we must consider Kb, since we're concerned with the extent of uptake of protons by the base NH3.
Kb = 1.8 x 10^-5 = [OH-][NH4+] / [NH3] = (2.3 x 10^-6)(2.3 x 10^-6) / (x - (2.3 x 10^-6))
Solving for x, we obtain x = 2.6 x 10^-6 mol/dm3.
Hence the molarity of NH3 must be GREATER than 2.6 x 10^-6 mol/dm3 in order to precipitate Fe(OH)2 (s) from a 0.003 mol/dm3 solution of FeCl2 (aq).
b)(i)
Mg(OH)2 (s) <--> Mg2+ (aq) + 2OH- (aq)
Let the molar solubility of MgOH2 (ie. [Mg2+] or 1/2[OH-] in saturated soln) be x.
Ksp = 2.0 x 10^-11 = [Mg2+][OH-]^2 = (x)(2x)^2
Ksp = 2.0 x 10^-11 = 4x^3
[Mg2+] = x = 1.71 x 10^-4 mol/dm3
b)(ii)
% left in solution = (1.71 x 10^-4)/(0.054) = 0.3167%
Hence % extracted = (100 - 0.3167) = 99.683%
Ans to Q37
Comment : notice that no. of mol (hence pressure at constant volume) on LHS is equal to no. of mol at RHS. Which means that no. of mol of gas at equilibrium = no. of mol of gas at initial conditions. Which also means that as long as volume is kept constant (which is usually the case under industry conditions, and for Kp equilibrium problems), equilibrium pressure = initial pressure. Temperature, of course, is always kept constant in this context (Ksp values are only constant for specific temperatures).
Alternative Working #1
H2 + I2 <--> 2HI
Initial (Pa) | 3/5x | 2/5x | 0
Change (Pa) | -0.42(3/5)x | -0.63(2/5)x | + 2(0.252)x
Equilibrium (Pa) | 0.58(3/5)x | 0.37(2/5)x | 0.504x
Total pressure of gas at equilibrium = 0.348x + 0.148x + 0.504x = x
Since the total pressure at equilibrium is given to be 5x10^7 Pa, x = 5x10^7 Pa
Hence, Equilibrium (Pa) | 17,400,000 | 7,400,000 | 25,200,000
Kp = P^2(HI) / P(H2) x P(I2) = 4.931966 = 4.93
Alternative Working #2
H2 + I2 <--> 2HI
Initial (mol) | 3x | 2x | 0
Change (mol) | -0.42(3)x | -0.63(2)x | + 2(1.26)x
Equilibrium (mol) | 0.58(3)x | 0.37(2)x | 2.52x
Total no. of mol of gases at equilibrium = 5.0x mol
Partial pressure = P( )
P(HI) = (2.52/5)(5 x 10^7) = (0.504)(5 x 10^7)
P(H2) = ((0.58)(3)/5)(5 x 10^7) = (0.348)(5 x 10^7)
P(I2) = ((0.37)(2)/5)(5 x 10^7) = (0.148)(5 x 10^7)
Kp = P^2(HI) / P(H2) x P(I2) = 4.931966 = 4.93
b)
Qc = [HI]^2 / [H2][I2] = (0.1)^2 / (0.4)(0.3) = 0.083
Since Qc < Kc, the position of equilibrium lies to the right, and the net reaction will proceed to the right (ie. reaction will continue to be product favoured) until equilibrium is achieved. Hence rate of forward reaction > rate of backward reaction, until equilibrium is achieved.
H2 + I2 <--> 2HI
Initial (mol) | 0.4 + x | 0.3 | 0.1
Change (mol) | -0.025 | -0.025 | +0.05
Equilibrium (mol) | 0.375+x | 0.275 | 0.15
Given Kc = 0.191 = [HI]^2 / [H2][I2] = (0.15)^2 / (0.375+x)(0.275)
x = 0.0533674 = 0.0534 mol
Ans to Q38
List of Mechanisms for common reactions at 'A' Level Organic Chemistry :
http://infinity.usanethosting.com/Tuition/OrganicChemistry_Mechanisms/index.htm