Review for SCH 4U1 Exam
Unit 1
1. Draw an electron dot diagram for XeF2 and XeF4. Derive the hybridization for each molecule and give the VSEPR shape, formula and name. Give the approximate angles for the F-Xe-F bonds in each molecule.
2. Describe the bonding differences between solid Potassium, Potassium chloride and Sulfur dichloride. Which one would conduct electricity in its solid form and explain why? Which would have the greater water solubility and explain why? Which would have the greater melting point and explain why? Which would be more malleable and explain why
Unit 2
3. Balance the following redox equation for the reaction occurring in an acidic solution.
2- 3+ Cr2O7 (aq) + CH3CH2OH (aq) Cr (aq) + CH3CO2H (aq)
4. For the reaction
M (s) + Cu2+ (aq) M2+ (aq) + Cu (s) E˚ cell = + 0.740 V at 25˚C
a) Determine the standard electrode potential for the reduction half reaction: b) Determine the identity of M. c) Draw the cell diagram.
Unit 3
5. Given the following reaction: C2H6 (g) + 7/2 O2 (g) 2 CO2 (g) + 3 H2O (l)
a) Predict the sign of ∆S for the above reaction. Explain your answer. b) Predict the value of ∆H for the above reaction. Explain your reasoning. c) Calculate the value of ∆H for the above reaction using value in the textbook. d) If the sign for ∆S is positive. What is sign of ∆G? Explain your answer.
6. The reaction 2 X + Y Z was studied and the following data were obtained:
Trial [X] (mol L-1) [Y] (mol L-1) Rate (mol L-1 s-1) 1 3.0 1.5 1.8 2 1.5 3.0 0.45 3 1.5 1.5 0.45
What is the rate law equation? What is the rate constant? Units for the rate constant? Determine a possible reaction mechanism.
Unit 4
7. At 500 K the following equilibrium is established: 2 NO (g) + Cl2 (g) 2 NOCl2 (g). An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl2 and NOCl2 respectively. Calculate Keq for this equation at 500 K.
-7 o o 8. Copper (II) iodide has a Ksp value of 1.08 x 10 at 20 C. Calculate its solubility at 20 C. Calculate its solubility in 1.0 M Cu(NO3)2 solution.
-5 9. Acetic acid, HC2H3O2, which is represented by HA, has an acid ionization constant Ka of 1.8 x 10 . + a) Calculate the [H3O ] in a 0.500 molar solution. b) Calculate the pH and the pOH. c) Calculate the % ionization. d) If 0.400 M sodium acetate is added to the solution, what will be the pH of the new solution? Unit 5
10. Draw and name an example of a 3 carbon compound in each of the functional groups. Place them in order of increasing boiling point.
11. Create N–ethyl–N–methyl 2–methylbutanamide using alkanes and an alkenes.
Try multiple choice at: http://www.tps.k12.mi.us/staff/eferwerda/ap_chemistry_quiz_index.htm Answers
1. a) XeF2 Xe [Kr] AX5 based F 5s2 5p6 Trigonal Xe [Kr] bipyramidal Xe sp3d 5s2 5p5 5d1 actual: Xe [Kr] F 3 0 sp d 5d AX2E3 linear, =180°
b) XeF4 Xe [Kr] AX6 based 6 5s2 F 5p F octahedral 3 2 Xe [Kr] Xe sp d 4 actual: 5s2 F 5p F 5d2 AX4E2 Xe [Kr] Square 3 2 0 sp d 5d planar, =90°
2. Potassium is a metallic crystal. It consists of positive metal kernals surrounded by and attracted to a “sea” of mobile or delocalized valence electrons. These mobile valence electrons conduct the current. This element will react with water, but will not dissolve in it. The packing is not dense and it will have a low melting point. It packing is body centered which is very soft and malleable.
Potassium chloride is an ionic crystal. The potassium has transferred its valence electron to chlorine creating oppositely charged ions. The opposite charges hold the ions rigidly in place preventing their move- ment. Since the electrons are also localized, there are no mobile charges, ions or elec- trons. Therefore there is no electrical conductivity.
This ionic compound has very high water solubility as the ions can be separated by H2O molecules. The ionic bond is very strong and this compound will have the highest melting point. It is brittle due to repulsions of like ions as the planes move over one another and split apart.
Sulfur dichloride is a molecular crystal. This compound is polar covalent with localized bonds and dipole - dipole intermolecu- lar forces. Due to the localized bonds, there are no mobile electrons and therefore no electrical conductivity. It will be water soluble due to the polar nature of the molecule. It is softer than an ionic compound as the weaker intermolecular forces allow for a little movement of neighboring molecules in the solid. The melting point will be the lowest due to the weaker intermolecular forces versus in- terparticle forces of the other two. Due to the weak intermolecular forces, it will be slightly “soft” but not very malleable. 3. Cr2O72- (aq) + CH3CH2OH (aq) Cr3+ (aq) + CH3CO2H (aq) +6 -1 +3 +3 - ox: 3(CH3CH2OH CH3CO2H + 4 e ) - red: 2(Cr2O7 2- + 6 e 2 Cr 3+ ) 2- + 3+ 2 Cr2O7 + 3 CH3CH2OH + 16 H 4 Cr + 3 CH3CO2H + 11 H2O
4. (a) Cu2+(aq) + 2e- Cu(s) E˚ = + 0.340 V
(b) M(s) M2+(aq) + 2e- E˚(ox) = ? V Cu2+(aq) + 2e- Cu(s) E˚ = + 0.340 V Cu2+(aq) + M(s) M2+(aq) + Cu(s)E˚ cell = + 0.740 V E˚(ox) = + 0.740 V - + 0.340 V = + 0.400 V M2+(aq) + 2e- M(s) E˚ = -0.400 V
M is Cd.
(c)
5. a) Entropy has a negative sign, because you are going from a gas to a liquid (more order). b) Enthalpy is negative, or exothermic reaction. Combustion reactions are exothermic and give off heat.
c) ∆H = ∆Hf(products) - ∆Hf(reactants) = (2 mol (-393.5 kJ/mol) + 3 mol (-285.8 kJ/mol)) – (1 mol (-84.7 kJ/mol) = -1 559.7 kJ d) Free energy will be negative at all temperatures. Both entropy and enthalpy are favorable.
6. a) Using Trial 3 and 2 m m rate2 [Y2 ] 0.45 [3.0] m m then m so 1 2 and m 0 rate wrt Y is zero order rate3 [Y3 ] 0.45 [1.5] Using Trials 1 and 2 1.8 [3.0]n so 4 2n and n 2 rate wrt X is second order 0.45 [1.5]n Rate k[X]2[Y]0 k[X]2 b) & c) Using Trial 1 Rate Rate k[X]2 ; k [X]2 1.8 mol L-1 s-1 k 0.20 mol-1 Ls-1 (3.0 molL-1)2
d) X + X X2
X2 + Y X2Y Let Z = X2Y 2X + Y Z
7. P C = ; R = 0.0821 atmLmol-1 K-1 RT RT = 0.0821 atmLmol-1 K-1 x 500 K = 41.05 atmLmol-1
[NO] = 0.002314 mol/L; [Cl2 ] = 0.004167 mol/L and [NOCl2 ] = 0.006821 mol/L 2 2 [NOCl2 ] (0.006821) Keq 2 2 2085.2 [NO] [Cl2 ] (0.002314) (0.004167) OR 2 2 pNOCl2 (0.28) Kp 2 2 50.8 pNO pCl2 (0.095) (0.171) n Keq = Kp RT where n # moles reactants - # moles products (3-1) Keq = 50.8 x 41.05
Keq = 2085.3
8. (a) 2 - 2 1- 2 -7 CuI2 Cu 2 I ; K sp = [Cu ][I ] 1.08 x 10 I CE x 2x (x)(2x)2 1.08 x 10-7 4 x 3 1.08 x 10-7 x 3.00 x 10-3 molar solubility 3.00 x 10-3 mol 317.35 g Solubility x 0.95 g/L L mol
(b) 2 - Cu(NO3 )2 Cu 2 NO3 1.0 M 2 - 2 1- 2 -7 CuI2 Cu 2 I ; Ksp = [Cu ][I ] 1.08 x 10 I CE 1.0 2x (1.0)(2x)2 1.08 x 10-7 4 x 2 1.08 x 10-7 x 1.64 x 10-4 molar solubility 1.64 x 10-4 mol 317.35 g Solubility x 0.052 g/L L mol [H O ][A-] HA H O H O A- ; K = 3 9. a) 2 3 a [HA] I CE 0.500 - x x x 1.8 x 10-5 100 Rule - applies -5 -3 x [H3O ] 0.500 x 1.8 x 10 3.0 x 10
-3 b) pH - log [H3O ] -log 3.0 x 10 2.52 pOH 14.00 - pH 14.00 - 2.52 11.48
-3 [H3O ]eq 3.0 x 10 c) % ionation x 100 x 100 0.60 % [HA]i 0.500 - - [H3O ][A ] d) HA H2O H3O A ; Ka = [HA] I CE 0.500 - x x 0.400 x 1.8 x 10-5 100 Rule - applies 0.500 x 1.8 x 10-5 x [H O ] 2.3 x 10-5 pH 4.64 3 0.400
nb -5 0.400 OR pH pKa log - log1.8 x 10 log na 0.500 4.64
10. CH2 = CHCH3 - 47 C C C
CH3CH2CH3 - 42 C C C
CH ≡ CCH3 - 23 C C C
CH3 – O – CH2CH3 11 C O C C
CH3CH2CH2NH2 48 C C C N
O CH3CH2CHO 49 C C C
O CH3COOCH3 53 C C O C
CH3 – CO – CH3 56 O C C C
CH CH CH OH 97 3 2 2 C C C OH
O CH3CH2COOH 141 C C C OH
O CH3CH2CONH2 213 C C C N
O 11. Make CH3 C C C C N
CH C 2 CH3
a) methylaminoethane H H H C H + Br Br H C Br + HBr H H
H H H H H C Br + H N H C N + HBr H H H H
H H H H C C + H Br H C C Br H H H H
H H H H H H CH3 H C C Br + N C H H C C N + HBr H H H H H H H b) 2 – methylbutanoic acid H H H H H H H H H C C C C H + Br Br H C C C C Br + HBr
H H CH3 H H H CH3 H H H H H H H H H H C C C C Br + H O H H C C C C OH + HBr
H H CH3 H H H CH3 H H H H H H H H O
H C C C C OH (O) H C C C C H + + H2O
H H CH3 H H H CH3 H H H O H H H O
H C C C C H + (O) H C C C C OH
H H CH3 H H CH3
C) N–ethyl–N–methyl 2–methylbutanamide
H H H O H H O H CH3 H C C C C OH N C C H C C C C N + + H2O H3C CH2 H H CH3 H H C CH3