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Module 1: Question 1 In the reaction of gaseous N 2 O 5 to yield NO 2 gas and O 2 gas as shown below, the following data tableis obtained: → 4 NO2 (g) + O2 (g)
- Using the [O 2 ] data from the table, show the calculation of the instantaneous rate early in the reaction (0 secs to 300 sec). 2 N 2 O 5 (g) Data Table # Time (sec) [N 2 O 5 ]^ [O 2 ] 0 0.300 M 0 300 0.272 M 0.014 M 600 0.224 M 0.038 M 900 0.204 M 0.048 M 1200 0.186 M 0.057 M 1800 0.156 M 0.072 M 2400 0.134 M 0.083 M 3000 0.120 M 0.090 M
- Using the [O 2 ] data from the table, show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs).
- Explain the relative values of the early instantaneous rate and the late instantaneous rate. Your Answer:
- rate = (0.014 - 0) / (300 - 0) = 4.67 x 10 -^5 mol/Ls
- rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10 -^5 mol/Ls
- The late instantaneous rate is smaller than the early instantaneous rate.
Your Answer: 0.693 = k t1/ 0.693 = k (5720) k = 1.21 x 10 -^4 ln [A] - ln [A] 0 = - k t ln 19.8 - ln 100 = - 1.21 x 10-^4 t t = 13, 384 years Question 4 Using the potential energy diagram below, state whether the reaction described by the diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer. Your Answer: The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it has relatively large Eact. Question 5 Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of COand 2.40 mole of H 2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of H 2 O and corresponding amounts of CO, H 2 , and CH 4. CO (^) (g) + 3 H 2 (g) CH4 (g) + H 2 O (^) (g) Your Answer: 0.309 mole of H 2 O formed = 0.309 mole of CH 4 formed
0.309 mole of H 2 O formed = 0.800 - 0.309 = 0.491 mole CO 0.309 mole of H 2 O formed = 3 x 0.309 mole H 2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H 2 [CO] = 0.491 mole / 8.00 L = 6.1375 x 10-^2 M [H2] = 1.473 mole / 8.00 L = 18.4125 x 10-^2 M [CH4] = 0.309 mole / 8.00 L = 3.8625 x 10-^2 M [H2O] = 0.309 mole / 8.00 L = 3.8625 x 10-^2 M Kc = [3.8625 x 10 -^2 ] [3.8625 x 10-^2 ] / [6.1375 x 10 -^2 ] [18.4125 x 10-^2 ]^3 Kc = 3. Question 6 Explain the terms substrate and active site in regard to an enzyme. Your Answer: Enzymes are large protein molecules that act as catalysts and increases the rate of a reaction. The catalysts act only on one type of substance to cause one type of reaction and this is called a substrate. Active sites are enzymes that are spherical in shape together ith a group of atoms on a surface of protein. These active sites bind with the substrate causing it to undergo a reaction. The substrate and active site forms a complex that creates a new pathway with lower activation energy and thus speed upa reaction. Question 7 The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants.Be sure to explain your answer. 2 O 3 (g) 3 O 2 (g) Kc = 2.54 x 10^12 Your Answer: Kc = 2.54 x 1012 is very large. Thus, the equilibrium mixture is made up of predominantly products.
H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] and H 2 = [0.186] with Kc = 3.62. If the concentration of CO at equilibrium is increased to [0.200], how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirmyour answer. CO (^) (g) + 3 H 2 (g) CH4 (g) + H 2 O (^) (g) Your Answer: Kc = 3. Qc = [0.200] [0.0380] / [0.0620] [0.186]^3 Qc = 19. Qc is greater than Kc and thus the equilibrium will shift to the left towards the direction of the reactants. Kc = 3.62 when H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] and H 2 = [0.186 M] When CO = [0.200], Q = [0.0380] [0.0380] = 1. [0.200] [0.186]^3 The reaction must shift briefly in the forward direction to decrease the [CO] to come back to equilibrium.This is in agreement with Qc < Kc which also predicts the reaction will proceed to theright. new CO concentration is 0. Question 9 The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is decreased from 6.00 liters to 2.00 liters, how and for what reason will the equilibrium shift? Be sure tocalculate the value of the reaction quotient, Q, and use this to confirm your answer. CO (^) (g) + 3 H 2 (g) CH4 (g) + H 2 O (^) (g) Your Answer: Initial volume = 6.00 L Final volume = 2.00 L Volume is reduced by 3. Pressure is inversely proportional to volume, thus pressure is tripled and concentration of gases is also tripled. Final concentration = Initial concentration x 3
Qc = [3CH 4 ] [3H 2 O] / [3CO] [3H 2 ]^3 Qc = Kc / 9 Qc is lesser than Kc and thus the equilibrium will shift to the right towards the direction of the products. When volume decreases from 6.00 to 2.00, the pressure triples and the concentration of all gases (CO, H 2 , CH 4 , and H 2 O) triples so: (at equilibrium) Qc =Kc = [CH 4 ] [H 2 O] = 3. [CO] [H 2 ]^3 (volume 1/3 = pressure tripled = conc tripled) Qc = [3 CH 4 ] [3 H 2 O] = Kc [3 CO] [3 H 2 ]^3 The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back to equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the products).
- Qc should be used to confirm your answer - for what reason will it shift this way? Question 10 The equilibrium reaction below has the Kc = 3.93 at 25oC. If the temperature of the system at equilibrium is decreased to 10 oC, how and for what reason will the equilibrium shift? Also show andexplain how and why the Kc value will change. CO (^) (g) + 3 H2 (g) CH 4 (g) + H 2 O (^) (g) ∆H^0 = - 206.2 kJ Your Answer: Temperature is decreased in the reaction at equilibrium that has a - ∆H^0 , thus reaction shifts in the forward direction. The reaction shift in the direction that produces some heatIn this reaction, the concentration of the product increases and the concentration of the reactants decreases and thus
4 4 4 2 4 3 3 4 Module 2: Question 1 Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula andexplain your answer. (1) Cr(OH) 3 (2) HAsO 4 (3) CoCO 3 Your Answer:
- base - contains Cr metal + OH polyatomic group
- acid - contains H + polyatomic group (AsO 4 )
- salt - contains Co metal + polyatomic group (CO 3 ) Question 2 For the Brønsted-Lowry acid base reactions shown below, list the stronger acid, stronger base, weaker acid and weaker base in the answer blanks provided: NH +^ + H PO -^ NH + H PO Stronger acid: Stronger base: Weaker acid: Weaker base: Your Answer: Stronger acid: H 3 PO 4 Stronger base: NH 3 Weaker acid: NH + Weaker base: H 2 PO - Question 3 For the compounds shown below, choose the stronger acid, explain why it is the stronger acid and write the formula of its conjugate base in the answer blank:
4 HClO 4 or HBrO 4 or HIO 4 Stronger acid is: Explanation: Formula of conjugate base of the stronger acid: Your Answer: Stronger acid is: HClO 4 > HBrO 4 > HIO 4 Cl is more electronegative than Br so HClO 4 is stronger acid than HBrO 4 and so is Br to I, Br is more electronegative than I so HBrO 4 is stronger than HIO 4. Stronger acid is: HClO 4 Explanation: Cl has the higher electronegativity (of Cl, Br or I) which makes the H-O bond of HClO 4 most polar and most likely to form H+ Formula of conjugate base of stronger acid: conjugate base of HClO 4 is ClO - Question 4 Show the calculation of the [H+] and pH of a 0.00350 M solution of the strong acid H 2 SO 4. Your Answer: [H+] = 2 x [H 2 SO 4 ] = 2 x (0.00350] [H+] = = 0.007 M pH = - log [H+] = - log ( 0.007)pH = 2. Question 5 In the titration of 15.0 mL of H 2 SO 4 of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.4 mL of 0.130 M NaOH is added. Show the calculation of the molarity of the H 2 SO 4.
2 3 2 2 3 2 4 pH = - log [H+] = - log (3.41x10-^3 )pH = 2. % ionization = ([C H O - ]/ [HC H O ]) x 100 % ionization = (3.41x10-^3 / 0.645) x 100 % ionization = 0.53% Question 7 Predict and explain whether a solution of LiF is acidic, basic or neutral. Your Answer: Basic since Li hydrolyzes to form strong base [LiOH]. LiF: Basic since F-^ hydrolyzes to form a weak acid (HF) and OH- Question 8 Show calculation of the pH of a buffer prepared by mixing 0.100 M NH 4 Cl and 0.0750 M NH 3. NH 3 + H 2 O (^) (liq) NH +^ + OH-^ Ka = 1.8 x 10-^5 Your Answer: NH 3 + H 2 O (^) (liq) NH 4 +^ + OH- 0.0750 0.100 0
- x +x +x 0.0750-x 0.100+x x Ka = 1.8 x 10-^5 Ka = [NH 4 +] [OH-] / [NH 3 ] 1.8 x 10-^5 = (0.100+x) (x) / (0.0750-x)
2 3 2 a 2 3 2 1.8 x 10-^5 = (0.100) (x) / (0.0750)x = 1.35x10-^5 = [OH-] pOH = - log [OH-] = - log (1.35x10-^5 ) pOH = 4. pH = 14 - pOH = 14 - 4. pH = 9. Question 9 Show the calculation of the pH of a solution obtained by adding 0.0100 mole of OH-^ to a buffer of pH 4.74 which originally contains 0.100 M NaC 2 H 3 O 2 and 0.100 M HC 2 H 3 O 2. HC 2 H 3 O 2 H+^ + C H O -^ K = 1.8 x 10 -^5 Your Answer: HC 2 H 3 O 2 H+^ + C H O - 0.100 0 0. +0.0100 mole OH- 0.090 0 0.
- x +x +x 0.090-x x 0.110+x Ka = 1.8 x 10-^5 Ka = [H+] [C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] 1.8 x 10 -^5 = (x) (0.110+x) / (0.090-x ) 1.8 x 10-^5 = (x) (0.110) / (0.090)x = 1.47x10-^5 = [H+] pH = - log [H+] = - log (1.47x10-^5 )pH = 4. pH went up slightly fron 4.74 to 4.83 when OH-^ was added.
initial 1.0x10-^6 = [s] [2s]^2 1.0x10-^6 = 4s^3 s = 6.3x10-^3 mol/L Question 2 Show the calculation of the Ksp of MnS if the solubility of MnS is 0.0001375 g/100 ml. MW of MnS = 87. MnS (s) Mn+2^ (aq) + S-^2 (aq) Your Answer: molar solubility of MnS = (0.0001375 g/100 ml) x (1000 mL) x (1 mol/87.01 g MnS)molar solubility of MnS = 1.580x10-^5 Ksp = [Mn+2] [S-^2 ] Ksp = (1.580x10-^5 ) (1.580x10-^5 ) = 2.496x10-^10 Question 3 Show the calculations and predict if a precipitate of PbI 2 will form if 1.0 x 10-^4 M Pb(NO 3 ) 2 is mixedwith 1. x 10-^3 M NaI? The Ksp of PbI 2 = 7.1 x 10-^9. Your Answer: Ksp = [Pb+2] [I-]^2 = 7.1 x 10 -^9 Qc = [Pb+2]initial [I-]^2 = [1.0 x 10-^4 ] [ 1.0 x 10-^3 ]^2 Qc = 1.0x10-^10 Qc (1.0x10-^10 ) < Ksp (7.1 x 10-^9 ) thus reaction should go in the forward direction (solution) and PbI 2 will not precipitate. Question 4 State and explain which has greater entropy: 1 mole of N 2 gas at 1 atm or 1 mole of N 2 gas at 2 atm. Your Answer:
1 mole of N 2 gas at 1 atm has greater entropy than 1 mole of N 2 gas at 2 atm because the increase in pressure will cause compression of the particles of nitrogen gas at 2 atm into a smaller space and decreases the entropy. Question 5 Explain whether this physical reaction results in an increase or decrease in entropy. H 2 O (s) → H 2 O (l) Your Answer: The reaction is from solid H 2 O to liquid H 2 O and liquid has greater entropy because it has more available motions than solid. Thus, reaction from solid to liquid increases the entropy. Question 6 Determine ΔS^0 for the following reaction using the data given. Explain whether this value agrees with prediction of the ΔS^0 value. ΔS^0 C 2 H 6 = 229.5 J/mole, ΔS^0 O 2 = 205.5 J/mole, ΔS^0 CO 2 = 213.7 J/mole, ΔS^0 H 2 O = 188.7 J/mole 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Your Answer: ΔS^0 = ΔS^0 products - ΔS^0 reactants ΔS^0 = [4 ΔS^0 CO 2 + 6 ΔS^0 H 2 O] - [2 ΔS^0 C 2 H 6 + 7 ΔS^0 O 2 ]ΔS^0 = [(4x213.7) + (6x188.7)] - [(2x229.5) + (7x205.5)] ΔS^0 = 89.5 J/mol Reaction has a higher entropy and system is more disorder because ΔS^0 is positive. Question 7 Determine ΔG^0 for the following reaction using the data given and use it to predict the spontaneity ofthe reaction. ΔGf^0 C 2 H 6 = - 32.9 kJ/mole, ΔGf^0 CO 2 = - 394.4 kJ/mole, ΔGf^0 H 2 O = - 228.6 kJ/mole 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Your Answer:
Your Answer: MnS (s) + O 2 (g) → Mn (s) + SO 2 (g) ΔG^0 = - 85.9 kJ Question 10 Using the table and images (A - F) below, predict, and explain by doing a Qc vs Ksp calculation, whichtube shows the result when colorless 1 x 10 -^6 M AgNO 3 solution was added to colorless 1 x 10 -^6 M NaBr solution. A B C D E F Substance Formula Color K sp Copper (II) carbonate CuCO 3 Blue (^) 1.4 x 10 -^10 Lead (II) chromate PbCrO 4 Yellow 2.8 x 10 -^13 Silver bromide Your Answer: AgBr White 5.0 x 10 -^13 Ksp = [Ag+] [Br-] = 5 .0 x 10-^6 Qc = [Ag+]initial [Br-]initial = [1.0 x 10-^6 ] [ 1.0 x 10 -^6 ]Qc = 1.0x10-^12 Qc (1.0x10-^12 ) < Ksp (5 .0 x 10-^6 ) thus the reaction should go in the forward direction (solution) and AgBr will not precipitate. Image B shows the result of when colorless AgNO 3 solution was added to colorless NaBr solution because it does not precipitate, its color is still colorless.
- E (^) cell = E - E Qc = (1 x 10 -^6 ) x (1.0 x 10 -^6 ) = 1 x 10 -^12 Ksp for AgBr = 5.0 x 10 -^13 Since Qc > Ksp a white precipitate will form = E Module 4: Question 1 For the cell described by the following cell diagram. Cr (s) | Cr+3^ (aq, 1 M) || Al+3^ (aq, 1 M) | Al (s) Al+3^ + 3e-^ → Al E^0 = - 1.66 v Cr+3^ + 3e-^ → Cr E^0 = - 0.73 v (1) The anode half reaction is (2) The cathode half reaction is (3) The overall cell reaction is (4) Show the calculation for the total cell potential (5) State and explain whether the cell is voltaic or electrolytic Your Answer:
- Cr (s) → Cr+3^ (aq) + 3e-
- Al+3^ (aq) + 3e-^ → Al (s)
- Cr (s) + Al+3^ (aq) → Cr+3^ (aq) + Al (s) 0 0 0 cathode anode E^0 cell = - 1.66 - (-0.73) E^0 cell = - 0.93 v
- Since the total cell potential is negative, cell will not occur spontaneously and would have to be electrolytic. Question 2 In the following reaction, identify the following and explain your answersZn
- 2 H+^ + 2 Cl-^ → Zn+2^ + 2 Cl-^ + H 2
