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CHM 2046 Exam #1 (2010.02.0203)
- A solution is prepared by dissolving 38.6 g sucrose (C 12
H
22
O
11 ) in 495 g of water. Determine the mole fraction of sucrose if the final volume of the solution is 508 mL. Answer: E A) 1.28 × 10
- 3 B) 7.23 × 10
- 2 C) 2.45 × 10
- 3 D) 7.80 × 10 -^2 E) 4.09 × 10 -^3 The MW of sucrose is 342.30 (this should take 30 seconds to calculate). So, we have 38.6/342.30 = 0.1128 mol sucrose (on things like this, I give one extra figure and round to the correct number of sig. figs. at the end). Similarly, we would have 495/18.02 = 27.48 moles of water. Thus, we have 27.58 moles total and the mole fraction of sucrose is 0.1128/27.58 = 0.00409 (to three sig. figs.).
- What data should be plotted to show that experimental concentration data fits a first-order reaction? Answer: E A) ln(k) vs. 1/T B) [reactant] vs. time C) ln(k) vs. Ea D) 1/[reactant] vs. time E) ln[reactant] vs. time This is straight out of the book and should require no explanation. The formula is ln[A] = -kt + ln[A] 0.
- The half-life for the second-order decomposition of HI is 15.4 s when the initial concentration of HI is 0.67 M. What is the rate constant for this reaction? Answer: E A) 4.5 × 10 -2^ M-1s- B) 2.2 × 10
M
s
C) 3.8 × 10 -2^ M-1s- D) 1.0 × 10 -2^ M-1s- E) 9.7 × 10
M
s
This is a simple rearrangement of the formula for the half-life. Writing the equation and plugging in gives k = 1/( t 1/
[A]
0 ) = 1/(15.4 × 0.67) = 0.097 M-1s-1.
- A solution is formed at room temperature by dissolving enough of the solid solute so that some solid remains at the bottom of the solution. Which Answer: C
statement below is TRUE? A) The solution is considered supersaturated. B) The solution is considered unsaturated. C) The solution is considered saturated. D) The solution would be considered unsaturated if it were cooled a bit to increase the solubility of the solid. E) None of the above are true. Simple definition right out of the book and lecture!
- If the activation energy for a given compound is found to be 42 kJ/mol, with a frequency factor of 8.0 × 10 10 s
- 1 , what is the rate constant for this reaction at 298 K? Answer: B A) 1.4 × 10 9 s
- 1 B) 3.5 × 103 s-^1 C) 7.4 × 10 -^4 s-^1 D) 4.6 × 10 5 s-^1 E) 2.9 × 10 -^4 s-^1 This is simple if you know the formula and are comfortable with units. Note that you must convert kJ to J before doing the full calculation. The equation and numbers are shown now. k = A·exp[- E a / RT ] = 8.0 × 1010 × exp[-42000/(8.31447215 × 298)] = 8.0 × 1010 × exp[-16.95] = 8.0 × 10 10 × 4.38 × 10
= 35 07 s
.
- What is the overall order of the following reaction, given the rate law? 2NO(g) + H 2 (g) → N 2 (g) + 2H 2 O(g) Rate = k[NO] 2 [H 2 ] Answer: E A) 4th order B) 1st order C) 0th order D) 2nd order E) 3rd order 2 + 1 = 3. I hope I do not have to say more!
- What are the units of k in the following rate law? Rate = k[X][Y] 1/ Answer: E A) M1/2s-^1 B) M/s C) M
E)
A quarter life of a first order reaction would be when the concentration becomes 1/4th of the original value. So, lnA = ln(A 0 /4) = -kt 1/
- ln(A 0 ). Rearranging gives ln[A 0
/(A
0 /4)] = ln(4)/k = 1.386/k.
- How many moles of KF are contained in 244 mL of 0.135 m KF solution? The density of the solution is 1.22 g/mL. Answer: C A) 3.29 × 10
- 2 mol KF B) 1.67 × 10
- 2 mol KF C) 3.99 × 10
- 2 mol KF D) 4.31 × 10
- 2 mol KF E) 2.32 × 10 -^2 mol KF First, the MW of KF is 58.10 g/mol. If we have a kg of solvent, then m = 0.135 means that we would have 58.10 × 0.135 = 7.844 g of KF in 1007.844 g of solution. The volume of this solution would be 1007.8/1.22 = 826.1 mL. This means that we would have 7.844 × 244/826. = 2.317 g of KF or, in other words 2.317/58.10 = 0.0399 moles of KF.
- (^) The rate constant for a second-order reaction is 0.54 M-^1 s-^1. What is the half-life of this reaction if the initial concentration is 0.27 M? Answer: B A) 5.0 s B) 6.9 s C) 2.0 s D) 1.3 s E) 1.7 s This is just a simple plug in. t 1/2 = 1/ kA 0 = 1/(0.54 × 0.27) = 6.9 s.
- Draw the Lewis structure for the molecule C 3
H
6 (not cyclic). How many sigma and pi bonds does it contain? Answer: A A) 8 sigma, 1 pi B) 7 sigma, 2 pi C) 9 sigma, 0 pi D) 8 sigma, 2 pi E) 9 sigma, 1 pi The molecule is CH 3 -CH=CH 2. We see 6 C-H bonds and a single C-C bond. Also, the first bond in a C=C double bond is a sigma bond. Counting, we thus have 8 sigma bonds. There is only one pi bond (the second bond of the double bond). So, you see 8 sigma and one pi and (A) is the answer!
- A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass of the unknown compound? Answer: E A) 294 g/mol B) 341 g/mol C) 448 g/mol D) 195 g/mol E) 223 g/mol The formula here is M = wRT/ II V = 0.0152(0.08206)(298)/{(8.44/760) × 0.1500} = 223 g/mol. (Units are left as an exercise for you!)
- Consider the molecule below. Determine the hybridization at each of the 2 labeled carbons. Answer: C A) C1 = sp^3 d, C2 = sp^3 d^2 B) C1 = sp, C2 = sp^2 C) C1 = sp 2 , C2 = sp 3 D) C1 = sp 3 , C2 = sp 3 d E) C1 = sp 2 , C2 = sp 3 d C1 has a double bond and is obviously sp 2
. C2 is surrounded by 4 single bonds and is thus easily seen to be sp^3. So, (C) is the answer.
- Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molarity of the HCl, if the solution has a density of 1.20 g/mL. Answer: D A) 9.35 M B) 13.9 M C) 7.79 M D) 12.8 M E) 10.7 M A liter of solutoin would weight 1200 g. 39% of this would be 468 g of HCl. The MW of HCl is 36.46 g/mol. Thus, we would have 468/36.46 mol/L = 12.8 M.
- Write a balanced reaction for which the following rate relationships are true. Rate = = = - Answer: C
D) trigonal planar E) tetrahedral Simple application of our definitions! sp is linear, sp^2 is trigonal planar, sp^3 is tetrahedral, sp^3 d is trigonal bipyramidal, and sp^3 d 2 is octahedral.
- Determine the solubility of N 2 in water exposed to air at 25°C if the atmospheric pressure is 1.2 atm. Assume that the mole fraction of nitrogen is 0.78 in air and the Henry's law constant for nitrogen in water at this temperature is 6.1 × 10
- 4 M/atm. Answer: B A) 3.6 × 10 -^4 M B) 5.7 × 10 -^4 M C) 6.5 × 10 -^4 M D) 1.8 × 10 -^4 M E) 1.5 × 10
- 4 M Simple application of Henry's law: c = K H P = 0.00061 × 0.78 × 1.2 = 0.00057 M.
- Which of the following ions should have the most exothermic enthalpy of hydration? Answer: B A) Mg2+ B) Al 3+ C) Ca 2+ D) Na
E) Sr 2+ This would be the result of a coulombic attraction between the ion and the negative end of the water molecule. The smallest ion, Al3+, also has the highest charge. Hence, it would have the most exothermic heat of hydration. Note that this was not explicitly in the book or notes or lecture--you were expected here to put ideas together that were in the book, notes, and lecture.
- Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. Answer: A A) 0.100 m AlCl 3 B) 0.100 m C 6 H 12 O 6 C) 0.100 m NaCl D) 0.100 m MgCl 2 E) They all have the same boiling point. The solution with the largest number of particles per mole is (A), with 4 ions per formula unit.
(B) has 1, (C) has 2, and (D) has 3. All have the same concentration. So, the one with the greatest van't Hoff factor is the winner!
- Place the following solutions in order of increasing osmotic pressure. I. 0.15 M C 2
H
6
O
2 II. 0.15 M NaCl III. 0.15 M CH 3
CO
2
H
Answer: D A) III < I < II B) II < III < I C) II < I < III D) I < III < II E) I < II < III I is a nonelectrolyte. III is a weak--only partially dissociated--electrolyte. II is a strong electrolyte. So, the number of particles released in soluton at the same concentration would follow I < III < II. This (D) would be the order of increasing osmotic pressure.
- The first-order decomposition of N 2 O at 1000 K has a rate constant of 0. s-^1. If the initial concentration of N 2 O is 10.9 M, what is the concentration of N 2 O after 9.6 s? Answer: D A) 1.4 × 10
- 3 M B) 8.7 × 10 -^3 M C) 1.0 × 10 -^3 M D) 7.4 × 10 -^3 M E) 3.6 × 10 -^3 M Just plug in to the equation for first-order decay: A = A 0 e-kt^ = 10.9 × exp[-0.76 × 9.6] = 10.9 × exp[-7.30] = 10.9 × 0.000678 = 0.0074 M. b
