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# Page 258

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1. Page 258 • At higher concentrations, the particles are closer together, and will collide with each other more often. More frequent collisions means more effective collisions and a higher reaction rate. • Mathematical rate laws: • r= k[H2SO4(aq)][Ca(OH)3(aq)] • r=k[I-(aq)]5 [H+(aq)]6 [IO3(aq)]3 • r=k[HCl(aq)]2 • r=k[N2O4(g)][H2O(g)]

2. Page 258 • The waiting time should be shorter, since the H2O2 concentration is higher, the reaction will be faster. • Calculate the number of moles of H2O2: (22g) ÷(34 g/mol) =0.647mol Calculate the concentration of H2O2: (0.647mol) ÷ (0.500 L) = 1.294 mol/L Set up rate law and solve: r= k[H2O2]2 =5.32x10-7 (1.294)2 = 8.91x10-7 mol/(L·s)

3. Page 258 • r=k[NO]2[Br2], so k= r ÷ [NO]2[Br2] or k= 1.60 mol/(L·s) ÷ (1 mol/L)2 (2.0 mol/L) k = 0.80 L2/mol2s • r=k[A][B], so k= r ÷ [A][B] k=7.3x10-3 mol/(L·s) ÷ (3.1 mol/L)(0.75 mol/L) k=3.13x10-2 L/(mol·s)

4. P 258 • Decreasing the volume of a system of gases by half without letting any gas escape, would effectively double the pressure of the system (Boyle’s Law). Since a gas’ pressure is effectively the same as its concentration, we would be doubling the concentration of both reactants. The rate would obviously increase, and we can estimate by how much it would increase: if the initial rate is k·[C]2·[D]3 and we double [C] and [D], then: The new rate is k·[2C]2 ·[2D]3 or simplified, k·4·8·[C]2·[D]3 or in other words, it is 32 times faster than the initial rate.

5. Page 267 1. Food can be preserved in a refrigerator because cold slows down the reactions that spoil the food (also, it prevents bacteria) 2. The red line (#3) represents higher temperature particles. • a) It would have little or no effect on the overall reaction. The first “peak” would be lower. b) The overall rate would increase. The second “peak” would be lower. c) It might slow down the reaction, but only if it caused the “peak” to rise above the 2nd peak d)It would cause the reaction rate to decrease. The second peak would become higher.

6. Page 276 • Normally a catalyst cannot increase the amount of product produced (the final concentration). The only exception is if the catalyst is required for the reaction to occur at all. • One would expect reaction #2 to be fastest, since it has fewer bonds to break. Reaction #3 would be slowest, as it has the most bonds to break

7. Page 276 • An inhibitor could raise the activation energy (move line Ea further to the right) so that fewer particles would have sufficient energy to react. In other words, it would slow the reaction. • a) Curve 3= with a catalyst Curve 2 = without a catalyst or inhibitor Curve 1 = with an inhibitor b) Curve 3 would be fastest for the reverse reaction, since a catalyst speeds up both the direct and the reverse reaction.

8. Page 272 • a) The particles with sufficient energy are located in part B, to the right side of line Ea b) The curve will change as follows: i) The temperature increase stretches the graph towards the right, so that the peak will move right. More particles will have sufficient energy, and the rate will increase. ii) Temperature decrease will squash the graph towards the left, moving the peak left. Fewer particles will have enough energy, and the rate will decrease iii) Adding a catalyst will move the line Ea towards the left. More particles will have sufficient energy and the rate will increase. iv) adding reactant particles will stretch the blue curve upwards, increasing the number of particles in area B and increasing the rate of reaction.

9. Page 272 • Sample answers (others are possible) a) Heating the water will cook the potato faster, cooling it will cook it more slowly b) Keeping it cold will slow spoilage, allowing it to get warmer will speed up spoilage c) Keeping air out, or adding an inhibitor, like lemon juice, will slow the browning d)The fish will spoil slower if you... Put it in the coolest part of the boat... or keep it in the water on a stringer line... or put it in an ice chest.

10. Page 272 • Mg(s) + H2SO4(aq)  MgSO4(aq) + H2(g) a) very little effect. The magnesium is solid, so its concentration will not change. Adding pieces could slightly increase the surface area causing a small increase in reaction rate. b) No effect. The concentration of the acid does not change, so the rate stays the same c) Increases Rate. Powdered magnesium has much higher surface area than magnesium pieces. d) Increases Rate. The increased concentration of the acid will raise the reaction rate.

11. Page 272 4. Judging reaction rates based only on the nature of the reaction is difficult. There are several factors to consider, and no way to fully weight them without experimentation. That said, the answers on the following slides are what I believe to be the best answers:

12. Page 272 4a) The fastest reaction will probably be #2. Reasons: all the reactants are aqueous (homogeneous reaction), the reactants are all ionic (faster than covalent) and all reactants have relatively few bonds to break. Arguments could be made in favor of #4 (all gases, few bonds), but the question specified at “room temperature”, and from experience I know that methane doesn’t burn spontaneously at room temperature.

13. Page 272 • 4b) The slowest reaction will probably be #1. Reasons: Since it is a heterogeneous reaction, the reaction can only occur along the surface of the piece of lithium. Arguments could be made for #3, since it has large molecules with many bonds to be broken, however, since it is homogeneous, the reaction can occur throughout the mixture, speeding it up significantly.

14. Page 272 • 4c) This question makes no sense, but if I had to chose an answer, I would go with #2 for the reasons stated in 4a)

15. Page 272 • 5 The difference between the two graphs is that the activation energy (line Ea) appears to be further to the right in graph b. The best explanation is that graph “a” is with a catalyst and graph “b” is without a catalyst. A catalyst would lower the activation energy.

16. Pages 272-273 • 6. The nature of the reactants (and perhaps because CaCl2 produces more solute particles than NaCl) • 7. The flour dust in the silo has a much greater surface area than in a bag of flour. • 9 a) Step 2, b) step 3, c) step 2, d) step 2 • 11. reduce the concentration of the aqueous solution, or use lumps of solid instead of a powder, or reduce the reaction environment temperature (eg. Use an ice bath), or look for an inhibitor for this reaction. • 13. answer: b) 2AB because the concentration of the reactant [A] decreases twice as fast as the product [B] increases.