1 A burning splint will burn more vigorously in pure oxygen than in air because

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1 Slide 1 / 38 1 burning splint will burn more vigorously in pure oxygen than in air because oxygen is a reactant in combustion and concentration of oxygen is higher in pure oxygen than is in air. oxygen is a catalyst for combustion. oxygen is a product of combustion. nitrogen is a product of combustion and the system reaches equilibrium at a lower temperature. nitrogen is a reactant in combustion and its low concentration in pure oxygen catalyzes the combustion.

2 Slide 2 / 38 2 s the temperature of a reaction is increased, the rate of the reaction increases because the. reactant molecules collide less frequently reactant molecules collide more frequently and with greater energy per collision activation energy is lowered reactant molecules collide less frequently and with greater energy per collision reactant molecules collide more frequently with less energy per collision

3 Slide 3 / 38 3 Which of the following will not affect the rate of a chemical reaction? increasing the concentration of a reactant decreasing the concentration of a reactant increasing the temperature at which the reaction is carried out performing the reaction in the presence of a catalyst ll of the above will affect the rate of a chemical reaction.

4 Slide 4 / 38 4 reaction was found to be first order in carbon monoxide concentration. The rate of the reaction if the [O] is doubled, with everything else kept the same. doubles remains unchanged triples increases by a factor of 4 is reduced by a factor of 2

5 Slide 5 / 38 5 reaction was found to be second order in carbon monoxide concentration. The rate of the reaction if the [O] is doubled, with everything else kept the same. doubles remains unchanged triples increases by a factor of 4 is reduced by a factor of 2

6 Slide 6 / 38 6 If the rate law for a reaction is first order in and second order in, then the rate law is k [][] k[] 2 [] 3 k[][] 2 k[] 2 [] k[] 2 [] 2

7 Slide 7 / 38 7 reaction was found to be third order in. Increasing the concentration of by a factor of 3 will cause the reaction rate to. remain constant increase by a factor of 27 increase by a factor of 9 triple decrease by a factor of the cube root of 3

8 Slide 8 / 38 8 It was experimentally determined that the rate for the reaction ( + P) increased by a factor of 9 when th concentration of was tripled. The reaction is order in. zero first second third one-half

9 Slide 9 / 38 9 The rate law for a reaction is rate = k[][] 2 Which one of the these statements is false? The reaction is first order in. The reaction is second order in. The reaction is second order overall. k is the reaction rate constant If [] is doubled, the reaction rate will increase by a factor of 4.

10 Slide 10 / The rate law of the overall reaction + Is rate=k[] 2 Which of the following will not increase the rate of the reaction? increasing the concentration of reactant increasing the concentration of reactant increasing the temperature of the reaction adding a catalyst for the reaction ll of these will increase the rate.

11 Slide 11 / Which energy difference in this energy profile corresponds to the activation energy for the forward reaction? x y x+y x-y y-x

12 Slide 12 / In the above drawing, what quantity is represented by the sum of x + y? heat of reaction, H activation energy of the forward reaction activation energy of the reverse reaction potential energy of the reactants potential energy of the products

13 Slide 13 / In the above drawing, what quantity is represented by y? heat of reaction, ΔH activation energy of the forward reaction activation energy of the reverse reaction potential energy of the reactants potential energy of the products

14 Slide 14 / In the above drawing, what quantity is not affected by the presence of a catalyst? x y Neither x nor y oth x and y are affected by the presence of a catalyst.

15 Slide 15 / In the energy profile of a reaction, the species that exists at the maximum on the curve is called the. product activated complex activation energy enthalpy of reaction atomic state

16 Slide 16 / catalyst can increase the rate of a reaction. by lowering the activation energy of the reverse reaction by increasing the overall activation energy (a ) of the reaction by providing an alternative pathway with a lower activation energy ll of these are ways that a catalyst might act to increase the rate of reaction.

17 Slide 17 / t equilibrium,. all chemical reactions have ceased the rates of the forward and reverse reactions are equal the rate constants of the forward and reverse reactions are equal the value of the equilibrium constant is 1 the limiting reagent has been consumed

18 Slide 18 / Which of the following expressions is the correct equilibrium-constant expression for the reaction below? (NH 4 ) 2 Se (s) 2NH 3 (g) + H 2 Se(g) [ NH 3 ] [H 2 Se] / [NH 4 ) 2 Se ] [NH 4 ) 2 Se ] / [ NH 3 ] 2 [H 2 Se] 1/ [NH 4 ) 2 Se ] [ NH 3 ] 2 [H 2 Se] [ NH 3 ] 2 [H 2 Se] / [NH 4 ) 2 Se ]

19 Slide 19 / Which of the following expressions is the correct equilibrium-constant expression for the reaction below? HF (aq) H 2 O(l) H 3 O + (aq) + F- (aq) [HF][ H 2 O]/ [H 3 O + ][F - ] 1/HF [H 3 O+][F - ] / [HF][ H 2 O] [H 3 O + ][F - ] / [HF] [F-]/[HF]

20 Slide 20 / The equilibrium constant for the gas phase reaction N 2 (g) +3H 2 (g) ( 2NH 3 (g) is Keq = 4.34x10-3 at t equilibrium. products predominate reactants predominate roughly equal amounts of products and reactants are present only products are present only reactants are present

21 Slide 21 / The equilibrium constant for the gas phase reaction 2NH 3 (g) N 2 (g) +3H 2 (g) is K eq =230 at t equilibrium, products predominate reactants predominate roughly equal amounts of products and reactants are present only products are present only reactants are present

22 Slide 22 / Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide? N 2 O 4 (g) 2NO 2 (g) [ NO 2 ] /[ N 2 O 4 ] [ NO 2 ] 2 / [ N 2 O 4 ] [ NO 2 ]/ [ N 2 O 4 ] 2 [NO 2 ][N 2 O 4 ] [NO 2 ] 2 [N 2 O 4 ]

23 Slide 23 / The equilibrium-constant expression for the reaction Ti(s) + 2l 2 (g) Til 4 (l) is given by Til 4 (l)/ [Ti(s)] + [l 2 (g)] [Ti(s)] [l 2 (g)] 2 / [ Til 4 (l)] [ Til 4 (l)] / [l 2 (g)] 2 [l 2 (g)] -2 [ Til 4 (l)] / [Ti(s)] [l 2 (g)] 2

24 Slide 24 / onsider the following equilibrium. 2SO 2 (g) + O 2 (g) 2SO 3 (g) The equilibrium cannot be established when is/are placed in a 1.0-L container mol SO 2 and 0.25 mol O mol SO mol SO 2 and 0.25 mol SO mol O 2 and 0.50 mol SO mol SO 3

25 Slide 25 / If the value for the equilibrium constant is much greater than 1, then the equilibrium mixture contains mostly. reactants products

26 Slide 26 / Pure and pure are excluded from equilibrium-constant expressions. gases and compounds solids and liquids liquids and elements ions and molecular compounds cids and bases

27 Slide 27 / Pure and pure are excluded from equilibrium-constant expressions. gases and compounds gases and liquids liquids and elements ions and molecular compounds cids and bases

28 Slide 28 / Of the following equilibria, only will shift to the left in response to a decrease in volume. H 2 (g) + l 2 (g) 2Hl (g) 2SO 3 (g) 2SO 2 (g) + O 2 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g) 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 2HI (g) H 2 (g) + I 2 (g)

29 Slide 29 / The reaction below is exothermic: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Le hatelier's Principle predicts that will result in an increase in the number of moles of SO 3 (g) in the reaction container. increasing the pressure decreasing the pressure increasing the temperature removing some oxygen increasing the volume of the container

30 Slide 30 / For the endothermic reaction ao 3 ao (s) + O 2 (g) hatelier's principle predicts that will result in an increase in the number of moles of O 2. increasing the temperature decreasing the temperature increasing the pressure removing some of the ao 3 none of the above

31 Slide 31 / In which of the following reactions would increasing pressure at constant temperature not change the concentrations of reactants and products, based on Le hatelier's principle? N 2 (g) + 3H 2 (g) 2NH 3 (g) N 2 O 4 (g) 2NO 2 (g) N 2 (g) + 2O 2 (g) 2NO2 (g) 2N 2 (g) +O 2 2N 2 O (g) N 2 (g) +O 2 (g) 2NO(g)

32 Slide 32 / onsider the following reaction at equilibrium: 2NH 3 (g) N 2 (g) + 3H 2 (g) ΔH = kj Le hatelier's principle predicts that adding N 2 (g) to the system at equilibrium will result in. a decrease in the concentration of NH 3 a decrease in the concentration of H 2 (g) an increase in the value of the equilibrium constant a lower partial pressure of N 2 removal of all of the H 2 (g)

33 Slide 33 / onsider the following reaction at equilibrium: 2O 2 (g) 2O(g) + O 2 (g) ΔH = -514 kj Le haelier's principle predicts that adding O 2 (g) to the reaction container will. increase the partial pressure of O (g) at equilibrium decrease the partial pressure of O 2 (g) at equilibrium increase the value of the equilibrium constant increase the partial pressure of O 2 (g) at equilibrium decrease the value of the equilibrium constant

34 Slide 34 / onsider the following reaction at equilibrium: 2O 2 (g) 2O(g) + O2 (g) H = -514 kj Le hatelier's principle predicts that an increase in temperature will. increase the partial pressure of O 2 (g) decrease the partial pressure of O 2 (g) decrease the value of the equilibrium constant increase the value of the equilibrium constant increase the partial pressure of O

35 Slide 35 / onsider the following reaction at equilibrium. 2O 2 (g) 2O(g) + O 2 (g) ΔH = -514 kj Le hatelier's principle predicts that the equilibrium partial pressure of O (g) can be maximized by carrying out the reaction. at high temperature and high pressure at high temperature and low pressure at low temperature and low pressure at low temperature and high pressure in the presence of solid carbon

36 Slide 36 / onsider the following reaction at equilibrium: 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔH = -99 kj Le hatelier's principle predicts that an increase in temperature will result in. a decrease in the partial pressure of SO 3 a decrease in the partial pressure of SO 2 an increase in K eq no changes in equilibrium partial pressures the partial pressure of O 2 will decrease

37 Slide 37 / The effect of a catalyst on an equilibrium is to increase the rate of the forward reaction only increase the equilibrium constant so that products are favored slow the reverse reaction only increase the rate at which achieved without change shift the equilibrium to the right

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