## vera_ewing one year ago chemistry question

1. vera_ewing

@rvc @TrojanPoem

2. rvc

do you know the rate formula

3. vera_ewing

Yeah. I think it's either A or B not really sure...

4. TrojanPoem

I think this question is chemical equilibrium related. I think (A) as Water is not counted but oxygen water is counted.

5. rvc

yep i agree with trojan

6. rvc

wait

7. rvc

im not sure :/

8. TrojanPoem

Give me a second, I am gonna check.

9. TrojanPoem

I found this : A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. [So H2o2 only]

10. rvc

oh okay

11. vera_ewing

So what would the answer be?

12. rvc

agree with trojan A

13. vera_ewing

What happens to the energy of the transition state when a catalyst is used? A. The energy increases. B. The energy decreases. C. The energy does not change. D. The energy varies during the catalyzed reaction.

14. rvc

The catalyst will decreease the energy of activation so as to increease the rate of the reaction

15. vera_ewing

This one?

16. rvc

17. sweetburger

I think the transition state is at the top of the curve which is represented by D on the graph.

18. TrojanPoem

it's D.

19. rvc

|dw:1435413759932:dw|

20. vera_ewing

Wait, so it's D. A or C. D ?

21. sweetburger

22. sweetburger

confusing lol

23. vera_ewing

Which of the following statements is true if HI is formed in the following reaction? H2(g) + I2(g) → 2HI(g) A. Some of the collisions of the reactants have enough energy to overcome the activation energy. B. Some of the collisions between reactants have enough energy to overcome the reaction mechanism. C. There are no collisions between the reactants in the reaction. D. All of the collisions between reactants have enough energy to overcome the activation energy.

24. sweetburger

It isnt

25. rvc

for product to form collision has to occur

26. sweetburger

In collision theory not all of the colliosions will result in a creation of a product. Only if the collision has the proper orientation and energy will it create a product.

27. TrojanPoem

So B,

28. rvc

yep the orientation matters too but nothing is mentioned about it

29. sweetburger

Yes, but never in any real reaction will all collision result in a product

30. rvc

oops B lol

31. sweetburger

I cant tell the difference between A and B they look worded the same

32. vera_ewing

It was A.

33. sweetburger

That question was strange

34. vera_ewing

35. vera_ewing

36. sweetburger

I think this is a zero order reaction due to the negative slope/straight line

37. sweetburger

Im not completely sure im a bit rusty with kinematics

38. vera_ewing

39. vera_ewing

Which of these is correct?

40. rvc

i think it is first order reaction

41. sweetburger

Yea, it is I was wrong.

42. sweetburger

I didnt that the it was the ln[X] and not just [X] on the y axis

43. sweetburger

didnt notice*

44. rvc

yep

45. vera_ewing

So which one is the answer?

46. rvc

need to solve this

47. rvc

@TrojanPoem what do you think?

48. rvc

well guys i need to go

49. rvc

Sorry to go like this in between :( all the best vera

50. TrojanPoem

cya @rvc

51. sweetburger

I think i have an Idea on how to solve this

52. sweetburger

We first say that y=-.0952x+.25 and then we plug in 10 for x and find the cooresponding y value

53. vera_ewing

What did you get when you did that?

54. sweetburger

nope I dont think that is going to work nm

55. abb0t

Water is not counted in the rate constant equation.

56. sweetburger

What?

57. abb0t

Understand that a catalyst does $$\sf \color{red}{NOT}$$ lower or decrease the energy of activation of a reaction, rather it finds a $$new$$ pathway so that it expends less energy, and proceeds faster and consequently, lowering the requirement of the transition energy for that same reaction to proceed.

58. abb0t

For instance, you have: $$\sf \color{blue}{A \rightarrow D}$$ without a catalyst, it might require 100 kj/mol but with the catalyst, it finds a new pathway that may only require 15 kj/mol.

59. vera_ewing

Ohh I got C for my answer at first, but now I think I did it wrong /:

60. vera_ewing

@abb0t What did you get?

61. sweetburger

@abb0t Could you help out with this question http://assets.openstudy.com/updates/attachments/558ea82ee4b0058b2bb701f0-vera_ewing-1435414099554-screenshot20150627at10.08.08am.png

62. abb0t

Well, what does the graph tell you? Is it zeroth order, first order, second order...

63. vera_ewing

First order right?

64. abb0t

Yep. If the graph is linear and has a negative slope, it's first order. Remember that.

65. abb0t

So, what equation should you use? Since it's linear, it will be something along the lines of: y = mx+ b which is the equation of a line.

66. vera_ewing

I'm not really sure. I just used the one that @sweetburger gave me...

67. abb0t

[A] = [A]$$_0$$ - kt

68. sweetburger

I cant tell if the equation should be ln[A]=-kt+ln[A0] or something like [A]=[A0]e^(-kt)

69. abb0t

oops, forgot the ln. this is first order.

70. sweetburger

that subscript function is useful i need to find that

71. abb0t

They are both the same. Think of the natural log function, but you can use either.

72. abb0t

I gtg. But i've clarified a lot. I'm sure sweetburger can take it from here.

73. sweetburger

Oh boy

74. vera_ewing

Thanks @abb0t

75. sweetburger

Oh i think I may have figured it out

76. vera_ewing

Okay...

77. sweetburger

I literally just need to find another concentration value at a specific time and this would be so easy.

78. vera_ewing

Anything?

79. sweetburger

Nope, closest to any of the answer i have gotten is like -.018s^-1. I'm sorry, but this question is just escaping me.

80. vera_ewing

Okay, if I were to guess, which should I choose?

81. sweetburger

I dont know how much time you have to complete this, but I could try this one more way

82. sweetburger

No that didnt work either

83. sweetburger

Idk maybe try A, but I really dont know

84. sweetburger

OH ITS B

85. sweetburger

I just needed to do this i think .25/-.0952 = -2.62sec^-1

86. vera_ewing

It was A!!

87. sweetburger

Very sorry :/

88. abb0t

C'mon guys, that's why you have the graph! for a first-order reaction, it gives you what you have there, a straight line with the slope of the line equal to $$\sf \color{red}{-k}$$ If you know that your slope is -k, plug it in, and then substitute your value into the proper equation to solve for the rate of reaction is the change in concentration of the reactants or the change in concentration of the products per unit time. Which means, you're looking for $$\sf \color{red}{\frac{M}{s}}$$

89. sweetburger

Im clueless @abb0t