vera_ewing
  • vera_ewing
chemistry question
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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vera_ewing
  • vera_ewing
@rvc @TrojanPoem
rvc
  • rvc
do you know the rate formula
vera_ewing
  • vera_ewing
Yeah. I think it's either A or B not really sure...

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More answers

TrojanPoem
  • TrojanPoem
I think this question is chemical equilibrium related. I think (A) as Water is not counted but oxygen water is counted.
rvc
  • rvc
yep i agree with trojan
rvc
  • rvc
wait
rvc
  • rvc
im not sure :/
TrojanPoem
  • TrojanPoem
Give me a second, I am gonna check.
TrojanPoem
  • 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]
rvc
  • rvc
oh okay
vera_ewing
  • vera_ewing
So what would the answer be?
rvc
  • rvc
agree with trojan A
vera_ewing
  • 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.
rvc
  • rvc
The catalyst will decreease the energy of activation so as to increease the rate of the reaction
vera_ewing
  • vera_ewing
rvc
  • rvc
i have already said about that
sweetburger
  • sweetburger
I think the transition state is at the top of the curve which is represented by D on the graph.
TrojanPoem
  • TrojanPoem
it's D.
rvc
  • rvc
|dw:1435413759932:dw|
vera_ewing
  • vera_ewing
Wait, so it's D. A or C. D ?
sweetburger
  • sweetburger
Answer is C. D
sweetburger
  • sweetburger
confusing lol
vera_ewing
  • 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.
sweetburger
  • sweetburger
It isnt
rvc
  • rvc
for product to form collision has to occur
sweetburger
  • 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.
TrojanPoem
  • TrojanPoem
So B,
rvc
  • rvc
yep the orientation matters too but nothing is mentioned about it
sweetburger
  • sweetburger
Yes, but never in any real reaction will all collision result in a product
rvc
  • rvc
oops B lol
sweetburger
  • sweetburger
I cant tell the difference between A and B they look worded the same
vera_ewing
  • vera_ewing
It was A.
sweetburger
  • sweetburger
That question was strange
vera_ewing
  • vera_ewing
vera_ewing
  • vera_ewing
How about this one?
sweetburger
  • sweetburger
I think this is a zero order reaction due to the negative slope/straight line
sweetburger
  • sweetburger
Im not completely sure im a bit rusty with kinematics
vera_ewing
  • vera_ewing
vera_ewing
  • vera_ewing
Which of these is correct?
rvc
  • rvc
i think it is first order reaction
sweetburger
  • sweetburger
Yea, it is I was wrong.
sweetburger
  • sweetburger
I didnt that the it was the ln[X] and not just [X] on the y axis
sweetburger
  • sweetburger
didnt notice*
rvc
  • rvc
yep
vera_ewing
  • vera_ewing
So which one is the answer?
rvc
  • rvc
need to solve this
rvc
  • rvc
@TrojanPoem what do you think?
rvc
  • rvc
well guys i need to go
rvc
  • rvc
Sorry to go like this in between :( all the best vera
TrojanPoem
  • TrojanPoem
cya @rvc
sweetburger
  • sweetburger
I think i have an Idea on how to solve this
sweetburger
  • sweetburger
We first say that y=-.0952x+.25 and then we plug in 10 for x and find the cooresponding y value
vera_ewing
  • vera_ewing
What did you get when you did that?
sweetburger
  • sweetburger
nope I dont think that is going to work nm
abb0t
  • abb0t
Water is not counted in the rate constant equation.
sweetburger
  • sweetburger
What?
abb0t
  • 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.
abb0t
  • 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.
vera_ewing
  • vera_ewing
Ohh I got C for my answer at first, but now I think I did it wrong /:
vera_ewing
  • vera_ewing
@abb0t What did you get?
sweetburger
  • sweetburger
@abb0t Could you help out with this question http://assets.openstudy.com/updates/attachments/558ea82ee4b0058b2bb701f0-vera_ewing-1435414099554-screenshot20150627at10.08.08am.png
abb0t
  • abb0t
Well, what does the graph tell you? Is it zeroth order, first order, second order...
vera_ewing
  • vera_ewing
First order right?
abb0t
  • abb0t
Yep. If the graph is linear and has a negative slope, it's first order. Remember that.
abb0t
  • 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.
vera_ewing
  • vera_ewing
I'm not really sure. I just used the one that @sweetburger gave me...
abb0t
  • abb0t
[A] = [A]\(_0\) - kt
sweetburger
  • sweetburger
I cant tell if the equation should be ln[A]=-kt+ln[A0] or something like [A]=[A0]e^(-kt)
abb0t
  • abb0t
oops, forgot the ln. this is first order.
sweetburger
  • sweetburger
that subscript function is useful i need to find that
abb0t
  • abb0t
They are both the same. Think of the natural log function, but you can use either.
abb0t
  • abb0t
I gtg. But i've clarified a lot. I'm sure sweetburger can take it from here.
sweetburger
  • sweetburger
Oh boy
vera_ewing
  • vera_ewing
Thanks @abb0t
sweetburger
  • sweetburger
Oh i think I may have figured it out
vera_ewing
  • vera_ewing
Okay...
sweetburger
  • sweetburger
I literally just need to find another concentration value at a specific time and this would be so easy.
vera_ewing
  • vera_ewing
Anything?
sweetburger
  • 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.
vera_ewing
  • vera_ewing
Okay, if I were to guess, which should I choose?
sweetburger
  • sweetburger
I dont know how much time you have to complete this, but I could try this one more way
sweetburger
  • sweetburger
No that didnt work either
sweetburger
  • sweetburger
Idk maybe try A, but I really dont know
sweetburger
  • sweetburger
OH ITS B
sweetburger
  • sweetburger
I just needed to do this i think .25/-.0952 = -2.62sec^-1
vera_ewing
  • vera_ewing
It was A!!
sweetburger
  • sweetburger
Very sorry :/
abb0t
  • 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 \(\sf \color{red}{\frac{M}{s}}\)
sweetburger
  • sweetburger
Im clueless @abb0t

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