Sapling Question week 9/10
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Sapling Question week 9/10
Hey guys! For this question, I was able to answer parts of it but did not understand some of the parts. Here is what I was able to answer:
For the reaction
2A(g)+2B(g)+C(g)⟶3G(g)+4F(g)
the initial rate data in the table was collected, where [A]0 , [B]0 , and [C]0 are the initial concentrations of A , B , and C , respectively.
Experiment [A]0 (mmol⋅L−1) [B]0 (mmol⋅L−1) [C]0 (mmol⋅L−1) Initial rate (mmol⋅L−1⋅s−1)
1 17.0 100.0 280.0 3.40
2 34.0 100.0 210.0 6.80
3 34.0 200.0 70.0 27.2
4 17.0 100.0 140.0 3.40
part 1: Identify the order of reactant A.
I was able to answer this correctly: first order reactant
part 2: Identify the order of reactant B.
I was able to answer this correctly: second order reactant
part 3: Identify the order of reactant A.
I was able to answer this correctly: zero order reactant
part 4:What is the overall order of the reaction?
I was able to answer this correctly: 3
part 5: Write the rate law for the reaction where k is the rate constant.
My only answer I could come up with was 170000, but it was wrong. Can someone please explain how to solve this?
part 6: Calculate the rate constant, k , and identify its units.
Can someone please explain how to solve this part as well? I was not able to approach this question correctly and have no idea how to solve it.
part 7: Select the units for the rate constant.
I was able to answer this correctly: L2·mmol–2·s–1
part 8: Determine the initial rate of the reaction when [A]0=4.70 mmol⋅L−1 , [B]0=0.183 mmol⋅L−1 , and [C]0=12.0 mmol⋅L−1.
Can someone please explain how to solve this part as well? I was not able to approach this question correctly and have no idea how to solve it.
Thank you in advance to whoever is able to help me.
For the reaction
2A(g)+2B(g)+C(g)⟶3G(g)+4F(g)
the initial rate data in the table was collected, where [A]0 , [B]0 , and [C]0 are the initial concentrations of A , B , and C , respectively.
Experiment [A]0 (mmol⋅L−1) [B]0 (mmol⋅L−1) [C]0 (mmol⋅L−1) Initial rate (mmol⋅L−1⋅s−1)
1 17.0 100.0 280.0 3.40
2 34.0 100.0 210.0 6.80
3 34.0 200.0 70.0 27.2
4 17.0 100.0 140.0 3.40
part 1: Identify the order of reactant A.
I was able to answer this correctly: first order reactant
part 2: Identify the order of reactant B.
I was able to answer this correctly: second order reactant
part 3: Identify the order of reactant A.
I was able to answer this correctly: zero order reactant
part 4:What is the overall order of the reaction?
I was able to answer this correctly: 3
part 5: Write the rate law for the reaction where k is the rate constant.
My only answer I could come up with was 170000, but it was wrong. Can someone please explain how to solve this?
part 6: Calculate the rate constant, k , and identify its units.
Can someone please explain how to solve this part as well? I was not able to approach this question correctly and have no idea how to solve it.
part 7: Select the units for the rate constant.
I was able to answer this correctly: L2·mmol–2·s–1
part 8: Determine the initial rate of the reaction when [A]0=4.70 mmol⋅L−1 , [B]0=0.183 mmol⋅L−1 , and [C]0=12.0 mmol⋅L−1.
Can someone please explain how to solve this part as well? I was not able to approach this question correctly and have no idea how to solve it.
Thank you in advance to whoever is able to help me.
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Re: Sapling Question week 9/10
Part 5 isn't asking for a numerical value. You use the order for each reactant (found in parts 1,2,3) to find the rate law for this reaction:
rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.)
Using this equation, you can use the rate and concentrations from any of the experiments in the table to calculate k in part 6.
For part 8, you simply use the rate law and the rate constant calculated in part 6, plug in the given values, and solve for the rate. I hope this helps!
rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.)
Using this equation, you can use the rate and concentrations from any of the experiments in the table to calculate k in part 6.
For part 8, you simply use the rate law and the rate constant calculated in part 6, plug in the given values, and solve for the rate. I hope this helps!
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Re: Sapling Question week 9/10
For part 8 we use our rate law: rate= k[A][B]^2 and just plug in the initial concentrations of A and B (we don't need to include C since it is to the power of 0 making it equal to 1).
Re: Sapling Question week 9/10
To find the rate constant you must find the order of all the reactants and plug in the values from the table.
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Re: Sapling Question week 9/10
Jasmine Ho 3I wrote:Part 5 isn't asking for a numerical value. You use the order for each reactant (found in parts 1,2,3) to find the rate law for this reaction:
rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.)
Using this equation, you can use the rate and concentrations from any of the experiments in the table to calculate k in part 6.
For part 8, you simply use the rate law and the rate constant calculated in part 6, plug in the given values, and solve for the rate. I hope this helps!
This was very helpful thank you.
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Re: Sapling Question week 9/10
Jasmine Ho 3I wrote:Part 5 isn't asking for a numerical value. You use the order for each reactant (found in parts 1,2,3) to find the rate law for this reaction:
rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.)
Using this equation, you can use the rate and concentrations from any of the experiments in the table to calculate k in part 6.
For part 8, you simply use the rate law and the rate constant calculated in part 6, plug in the given values, and solve for the rate. I hope this helps!
Can you explain more in depth how to calculate part 6?
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Re: Sapling Question week 9/10
Lusin_Yengibaryan_3B wrote:Jasmine Ho 3I wrote:Part 5 isn't asking for a numerical value. You use the order for each reactant (found in parts 1,2,3) to find the rate law for this reaction:
rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.)
Using this equation, you can use the rate and concentrations from any of the experiments in the table to calculate k in part 6.
For part 8, you simply use the rate law and the rate constant calculated in part 6, plug in the given values, and solve for the rate. I hope this helps!
Can you explain more in depth how to calculate part 6?
In part 5, you got the rate law with constant k. The table gives values for initial concentrations [A], [B], [C], and the initial rates. You can choose any experiment and plug in whatever values are part of your rate law, and your only remaining variable will be k, so you can just solve!
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Re: Sapling Question week 9/10
Hi! For part 5, the generic rate law can be rewritten in terms of just A and B since the reaction is zero order in C and the reaction rate is independent of [C]. Therefore, rate = k[A]^a[B]^b. To solve for a and b, determine the orders of each reactant. Once you find the final equation, you just input the values for part 6.
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Re: Sapling Question week 9/10
For part six are the initial concentrations and our answer from part five supposed to be used? Because I couldn't seem to come up with an answer to this part
Re: Sapling Question week 9/10
For part 6 are we supposed to use the set up from part 5, because when I do that I keep getting 50000 which is wrong
Re: Sapling Question week 9/10
these helped me alot but I am still confused if part 5 should be numeric or not
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Re: Sapling Question week 9/10
To clear any misunderstanding with the homework, Part 5 should not be numeric, you just input the rate law, but it has to be simplified. For example, if it's [A]1,[B]2.[C]0, then it should be simplified into [A],[B]2.
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Re: Sapling Question week 9/10
Can someone explain why B is second order? When I tried solving for it I compared experiments 2 and 3 and the difference for B was x2 (which made me think it was first order) but the initial rates 4.80 and 19.2 don't have a difference of x2 so I don't get how its first or second order.
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Re: Sapling Question week 9/10
Part 5: This question is asking for the rate law which is represented in the non-numerical form of Rate=k[A]^a[B]^b[C]^c where you would use the order of each reactant that you found in parts 1, 2, & 3 and plug those orders in as the exponents a, b, & c, respectively. (Note: If a reactant is zero order, it would not appear in the rate law because its concentration does not affect the rate of reaction.)
Part 6: Here, you would use the rate law equation that you found in part 5 and choose an experiment from the table. If you were to pick experiment 1 for example, [A] of your rate law would be 17, [B] would be 100, [C] would be 280, and Rate would be the given initial rate for this reaction which is 3.40. Plug in these values and solve for k.
Part 8: Once you have determined your rate law in part 6, you would just plug and chug the values given in part 8 into the rate law you found. k would be the rate constant (k) that you should have calculated in part 6.
Hope this helps!
Part 6: Here, you would use the rate law equation that you found in part 5 and choose an experiment from the table. If you were to pick experiment 1 for example, [A] of your rate law would be 17, [B] would be 100, [C] would be 280, and Rate would be the given initial rate for this reaction which is 3.40. Plug in these values and solve for k.
Part 8: Once you have determined your rate law in part 6, you would just plug and chug the values given in part 8 into the rate law you found. k would be the rate constant (k) that you should have calculated in part 6.
Hope this helps!
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Re: Sapling Question week 9/10
Sarah Khan wrote:Can someone explain why B is second order? When I tried solving for it I compared experiments 2 and 3 and the difference for B was x2 (which made me think it was first order) but the initial rates 4.80 and 19.2 don't have a difference of x2 so I don't get how its first or second order.
Hello Sarah,
In order to solve for B it would be easier to first determine the order of C (as none of the values of C are identical for any of the experiments). If done correctly, you should get that C is zero order which means that its concentration will not affect the initial rate of the experiment. Therefore, when finding the orders of A and B, you no longer have to include C in your calculations because you already know that C doesn't affect the value at all. This makes calculating A and B a whole lot easier. The calculation should look like:(
k (20)^a (100)^b 4
______________ = ___
k (20)^a (200)^b 16
[Btw the "." behind the b doesn't mean anything, it is so that 4/16 visually ends up on the right side of the equation on chemistry community] Before beginning dividing, k and 20^a cancel out, so now you're just left with:
(100)^b 4
______ = ___
(200)^b 16
Which equals:
.5^b = .25
Now you just solve for the exponent (b) by taking the natural log of each side. Therefore, b should equal 2, meaning that B has an order of 2.
Hope this helps!
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