Sapling Week 9/10 Question 1

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Sean Phen
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Joined: Wed Sep 30, 2020 10:01 pm

Sapling Week 9/10 Question 1

Postby Sean Phen » Sun Mar 07, 2021 11:03 pm

"The rate of the given reaction is 0.600 M/s.
A+4B⟶2C

What is the relative rate of change of each species in the reaction?"

What is the easiest method of solving for the relative rate of change? I get confused on how to do it.

Tanya Bearson 2K
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Joined: Thu Dec 17, 2020 12:18 am

Re: Sapling Week 9/10 Question 1

Postby Tanya Bearson 2K » Sun Mar 07, 2021 11:14 pm

You can use the equation: rate = (-1/coefficient of species)*(change in concentration of species/change in time) to find the relative rate of each species which is the (change in concentration of species/change in time) part of the equation. For example for A you would use the equation .600 M/s = (-1/1)*(delta[A]/delta t) to find that (delta[A]/delta t) = -.600 M/s
Last edited by Tanya Bearson 2K on Sun Mar 07, 2021 11:17 pm, edited 1 time in total.

Stacey Phan 2I
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Re: Sapling Week 9/10 Question 1

Postby Stacey Phan 2I » Sun Mar 07, 2021 11:15 pm

I think you just multiply the rate by the coefficient. Reactants have negative rates since they are being consumed.

Melanie Lin 3E
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Joined: Wed Sep 30, 2020 9:38 pm

Re: Sapling Week 9/10 Question 1

Postby Melanie Lin 3E » Sun Mar 07, 2021 11:16 pm

Hi! I honestly suck at explaining but I'll do my best. I usually just remember its always (-1/a)*(d[A]/dt)=rate or any of the reactants (make sure to have that negative sign because they technically lose reactants as the forward reaction goes). As for products, it'll be (1/c)*(d[C]/dt)=rate (it'll be positive because we gain products). For example, (-1/b)*(d[B]/dt)=rate so d[B]/dt=(0.600 M/s)*(-4). Hope this helps!

Christine Ma 3L
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Re: Sapling Week 9/10 Question 1

Postby Christine Ma 3L » Sun Mar 07, 2021 11:19 pm

In general the rate of a reaction of the form is equal to = .

The solve for the relative rate of change of each reactant, you just multiply the overall reaction rate by its negative coefficient (since reactants are decreasing).





For the relative rate of change of a product, you just multiply the reaction rate by its coefficient.



Ven Chavez 2K
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Re: Sapling Week 9/10 Question 1

Postby Ven Chavez 2K » Sat Mar 13, 2021 11:03 am

The stoichiometric coefficients provide the information needed to solve the relative rate change. I think of it as 1A is consumed, 4B is consumed, and 2C is made. Therefore you multiply the rate of the reaction by how much each reactant is used and how much product is created. So you multiply the rate by 1 for A, multiply by 4 for B, and multiply by 2 for C. Since reactants are being consumed and their concentration decreases, the reactants have a negative rate. Since products are being formed and their concentration increases, the products have a positive rate.

SashaAnand2J
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Joined: Wed Sep 30, 2020 9:37 pm

Re: Sapling Week 9/10 Question 1

Postby SashaAnand2J » Sat Mar 13, 2021 9:29 pm

Hi Sean!
The unique rates per product and reactant are based upon stoichiometric coefficients. The inverse of the stoichiometric coefficient multiplied by the change in concentration over time of a particular substance is equal to the overall rate of the reaction. This should also help you relate the reactants to the products! Hope that helps.

Daniela Santana 2L
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Re: Sapling Week 9/10 Question 1

Postby Daniela Santana 2L » Sun Mar 14, 2021 10:56 pm

Hi! I solved this question by looking at the reaction and taking note of the coefficients in front of each substance. I would essentially find the reciprocal of the coefficient and put that in instead of the coefficient thats given. For example, the 4 in 4B would become 1/4. I would rewrite the whole reaction this way and set it equal to the rate (.600). You need to find the value of which you can multiply by the coefficient for each substance and get .600. I hope this helped!

205678283
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Joined: Fri Sep 24, 2021 5:44 am

Re: Sapling Week 9/10 Question 1

Postby 205678283 » Wed Mar 02, 2022 2:02 pm

all you have to do is multiply the rate they give you by the coefficient in the chemical equation! :)


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