## How to do 15.11 6th edition?

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Kobe_Wright
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Joined: Fri Sep 28, 2018 12:16 am

### How to do 15.11 6th edition?

I cannot figure out how to solve this problem or how heat is even factored into it?

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Joined: Fri Sep 28, 2018 12:19 am

### Re: How to do 15.11 6th edition?

I assume that because you're only looking for the initial rate of decomposition, the problem doesn't consider "being heated to 65*C" as factoring into the rate because it happens after the N2O5 is added.

Ashley Zhu 1A
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am

### Re: How to do 15.11 6th edition?

The problem tells you that N2O5 decomposes by a first order reaction, so the rate law will have the form rate = k[A] where A is the reactant, N2O5, and the it is raised to the first power since it is a first order reaction. You are also given k = 5.2 x 10^-3. As mentioned before, you don't need to do anything with the temperature here; the problem is just telling you that because k is different depending on what the temperature is. Since the "rate of decomposition" basically means the rate at which the reactants form products, all you need to do convert the 3.45 g of N2O5 into moles and divide by the 0.750L to get [N2O5] and then plug that and the k value into the equation rate = k[N2O5] to get the answer.

Kobe_Wright
Posts: 83
Joined: Fri Sep 28, 2018 12:16 am

### Re: How to do 15.11 6th edition?

Ashley Zhu 1A wrote:The problem tells you that N2O5 decomposes by a first order reaction, so the rate law will have the form rate = k[A] where A is the reactant, N2O5, and the it is raised to the first power since it is a first order reaction. You are also given k = 5.2 x 10^-3. As mentioned before, you don't need to do anything with the temperature here; the problem is just telling you that because k is different depending on what the temperature is. Since the "rate of decomposition" basically means the rate at which the reactants form products, all you need to do convert the 3.45 g of N2O5 into moles and divide by the 0.750L to get [N2O5] and then plug that and the k value into the equation rate = k[N2O5] to get the answer.

Okay I got this but how would you solve a second order problem where there is more than one reactant? What is the concentration now?

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Joined: Fri Sep 28, 2018 12:28 am
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### Re: How to do 15.11 6th edition?

Kobe_Wright wrote:
Ashley Zhu 1A wrote:The problem tells you that N2O5 decomposes by a first order reaction, so the rate law will have the form rate = k[A] where A is the reactant, N2O5, and the it is raised to the first power since it is a first order reaction. You are also given k = 5.2 x 10^-3. As mentioned before, you don't need to do anything with the temperature here; the problem is just telling you that because k is different depending on what the temperature is. Since the "rate of decomposition" basically means the rate at which the reactants form products, all you need to do convert the 3.45 g of N2O5 into moles and divide by the 0.750L to get [N2O5] and then plug that and the k value into the equation rate = k[N2O5] to get the answer.

Okay I got this but how would you solve a second order problem where there is more than one reactant? What is the concentration now?

You would need to know the concentration of all reactants in order to solve the problem