## Integrated Rate Laws (7b.1/ 15.21)

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

KarlaArevalo2F
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Joined: Tue Nov 14, 2017 3:01 am
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### Integrated Rate Laws (7b.1/ 15.21)

We are given that this is a first order reaction, 2A -> B + C, given that [A]0 = 0.153 mol/L and that after 115sec the conc. of B rises to 0.034 mol/L. How would we go about solving for [A]?

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

### Re: Integrated Rate Laws (7b.1/ 15.21)

Using the balanced mock chemical reaction, you can use stoichiometry to convert from [B] to [A]. Then, using the concentration of A that you solve in this manner, you can use the integrated rate law of the first order. Hope this helps.

Angela Grant 1D
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Joined: Fri Sep 28, 2018 12:25 am

### Re: Integrated Rate Laws (7b.1/ 15.21)

^to add on, you have to subtract the new [A] from the initial [A]. so when you use the molar ratio to get [A] when [B] = .034 mol/L, you should get [A] = .068 mol/L. then you have to subtract from the initial: .153 - .068 = .085 mol/L. this is the [A] you should use for your computation

KarlaArevalo2F
Posts: 67
Joined: Tue Nov 14, 2017 3:01 am
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thank you both!