## 7B9 part a

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

rachelle1K
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Joined: Sat Sep 07, 2019 12:16 am

### 7B9 part a

7B.9 For the first-order reaction A -> 3B + C, when [A]0 = 0.015 mol.L-1, the concentration of B increases to 0.018 mol.L-1 in 3.0 min. (a) What is the rate constant for the reaction expressed as the rate of loss of A?

For part a, I did
.018/3 = .006 to get the amount that C would increase as well, then subtracted the initial amount (.015) by .012. Why should we not take into account product C when determining the final concentration of A?

Chem_Mod
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### Re: 7B9 part a

We typically express the rate in terms of rate of loss of reactants, not products. The question also explicitly stated that the reaction rate is expressed as the rate of loss of A. To find the new concentration of A, you would subtract 0.006 from the initial concentration of A because for each mole A of A consumed, 3 moles of B are formed.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: 7B9 part a

think of this being like an ICE chart (except we aren't really talking about equilibrium, just a helpful model to figure out proportions):

A B C
I: .015M 0M ...
C: -.006M +.018M ....
E: .009M

Since we are only considering the rate in terms of the consumption of A, then we only need the final concentration of A. we only use B(product) as proportional relationship to A(reactant), so C is not even necessary to find in solving the problem.