## Problem 7A. 3 7th edition

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

Cecilia Jardon 1I
Posts: 74
Joined: Fri Sep 28, 2018 12:16 am

### Problem 7A. 3 7th edition

the equation C2H4(g) + 3O2(g)--> 2CO2(g) + 2H2O(g) and unique rate of reaction = 0.44 mol L^-1 s^-1 is given
it asks to find the rate at which O2 reacts.
The solutions manual says to multiply 0.44 mol L^-1 s^-1(3) which gives you 1.3 mol L^-1 s^-1.
I thought that for a unique average rate the equation was -1/a ((delta [A])/ delta t)).
so then you would do 0.44 x (-1/3) since your a= 3. Why is this wrong?

Kyleigh Follis 2H
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Joined: Fri Sep 28, 2018 12:28 am
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### Re: Problem 7A. 3 7th edition

The unique reaction rate is equal to -1/a of the reaction rate of O2. So, given the unique reaction rate = 0.44, then 0.44 = -1/3 (rate of O2). So you would multiply both sides by 3, which gives you 1.3 for the rate of O2.

Cecilia Jardon 1I
Posts: 74
Joined: Fri Sep 28, 2018 12:16 am

### Re: Problem 7A. 3 7th edition

Oh I see what you mean. Just what happens to the negative?

AhYeon_Kwon_2H
Posts: 48
Joined: Fri Sep 28, 2018 12:27 am

### Re: Problem 7A. 3 7th edition

Is it because we're dealing with rates that the negative isn't present?