## 7A.3A Reaction Concentration

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

J Medina 2I
Posts: 102
Joined: Wed Sep 25, 2019 12:17 am

### 7A.3A Reaction Concentration

I'm confused on the following question and was wondering if someone could provide a mental walkthrough of how they got the answer. (The answer is given by the textbook: (a) Second order in NO, first order in O2 )

"When the NO concentration is doubled, the rate of the reaction 2NO (g) + O2 (g) ---> 2NO2 (g) increases by a factor of 4. When both the O2 and the NO concentrations are doubled, the rate increases by a factor of 8.What are (a) the reaction order with respect to each reactant,"

Daniel Honeychurch1C
Posts: 109
Joined: Thu Jul 11, 2019 12:15 am

### Re: 7A.3A Reaction Concentration

When NO is doubled, the reaction rate increases by a factor of 4: the reaction order with respect to NO must be 2nd order (2^2 = 4)

When both NO and O2 are doubled the reaction rate increases by a factor of 8: we already know that NO is 2nd order, so its doubling alone results in the reaction order increasing by a factor of 4. Therefore, O2 must be first order, so that its doubling alone results in the reaction rate doubling. (Overall: 2^2 * 2^1 = 4 * 2 = 8)

Leonardo Le Merle 1D
Posts: 56
Joined: Wed Feb 27, 2019 12:16 am

### Re: 7A.3A Reaction Concentration

It helps me to isolate the reactant's effects on rate in an algebraic format.
2A = 4
2A*2B=8
Where A and B are the reactant's orders.
Therefore A must equal 2 and B must equal 1.

Tanmay Singhal 1H
Posts: 143
Joined: Sat Jul 20, 2019 12:16 am

### Re: 7A.3A Reaction Concentration

you have to find the order equation and mentally do the math