## Changing Differential Rate Law to Integrated Rate Law

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

Jessica Dang 1E
Posts: 11
Joined: Fri Sep 25, 2015 3:00 am

### Changing Differential Rate Law to Integrated Rate Law

Rate = $\frac{-1}{a} \frac{d[A]}{dt} = k[A]^n = k [A]^1$
and n = 1 because it's a first order reaction.

I had quick question as to what "1/a" represents because in the textbook the rate law for the consumption of A just says -1? (page 623 of textbook)

KVu 3G
Posts: 12
Joined: Fri Sep 25, 2015 3:00 am

### Re: Changing Differential Rate Law to Integrated Rate Law

a refers to the coefficient in a general equation $aA+ bB \rightarrow ...$

The term -1/a defines how fast A is being consumed relative to the other reactants and products in the equation. For example B may not be consumed at the same rate A does, it's rate of consumption is dependent on b.