## Psuedo First Order Reaction

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

Sarah Sharma 2J
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

### Psuedo First Order Reaction

Hello! Can someone please explain the purpose of the Pseudo first order reaction? I understand that it changes the way the equation is visualized so that we can only focus on one reactant however, how would one go about solving the substitution?

Johann Park 2B
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Joined: Thu Jul 27, 2017 3:01 am
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### Re: Psuedo First Order Reaction

So when we have a reaction with multiple reactants changing, such as A + B + C -> P, we make one [R] small and the others very large. In practice, we would thus have: k' = [A]N, just like in a normal first-order rate problem (except the fact that we denote k as k'). Then you would do the same by making the other reactant concentrations large. k' is found by the slope of the linear plot

Michelle Dong 1F
Posts: 110
Joined: Fri Sep 29, 2017 7:04 am

### Re: Psuedo First Order Reaction

Because we set the other reactants to be very large, they are essentially constant and the concentrations do not change significantly throughout the reaction. So, they don't have much impact on the overall rate. For example, if you have A+ B+ C --> P, and you set A to be really small and B and C to be really large and A, B, and C were all first order (overall reaction would be 3rd order), the new rate law for pseudo first order would just be k'[A], with k'=k[B][C]. If they give us the concentrations of B and C and we can calculate k' from the slope of the ln[A] vs time plot, then we would plug k', [B], and [C] into k'=k[B][C] to find the value of k.