## Rate Law Slopes

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

Aiden Metzner 2C
Posts: 104
Joined: Wed Sep 18, 2019 12:21 am

### Rate Law Slopes

In lecture it seemed like professor said that k is always the slope of the graph of the rate. Is this true for zero order, first order, and second order reactions. Or is it only true for zero order?

Katie Bart 1I
Posts: 104
Joined: Sat Aug 24, 2019 12:16 am

### Re: Rate Law Slopes

The second order graph has a slope of k, and the zero order graph has a slope of -k. For the first order, sometimes the graph is exponential, but the slope is equal to k if ln(A) is on the y axis.

Jessica Li 4F
Posts: 115
Joined: Fri Aug 09, 2019 12:16 am

### Re: Rate Law Slopes

For a zero order reaction, if you just plot [A] over time, it should be a straight line, and -k will be the slope. For other order reactions, the graphs will not be straight lines, so you cannot determine k (the slope) unless you take either the ln[A] of the reaction for a first order reaction, in which k will be the slope, or 1/[A] for a second order reaction.

This helps you tell what order a reaction is.

Mariah
Posts: 104
Joined: Fri Aug 02, 2019 12:16 am

### Re: Rate Law Slopes

Aiden Metzner 2C wrote:In lecture it seemed like professor said that k is always the slope of the graph of the rate. Is this true for zero order, first order, and second order reactions. Or is it only true for zero order?

There is a great diagram in the book explaining this, I believe it is on page 608.