Isolating x in Rate of Reaction?

  • Context: Undergrad 
  • Thread starter Thread starter mwall
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around isolating the variable x in an equation related to the rate of reaction. Participants explore methods for rearranging the equation and simplifying the expression, focusing on mathematical manipulation rather than the underlying chemistry.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents an equation involving exponential functions and seeks assistance in isolating x.
  • Another participant suggests "folding" the constants together to simplify the expression, proposing that it can be expressed in the form e^{ax} = b.
  • A further reply emphasizes the importance of taking logarithms to isolate x, providing a formula for x in terms of logarithmic functions.
  • Another participant reiterates the method of "folding" constants, referencing the property of exponents.

Areas of Agreement / Disagreement

Participants generally agree on the approach of simplifying the equation through mathematical manipulation, but there is no consensus on the specific steps or methods to be used.

Contextual Notes

Some assumptions regarding the properties of exponents and logarithms are present, but not explicitly stated. The discussion does not resolve the mathematical steps involved in isolating x.

Who May Find This Useful

Individuals interested in mathematical techniques for rearranging equations, particularly in the context of chemical kinetics and reaction rates.

mwall
Messages
3
Reaction score
0
I am working on a problem relating to rate of reaction. I am not sure how to isolate the x in the following equation.

e^-45/(8.31)(x)
e^-45/(8.31)(353)

mwall
 
Mathematics news on Phys.org
Sorry, I forgot the rest of the equation.

.072 = e^-(45/8.31*x)
.002 e^-(45/8.31*352)

mwall
 
Whenever you're not sure how to proceed with rearranging an expression like that one the first thing you should do is to "fold" the constants together. If you do so then it is simply expressed as,

e^{ax} = b.

So obviously you just need to take logs of both sides to get,

x = \frac{\log(b)}{a}
 
Last edited:
the first thing you should do is to "fold" the constants together

This can be done by remembering e^a / e^b = e^{a-b}
 
Thanks for your help.

mwall
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
1K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
2K
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K