Rate Constant Problem: Find n & Rate Constant

  • Thread starter Thread starter grint
  • Start date Start date
  • Tags Tags
    Constant Rate
Click For Summary
SUMMARY

The discussion focuses on determining the rate constant (k) and the reaction order (n) from the equation k' = k[OH-]^n. By applying logarithmic transformations, the equation is reformulated to log k' = n log([OH-]) + log k, which resembles the linear equation y = mx + b. The slope (m) of the resulting line corresponds to the reaction order n, which is calculated to be 1.04. Subsequently, the value of k can be derived using the calculated n and the measured k'.

PREREQUISITES
  • Understanding of chemical kinetics and rate laws
  • Familiarity with logarithmic functions and transformations
  • Basic knowledge of graphing linear equations
  • Experience with experimental data collection and analysis
NEXT STEPS
  • Study the derivation of rate laws in chemical kinetics
  • Learn about the significance of reaction orders in rate equations
  • Explore methods for calculating rate constants from experimental data
  • Investigate the use of linear regression to analyze kinetic data
USEFUL FOR

Chemistry students, researchers in chemical kinetics, and professionals involved in reaction mechanism studies will benefit from this discussion.

grint
Messages
3
Reaction score
0
http://imgur.com/IgOdMW3

Question in the link. How do I go from here and get the rate constant and n values?
 
Physics news on Phys.org
Equation: k' = k[OH- ]^n

k' and [OH-] measured / calculated

Log both sides of the equation above

log k' = log (k[OH- ]^n )
log k' = log k + n log([OH- ]

which is the same as

log k' = n log([OH- ] + log k
Which you can look at as y = mx + b

m = n, therefore slope = n, n = 1.04
calculate k after that
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
32
Views
16K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K