January 15, 2025, 01:05:03 PM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1]   Go Down

Topic: Rate Law  (Read 6815 times)

0 Members and 1 Guest are viewing this topic.

Offline Violet89

  • Regular Member
  • ***
  • Posts: 58
  • Mole Snacks: +1/-2
Rate Law
« on: February 15, 2013, 05:51:44 PM »
Consider the reaction with the rate law, Rate = k[BrO3-][Br-][H+]2

By what factor does the rate change if the concentration of BrO3- is doubled, that of Br- is decreased by a factor of 2 and that of H+ is decreased by a factor of 2?

[BrO3-] = 1
1 * 2 = 2

[Br-] = 1
???

[H+] = 2
([H+]^2) / (2^2) = ([1]^2) / (2^2) = 1 / 4


I know that the final answer is 1/4.

I'm confused on how to solve for Br-. I'd appreciated any help.
Logged

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27897
  • Mole Snacks: +1816/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Rate Law
« Reply #1 on: February 15, 2013, 06:51:33 PM »
Not sure what the problem is - how is it different from the approach you used for two other concentrations?
Logged
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Stovn0611

  • Regular Member
  • ***
  • Posts: 34
  • Mole Snacks: +4/-1
  • Gender: Male
Re: Rate Law
« Reply #2 on: February 16, 2013, 02:20:38 PM »
[Br-] starts at 1 and then decreases by a factor of 2. Think about what happens when you decrease 1 by a factor of 2 to get the new [Br-]
Logged
Pages: [1]   Go Up
« previous next »
Sponsored Links