Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rates and Hess' law

  1. Jun 6, 2004 #1
    I have two questions first on is on Rates and the second one is ont Hess' law.

    1.) If [A]o = .60 M and o = .80 M, what are the concencentrations of A and B after 2 minutes?

    Heres the reaction: A + 2B --> C
    The Rate law for this reaction is R=k[A]
    k = .00625s-1

    I found the concentration of A which is a direct plugin to the rate law, but i dont know how to get the concentration of B. The concentration of B at two mintues should be two time less than that of A, since two B's are being used up for every 1 A. I dont know how to go from there. Thanks

    Second Question:
    This problem involves Hess' law. The problem i am having is writing the equation for one of the sub reactions. I am given the heat of formation for
    Al2O3(s). so Al2O3(s) comes from Al(s) and O2 right.

    2Al(s) + (3/2)O2(g) ---> Al2O3(s)

    I think i am not doing this eqaution right that is why i am having a hard time with the overall problem.

  2. jcsd
  3. Jun 6, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    1) This is first order reaction kinetics
    d[A]/dt = -k[A] ,
    =>d[A]/[A] = -kdt. Now integrate this to get
    ln([A]/[Ao]) = -kt is the integral rate equation, which can also be written as
    [A] = [Ao] exp(-kt)

    Now it's just a matter of plugging in numbers to get [A]. Ooops, just realized that you've done this part already.

    Like you said (but not exactly), for every mole of A that reacts, 2 moles of B disappear. So find [Ao] - [A] and multiply this by 2. This gives you the change in conc. of B. If you subtract this from [Bo], you have your answer.

    2) There's nothing wrong with what you've written so far. Why do you think it's wrong ? What are you trying to calculate in this problem ? Perhaps explaining the complete problem might help.
  4. Jun 6, 2004 #3
    Thanks for part 1, I realized that i dont need help for part 2 i just figured it out. It was a hess' law problem and i kept trying to do it in my head and each time i tried it, it seemed like i had some oxygen's left over. I worte it on paper and it all worked out, so i dont need help on part two anymore. But thanks for the help on part 1.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Rates and Hess' law
  1. Hess's law (Replies: 4)

  2. Hess's Law (Replies: 2)

  3. Hess's Law (Replies: 0)

  4. Hess's Law (Replies: 6)

  5. Hess' Law (Replies: 3)