# Rates and Hess' law

## Main Question or Discussion Point

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.

Thanks

Gokul43201
Staff Emeritus
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.

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.