Under some set of conditions, ammonia
gas and fluorine gas react to form nitrogen trifluoride gas and
hydrogen gas at 8.73ºC.
What is the correct rate law expression (including the rate law
constant) given the following data:
- Rxn #[NH3]0[F2]0
Rate (M/min) 10.2740.2181.841x10-2 20.2740.4367.362x10-2 30.8220.4362.209x10-1
The rate law expression for this
reaction is:
Rate0 =
k[NH3]0x [F2]0y
Comparing Rxn #1 and #2:
y = 2, the reaction is 2nd
order with respect to fluorine concentration
Comparing Rxn #3 and #2:
x = 1, the reaction is 1st
order w.r.t. ammonia concentration
Plugging in values from Rxn #1 and
solving for k...
1.841x10-2
M/min = k (0.274M)1 (0.218M)2
k = 1.42 M-2min-1
In Rxn #2, how much time must pass
before [F2] = 0.324M?
Plugging in to the 2nd order
integrated rate law expression...
t = 0.558 minutes
In Rxn #1, how much time must pass
before [NH3] = 0.217M?
Plugging in to the 1st order
integrated rate law expression...
t = 0.164 minutes
I was wondering if what we learned today (July 15) is going to be part of our exam tomorrow. Thank you!
ReplyDeleteThe material that was specifically acids & bases, no. BUT we talked about a lot of equilibrium in class today, and your exam is on equilibrium, so some of the concepts we talked about today might be related to material on the exam. Good luck.
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