Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Brian Lin on December 29, 2013, 01:01:04 AM

This question took me a long time, but I think I got it.
The decomposition of NO2(g) occurs by the following bimolecular elementary reaction:
2NO2(g) :rarrow: 2NO(g) + O2(g)
The rate constant at 273K is 2.3*10^12 L/mol·s, and the activation energy is 111 kJ/mol. How long will it take for the concentration of NO2(g) to decrease from an initial partial pressure of 2.5 atm to 1.5 atm at 500.K? Assume ideal gas behavior.
First, I divided the question into 2 parts. Finding the K rate law for 500.K, not 273 K since the reaction occurs at 500.K. Thus, you use the Arrhenius equation
Ln(K_{2}/K_{1})=Ea/R(1/T_{1}1/T_{2})
where sub 2 is WHAT I WANT FOR 500.K and sub 1 is WHAT I WAS GIVEN (2.3*10^12L/mol·s)
Plugging and chugging..
K2 is 1.0*10^2 L/mol·s for 500.K
(You can take off the sub 2 now, since we will be dealing with only one k)
Now, attacking the 2nd part of the question. We need the time it takes.
Initially, I tried to make up my own concentrations for NO2 but I realized you get different answers. So, I assumed that you do NOT need to know initial concentrations. However, you are given pressures.
PV=nRT :rarrow: P=(n/v)RT :rarrow: P= MRT :rarrow: (P/RT)=M = [NO2] right?
Now, I used the rearranged form of 2nd Order Integrated Rate Law:
((1/[NO2])(1/[NO2]_{initial})) divided by K(rate law I got previously)= Time
((RT/(P_{final})(RT/P_{initial})) divided by K(rate law)= time
((0.08206*500./1.5)  (0.08206*500./2.5)) divided by 1.0*10^2 = time
Time= 1.1*10^3 seconds
I did not include units in the above calculations but for my paper work, I included and they canceled out perfectly.
What kept me stuck was that I did kept on setting up ratio proportions with P and n with the PV=nRT. Initially, I did not see that Molarity could be represented by the pressures and gas constant and temperatures (I kept on canceling them out since they are "constant" from initial to final pressure)
Thanks.