Question

SO2 mixed with O2, 800K is in the 2.00L metal container and it was 1.9atm. The reaction occur and than pressure of container dropped to 1.65atm. How many SO3 was produced?

Answer 1

start off with PV=nRT solve for n, number moles of SO2+O2
then, solve for final nf on the second condition. Nf represents the total number of moles of so2, so3,O2. Now you know for each SO3 produced, the original number of SO2 is reduced at the same rate, and for each SO3 produced, the number of O2 is reduced by half that rate.

ni=SO2i+O2i
nf=SO2f+O2f+SO3=SO2i-SO3 + O2i -1/2 SO3+ SO3
= SO2i+O2i -1/2 SO3=ni -1/2 SO3
or SO3=2(ni-nf) check my thinking.

Answer 2

My solution:
2SO2 + O2 -> 2SO3
When a reaction occurred once, 2 molecule of SO2 and 1 molecule of O2 are used, 2 molecule of SO2 is produced. This mean every one reaction loss one molecule on total and produce 2 molecule of SO3.

N1=(1.9atm)(2.0L)/800KR
=5.79x10^-2mol
N2=(1.65atm)(2.0L)/800KR
=5.03x10^-2mol
N1-N2=0.76×10^-2mol

So reaction occured 0.76×10^-2mol times and every reaction produce 2 molecule of SO3.

Conclusion: 0.76×10^-2molx2=1.52×10^-2 of SO3 was produced.

Am I right?

Answer 3

No. N2 includes SO3, leftover SO2, leftover O2. You assumed N1-N2 is the number of reactions occured.

Answer 4

You mean N1-N2 isnt number of reactions??

Answer 5

Yes it is, but I disagree with your counting of N2. Read my original post. Read what I just posted.