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?
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.
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?
No. N2 includes SO3, leftover SO2, leftover O2. You assumed N1-N2 is the number of reactions occured.
You mean N1-N2 isnt number of reactions??
Yes it is, but I disagree with your counting of N2. Read my original post. Read what I just posted.
To determine the number of moles of SO3 produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's calculate the initial number of moles of SO2 in the container. We can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Given:
P1 = initial pressure = 1.9 atm
V = volume = 2.00 L
R = 0.0821 L.atm/mol.K
T = temperature = 800 K
Plugging in the values:
n1 = (1.9 atm * 2.00 L) / (0.0821 L.atm/mol.K * 800 K)
n1 ≈ 0.115 mol
Next, let's calculate the final number of moles of SO2 in the container. The volume and temperature remain constant, so we can use the initial moles of SO2 to determine the final number of moles using the combined gas law equation:
P1V1 / n1 = P2V2 / n2
Given:
P1 = initial pressure = 1.9 atm
V = volume = 2.00 L
n1 = initial number of moles = 0.115 mol
P2 = final pressure = 1.65 atm
Plugging in the values:
(1.9 atm * 2.00 L) / (0.115 mol) = (1.65 atm * 2.00 L) / n2
n2 ≈ 0.093 mol
Now, we need to determine the change in the number of moles of SO2 to find the number of moles of SO3 produced. Since the balanced chemical equation for the reaction is:
2SO2 + O2 → 2SO3
We can see that for every 2 moles of SO2, we produce 2 moles of SO3. Therefore, the change in the number of moles of SO2 is equal to the number of moles of SO3 produced.
Change in moles of SO2 = n1 - n2
= 0.115 mol - 0.093 mol
= 0.022 mol
Hence, 0.022 moles of SO3 was produced during the reaction.