Aug 27, 2012
5
0
Status
Pre-Medical
The question is:
"Which graph depicts partial pressures as a function of time after 0.75 atm SO2 is mixed with 0.50 atm O2?"
and is based of of this equation
2 S02(g) + 1 02(g) → 2 S03(g)
Kp =5.82 x10^2 atm.-1 at 500C
And the solution states that
Choice C is correct. Sulfur dioxide and oxygen are both reactants, according to Reaction 1. When mixing 0.75 atm. SO2 with 0.50 atm. O2, Reaction 1 proceeds in the forward direction to establish equilibrium. The value of Keq for this reaction (582), as listed in the passage, is well above one. At equilibrium, there is significantly more SO3 than both SO2 and O2. This eliminates choices Band D. The SO2 should start higher than the O2, but it diminishes twice as fast as O2, thus the SO2 line should drop below the O2 line at equilibrium. If SO2 is completely depleted (all 0.75 atm. are consumed), 0.375 atm. of O2 would be consumed, leaving 0.125 atm. O2. This is because SO2 is the limiting reagent, if the reaction goes to completion. This makes choice C the best answer.
Reading of the graphs is not hard but what I don't understand is how .75 atm of SO2 reacts with .375 atm O2. I know the ratio is 2 SO2 to 1 O2 but how does that translate into pressures of .75 and .375? I know I am missing something simple but any explanation would be apprciated. Thanks.
 

TheRhymenocerous

2+ Year Member
Oct 4, 2014
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Moles are proportional to pressures for gases, so it's just the mole ratio – 2 moles/atm of SO2 react for every 1 mole/atm of O2. Half of 0.75 is 0.375.
 

chemtopper

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The question is:
"Which graph depicts partial pressures as a function of time after 0.75 atm SO2 is mixed with 0.50 atm O2?"
and is based of of this equation
2 S02(g) + 1 02(g) → 2 S03(g)
Kp =5.82 x10^2 atm.-1 at 500C
And the solution states that
Choice C is correct. Sulfur dioxide and oxygen are both reactants, according to Reaction 1. When mixing 0.75 atm. SO2 with 0.50 atm. O2, Reaction 1 proceeds in the forward direction to establish equilibrium. The value of Keq for this reaction (582), as listed in the passage, is well above one. At equilibrium, there is significantly more SO3 than both SO2 and O2. This eliminates choices Band D. The SO2 should start higher than the O2, but it diminishes twice as fast as O2, thus the SO2 line should drop below the O2 line at equilibrium. If SO2 is completely depleted (all 0.75 atm. are consumed), 0.375 atm. of O2 would be consumed, leaving 0.125 atm. O2. This is because SO2 is the limiting reagent, if the reaction goes to completion. This makes choice C the best answer.
Reading of the graphs is not hard but what I don't understand is how .75 atm of SO2 reacts with .375 atm O2. I know the ratio is 2 SO2 to 1 O2 but how does that translate into pressures of .75 and .375? I know I am missing something simple but any explanation would be apprciated. Thanks.
SO2, O2 and SO3 all are gases and present in the same container means at the same temperature and volume.
PV=nRT
n=PV/RT
if V and RT are constant, then P is proportional to n .Amount of gas is proportional to n and P
 
OP
N
Aug 27, 2012
5
0
Status
Pre-Medical
Thanks guys. After reading your answers I wanted to try to figure it out for myself and the way I understood it was if the mole ratio of SO2/O2 is 2 and the partial pressure ratio is .75/.5 is 1.5 which is smaller than 2 so that means SO2 is limiting. I just was not connecting the fact that if V, R and T are constant P does not equal n but is proportional to it. Thanks again.
 

helpmewithochem

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Oct 16, 2014
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Thanks guys. After reading your answers I wanted to try to figure it out for myself and the way I understood it was if the mole ratio of SO2/O2 is 2 and the partial pressure ratio is .75/.5 is 1.5 which is smaller than 2 so that means SO2 is limiting. I just was not connecting the fact that if V, R and T are constant P does not equal n but is proportional to it. Thanks again.
Can you please explain based on the data, why SO2 is limiting and not O2? I understand that common sense would dictate that O2 gas would be in excess, thus rendering SO2 the limiting reagent, but there are clearly more moles of SO2 and the experiment even stated that more SO2 was mixed than O2 (0.75 atm vs 0.5 atm). Thank you!
 

chemtopper

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Can you please explain based on the data, why SO2 is limiting and not O2? I understand that common sense would dictate that O2 gas would be in excess, thus rendering SO2 the limiting reagent, but there are clearly more moles of SO2 and the experiment even stated that more SO2 was mixed than O2 (0.75 atm vs 0.5 atm). Thank you!
Limiting reactant is not determined just by looking at the moles of the reactants available.Convert moles of SO2 to SO3 and moles of O2 to moles of SO3 using balanced equation.Now check which reactant gives you less number of moles of the product and that reactant is the limiting reactant.