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- Aug 27, 2012
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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.
"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.