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Madhumathy posted an Question

August 04, 2020 • 01:01 am

30 points

IIT JAM
Chemistry (CY)

Actual question: 2so3-->2so2 but the equation is considered to be 2so3-->2so2+o2

my doubt is that in what basis do we consider the formation of O2 and how concentration of O2 became 0.3

2 Answer(s)

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Ashutosh singh

Dear Madhu, read this carefully! Let's say that in this balanced reaction, 2 moles of SO3 decompose to give 2 moles of SO2 and 1 mole of O2. It means if 2 moles of SO2 is decomposed at the reactant side, 2 moles of SO2 and 1 mole of O2 will be produced at the product side. Similarly, if 1 mole of SO2 is decomposed at the reactant side, only 1 mole of SO2 will be produced and 0.5 mole of O2 will be produced. It's simple molar calculations. Coming to your question, as you can clearly see that, 0.6 moles of SO2 is being dissociated...that's why 0.6 and 0.3 moles of SO2 and O2 will be formed at the product side.
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I hope you have understood this in a lucid way. 🙏😊 If any doubt, you can ask me here.
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Accept my answer. 🙏
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Lingareddy 1

given equation is correct . 2moles So3 is 0.6. So one mole of oxygen is 0.3
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Madhumathy

here it is sir
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here question is wrong...it must be given the whole equation .....may be misprint
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Priyanshu kumar Best Answer

can you send full question i think it is given in the question 2SO3-2SO2 +O2
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mistakes is in question
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and check how 0.3 comes👆
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but why only SO2's Concentration is taken in finding the concentration of SO3 sir? we must subtract (0.6+0.3). isn't this right?
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no when you take equilibrium condition you have to proceed like this...taking X if one mole if 2 mole then 2x
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is this correct sir?
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like this
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simply proceed like this👆
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why only 2x is taken sir😫
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2 mole is there madhumathy..so 2x
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if one mole then only x like in O2
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everything is given madhumathy initially 1M of SO3
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thankyou sir🙋🏻‍♀️
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you are welcome😊
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