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Abhisar

  • one year ago

Rate of formation of \(\sf SO_3\) according to the reaction \(\sf 2SO_2+O_2 \rightarrow 2SO_3\) is \(\sf 1.6 \times 10^{-3}~Kg.min^{-1}\). Rate at which \(\sf SO_2\) reacts is?

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  1. Abhisar
    • one year ago
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    @Rushwr can you help?

  2. Abhisar
    • one year ago
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    @Woodward Help pwease... c:

  3. Abhisar
    • one year ago
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    According to me it should be either \(\sf 1.6 \times 10^{-3} or ~8 \times 10^{-4}\)

  4. Abhisar
    • one year ago
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    \(\sf \Large \frac{1}{2}-\frac{d[SO_2]}{dt}=\frac{1}{2}\frac{d[SO_3]}{dt} \), isn't that correct?

  5. Abhisar
    • one year ago
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    Umm...but the answer key given says it should be \(\sf 1.28 \times 10^{-3}\)

  6. Abhisar
    • one year ago
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    I dunno how :O

  7. Abhisar
    • one year ago
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    \[\frac{- d[SO _{2}] }{ dt }= \frac{ 1 }{ 2 }\frac{ d[SO _{3}] }{ dt }\] right? So I think u are correct ! How? Isn't the rate of formation of SO3 equal to rate of disappearance of SO2 ?

  8. Rushwr
    • one year ago
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    wait what's the equation ! I think I got i wrong!

  9. Abhisar
    • one year ago
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    \(\sf 2SO_2+O_2 \rightarrow 2SO_3\)

  10. Rushwr
    • one year ago
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    I have no idea !!!!!!!!!! Still wondering how it can be !!!!!!!!!!!!!!!!!!!!!

  11. Abhisar
    • one year ago
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    Ok, Thnks. I think the answer key is incorrect!

  12. Abhisar
    • one year ago
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    But can you reconfirm what the rate equation will be ? Wil it be \(\sf \Large \frac{1}{2}\frac{-d[SO_2]}{dt}=\frac{1}{2}\frac{d[SO_3]}{dt} \) or the one you wrote?

  13. Rushwr
    • one year ago
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    this one is correct I agree ! But The answer won't be the same isn't it !

  14. Abhisar
    • one year ago
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    If they are asking rate of disappearance of SO2 then it would be same as that of SO3 but if they mean the rate of reaction it should be half i.e. 8 X !0^-4. that's what I think.

  15. Rushwr
    • one year ago
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    I think it would have the same value. 1.6

  16. Abhisar
    • one year ago
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    Yeah, cz they are asking the rate at which SO2 reacts which sounds more like its individual rate. All right thanks a bunch for your time. really appreciate it c;

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