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Photon336

  • one year ago

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  1. Photon336
    • one year ago
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  2. Photon336
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    @sweetburger @Rushwr

  3. Rushwr
    • one year ago
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    is it C ?

  4. Photon336
    • one year ago
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    \[2CH _{4} + 3O _{2} --> 2CO + 4H _{2}O\]

  5. Photon336
    • one year ago
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    Like i though that for this reaction we have entropy goes up because we have on the reactant side 2+3 moles = 5 moles and on the product side we have 2+4 = 6 moles. So we have more moles on the product side so i thought more moles = more disorder.

  6. Photon336
    • one year ago
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    \[\Delta G = \Delta H - T \Delta S\]

  7. Photon336
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    wait... entropy >0 and if it's spontaneous G<0 so h <0

  8. Rushwr
    • one year ago
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    aaaah yes

  9. Rushwr
    • one year ago
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    okai since the product side had more moles than the reactants we took \[\Delta S\] as pistive right? Okai so anyhow the T\[\Delta S\] we consider the whole part as negative. But to find the enthalpy change if its +ve or -ve we need to know if the reaction is spontaneous or not right? But they haven't given that here.

  10. Photon336
    • one year ago
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    @Rushwr I agree, I had trouble with that, actually I just assumed that the reaction was spontaneous. b/c then I knew okay delta S was positive so then I thought that delta H had to be negative the book doesnt really give an explanation as to how you can figure out that it's exothermic, like it just says oh " this type of reaction is exothermic"

  11. Rushwr
    • one year ago
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    yeah so this reaction is exothermic

  12. Rushwr
    • one year ago
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    bond formation is always exothermic while bond dissociation is always endothermic. hence this reaction is exothermic. In exothermic reactions delta H is positive. Cuz it produces heat while in endothermic reactions delta H is negative cuz they absorb heat ! So that is why I choose that one

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