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    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #1

    Dec 9, 2010, 02:31 AM
    Is the entropy become Larger or smaller ?
    CH4(g)+2S2(g)---> CS2(g)+2H2S(g)

    Is here the entropy become smaller ?


    Thanks for help :)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Dec 9, 2010, 07:08 AM

    Tell me if I'm wrong, entropy is some kind of measure of disorder, and is directly proportional to the number of particles, right?

    So?
    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #3

    Dec 9, 2010, 07:47 AM
    Yes truth so is become Larger in the Right side yes?

    But entropy is also Dependent in the State of matter. (g)>(l)>(s)

    Like here:H2SO4(l) - H2SO4(aq) so here where will be more entropy ?

    Thanks.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Dec 9, 2010, 07:51 AM

    In your question, the entropy is the same on both sides.

    In this:
    H2SO4(l) - H2SO4(aq)

    There is more entropy on the left side.

    Unless entropy is also affected by intermolecular bonding, I will stay with my answer that the entropy is the same on both sides for the initial question. (I never did entropy in High School, it was not in the syllabus)
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    #5

    Dec 9, 2010, 08:11 AM
    OK is start to be more clear but look if I have 4O(l)--->3O(g) so where will be more entropy in the Right side Because you can arrange the particle in more Different ways and the Disorder become More height.
    While in Liquid There is some Order.

    IF I Right so the NO of the particle is not all the time say where will be more entropy .

    Hope you can Understand my English :)
    Thanks.

    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Dec 9, 2010, 08:40 AM

    I'm trying to :p

    When there are more particles, there is greater entropy, and the entropy is highest in gas, in the middle for liquid and lowest in solid.
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    pop000 Posts: 352, Reputation: 6
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    #7

    Dec 9, 2010, 09:11 AM
    Comment on Unknown008's post
    OK thanks for helping :)

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