Anonymous
Anonymous asked in Science & MathematicsChemistry · 6 months ago

At a certain temperature, 0.325 mol CH4 and 0.664 mol H2S are placed in a 2.50 L container. CH4(g)+2H2S(g)↽−−⇀CS2(g)+4H2(g)?

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  • Dr W
    Lv 7
    6 months ago
    Favorite Answer

    reaction

    .. 1 CH4 + 2 H2S <---> 1 CS2 + 4 H2

    ice table.. for MOLES

    ... .. ... .... .... ..... CH4.. .. .... .. H2S.. .. .. .CS2.. .. . .H2

    .. initial.. .. . ... .. 0.325.. .... ... 0.664.. .. . . . ..0.. .. . .. .0

    .. change.. .. .. . ... X.. .. .. ... ... 2X.. .. .. ... . .X.. .. ...4X

    .. equilibrium.. (0.325-X).. .(0.664+2x).... .. . X.. .. . .4X

    and we know X = moles CS2 = 14.5g * (1 mol / 76.14g) = 0.1904

    so we can add that data to our ice table to get

    ... .. ... .... .... ..... CH4.. .. .... .. H2S.. .. .. .CS2.. .. . .H2

    .. initial.. .. . ... .. 0.325.. .... ... 0.664.. .. . . . ..0.. .. . .. .0

    .. change.. .. .. . .0.1904.. .. .. 0.3808.. .... 0.1904... 0.7616

    .. equilibrium.. .. .0.1346... .. . 0.2832... .. 0.1904.. . 0.7616

    and we know that for a reaction of this form

    .. aA + bB <----> cC + dD

    the equilibrium constant is

    .... . ... .. [C]^c * [D]^d

    . Kc = ---- ----- ---- ----

    .. ... .. ... [A]^a * [B]^b

    where those [ ] indicate "molarity"

    and we know this is a 2.50L container so..

    .... . ... .. [0.1904/2.50] * [0.7616/2.50]^4

    . Kc = ---- ----- ---- ---- ---- ---- ---- ---- ---- = ... . .you get to finish

    .. ... .. ... [0.1346/2.50] * [0.1904/2.50]^2

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  • Anonymous
    6 months ago

    At equilibrium, 14.5 g CS2 is present. Calculate 𝐾c .

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