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Please help me with this chemistry question!!!?

33. Given the following data: 2A + 3B ---> C + D ∆Ho = -64 kJ

E + A ----> C ∆Ho = 100 kJ

2A + D -----> 2B ∆Ho = 148 kJ

Calculate

∆Ho for the overall reaction:

2E 2A + B + C

Update:

urgenttt!!! pleaaase!!

Update 2:

i have no ideaaaa...i need an answer pleaaaase ://

1 Answer

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  • Anonymous
    1 decade ago
    Favorite Answer

    Ho = -64 kj = hc +Hd - ( 2HA + 3 HB)................i

    HO = HEAT OF PRODUCTS -H OF REACTANTS

    HC - HA -HE = 100 ......................ii

    2HB - 2HA - HD = 148........................iii

    2 HA + HB + HC - 2 HE =X.............iv

    multiply ii eqn by 2

    2C - 2E - 2A =200-----

    SO, 2E = 2C - 2A -200.............V

    PUT IN IV

    2A + B - (2C -2A -200) = X

    4A + B+ 200 -C = X.........................VI

    ADD i and iii eqn

    C - 2A -B -2A =-64 + 148 =84

    C - 4A -B =84

    SO , 4A + B = -84 +C ...............VII

    PUT VII IN VI

    -84 + C +200 -C =X

    X= 116 KJ

    WHERE X = 2A +B +C -2E ENERGY FOR REACTION

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