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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
urgenttt!!! pleaaase!!
i have no ideaaaa...i need an answer pleaaaase ://
1 Answer
- Anonymous1 decade agoFavorite 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