Given the following thermochemical equations: X2+3Y2--> 2XY3 H= -320 kj X2+2Z2 -- > 2Xz2 H2= -110...

Given the following thermochemical equations: X2+3Y2--> 2XY3 H= -320 kj X2+2Z2 -- > 2Xz2 H2= -110 kj 2Y2+z2 ---> 2Y2Z H3= -270 kj Calcuate the change in enthalpy for: 4XY3+7Z2--> 6Y2Z+4XZ2 H=?

Solutions

Expert Solution

X2+3Y2--> 2XY3 H= -320 kj --------------->1

X2+2Z2 -- > 2Xz2 H2= -110 kj ---------------->2

2Y2+z2 ---> 2Y2Z H3= -270 kj ------------------>3

2 is multifly with 2

3 is multifly with 3

2X2+4Z2 -- > 4Xz2 H2= -220 kj

6Y2+3z2 ---> 6Y2Z H3= -810 kj

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2X2 +6Y2 + 7Z2 ------> 6Y2Z+4XZ2 H4 = -1030KJ

2X2+6Y2--> 4XY3 H1= -1280 kj

(-) (-) (-) (+)

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4XY3+7Z2--> 6Y2Z+4XZ2 H5 = 250KJ

queen_honey_blossom answered 1 year ago

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