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Given the following thermochemical equations: X2+3Y2 equals 2XY3 and Delta H =-340 kJ X2+2 Z2 equals...

Given the following thermochemical equations:

X₂+3Y₂ equals 2XY₃ and Delta H =-340 kJ

X₂+2 Z₂ equals 2XZ₂ and Delta H = -170 kJ

2Y₂ + Z₂ equals 2Y₂Z and Delta H=-260kJ

Calculate the change in enthalpy for the following reaction

4XY3+7 Z₂ equals 6Y₂Z+4XZ₂

Expert Solution

inverting equation #1..
2 XY3 -----> X2 + 3Y2... dH = +340 kJ.

and then multiplying by 2
4 XY3 ----> 2X2 + 6 Y2... dH = +680kJ =============>1

now you have 4 XY3 on the left.. as you need..

now you need to get rid of the 2 X2 on the right

2 x equation #2 will do that

2X₂+4 Z₂ -------> 4XZ₂ and Delta H = -340 kJ=============>2

3 x equation #3

6Y₂ +3Z₂ ,------>6Y₂Z and Delta H=-780kJ===========>3

--------------------------------------------------------------------------------------------------------------------------------------------------------

adding eqns

4 XY3 ----> 2X2 + 6Y2... dH = +680kJ ======>1

2X₂+4 Z₂ -------> 4XZ₂ and Delta H = -340 kJ===>2

6Y₂ +3Z₂ ,------>6Y₂Z and Delta H=-780kJ===>3

-----------------------------------------------------------------------------------------------

4 XY3 +7Z₂ ----------->6Y₂Z+ 4XZ₂

_{--------------------------------------------------------------------------------------}

+680-340-780

-440kJ

queen_honey_blossom answered 2 hours ago

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Given the following thermochemical equations: X2+3Y2 equals 2XY3 and Delta H =-340 kJ X2+2 Z2 equals...

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