Given the thermochemical equations X2+3Y2⟶2XY3 ΔH1=−320 kJ X2+2Z2⟶2XZ2 ΔH2=−170 kJ 2Y2+Z2⟶2Y2Z ΔH3=−290 kJ Calculate the change...

Question

Question

Given the thermochemical equations

X2+3Y2⟶2XY3 ΔH1=−320 kJ

X2+2Z2⟶2XZ2 ΔH2=−170 kJ

2Y2+Z2⟶2Y2Z ΔH3=−290 kJ

Calculate the change in enthalpy for the reaction.

4XY3+7Z2⟶6Y2Z+4XZ2

Answer 1

Answer #1

Given reactions are +34 — 2XY, AH = - 320 _0 *+ 22, 2X2, A Hy = -17045 a 4x + 2 = 2X22 sty = -29055 Then 4X4% +734 69642+ 4X,

Answer 2

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