Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...

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Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...

Given the thermochemical equations

X2+3Y2⟶2XY3Δ?1=−340 kJ

X2+2Z2⟶2XZ2Δ?2=−120 kJ

2Y2+Z2⟶2Y2ZΔ?3=−260 kJ

Calculate the change in enthalpy for the reaction.

4XY3+7Z2⟶6Y2Z+4XZ2

Δ?=__________kJ

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Answer 1

Answer #1

Lets number the reaction as 1, 2, 3, 4 from top to bottom
required reaction should be written in terms of other reaction
This is Hess Law
required reaction can be written as:
reaction 4 = -2 * (reaction 1) +2 * (reaction 2) +3 * (reaction 3)

So, ΔHo rxn for required reaction will be:
ΔHo rxn = -2 * ΔHo rxn(reaction 1) +2 * ΔHo rxn(reaction 2) +3 * ΔHo rxn(reaction 3)
= -2 * (-340) +2 * (-120) +3 * (-260)
= -340 KJ
Answer: -340 KJ

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Answer 2

Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...

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Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...

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Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−190 kJ 2Y2+Z2⟶2Y2ZΔ?3=−220 kJ Calculate the change in enthalpy for...

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Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...

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Add Answer to: Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...

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Add Answer to:
Given the thermochemical equations X2+3Y2⟶2XY3Δ?1=−340 kJ X2+2Z2⟶2XZ2Δ?2=−120 kJ 2Y2+Z2⟶2Y2ZΔ?3=−260 kJ Calculate the change in enthalpy for...