calculating enthalpy

Question

Given the thermochemical equations

X2+3Y2⟶2XY3Δ𝐻1=−370 kJX2+3Y2⟶2XY3Δ�1=−370 kJ

X2+2Z2⟶2XZ2Δ𝐻2=−130 kJX2+2Z2⟶2XZ2Δ�2=−130 kJ

2Y2+Z2⟶2Y2ZΔ𝐻3=−220 kJ2Y2+Z2⟶2Y2ZΔ�3=−220 kJ

Calculate the change in enthalpy for the reaction.

4XY3+7Z2⟶6Y2Z+4XZ2

Answer 1

For this type of question, you will need to manipulate the given half-equations to re-create the final reaction equation. You can do this by flipping the equation around (changing the arrow and the negative enthalpy value to a positive one). You can also "double up" an equation and multiple the enthalpy by 2 in order to get the correct number of coefficients.

4 XY₃ + 7 Z₂ -------> 6 Y₂Z + 4 XZ₂

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6 Y₂ + 3 Z₂ -------> 6 Y₂Z -660 kJ (multiplied it by 3 to get 6 Y₂Z on the right side)

2 X₂ + 4 Z₂ -------> 4 XZ₂ -260 kJ (multiplied by 2 to get 4 XZ₂ on the right side)

4 XY₃ -------> 2 X₂ + 6 Y₂ +740 kJ (flipped it & multiplied by 2 to get 4 XY₃ on the left side)

You will notice that there were extra elements not seen in the final equation (Y₂ and X₂). Because there is 6 Y₂ as a reactant on the 1st line and as a product on the 3rd line, they will cancel. Same for the 2 X₂ on lines 2 and 3. Now, you just add the 3 half-equations to get the same final equation. Add up the enthalpy values.

-660 + (-260) + 740 = -180 kJ