Question #35788

Given the following thermochemical equations:
X2 + 3Y2 ----> 2XY3 ^H1= -370 kJ
X2 + 2Z2 ----> 2XZ2 ^H2= -170 kJ
2Y2 + 2Y2Z -----> 2Y2Z ^H3= -280 kJ
Calculate the change in enthalpy for the following reaction:
4XY3 + 7Z2 -----> 6Y2Z + 4XZ2
change H= ________ kJ

Expert's answer

The change in enthalpy for the reaction equals: delta H = sum (delta H) of products - sum (delta H) of reagents.

For reagents and products of the following reaction:

delta H (XY3) = -370 kJ / 2 = -185 kJ.

delta H (XZ2) = -170 kJ / 2 = -85 kJ.

delta H (Y2Z) = -280 kJ / 2 = -140 kJ.

delta H (Z2) = 0

delta H (for the reaction) = 6*(-140 kJ) + 4*(-85 kJ) - 4*(-185)-0 = -840 kJ - 340 kJ + 740 kJ = -440 kJ.

Answer: -440 kJ.

For reagents and products of the following reaction:

delta H (XY3) = -370 kJ / 2 = -185 kJ.

delta H (XZ2) = -170 kJ / 2 = -85 kJ.

delta H (Y2Z) = -280 kJ / 2 = -140 kJ.

delta H (Z2) = 0

delta H (for the reaction) = 6*(-140 kJ) + 4*(-85 kJ) - 4*(-185)-0 = -840 kJ - 340 kJ + 740 kJ = -440 kJ.

Answer: -440 kJ.

## Comments

Assignment Expert07.03.19, 14:16Dear Jaime,

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Jaime04.03.19, 04:11Why is delta Z2 = 0??

Assignment Expert14.09.15, 16:27According to Hess's Law the change in enthalpy for the reaction equals: delta H = sum (delta H) of products - sum (delta H) of reagents.

For reagents and products of the following reaction:

delta H (XY3) = -370 kJ / 2 = -185 kJ.

delta H (XZ2) = -170 kJ / 2 = -85 kJ.

delta H (Y2Z) = -280 kJ / 2 = -140 kJ.

delta H (Z2) = 0

delta H (for the reaction) = 6*(-140 kJ) + 4*(-85 kJ) - 4*(-185)-0 = -840 kJ - 340 kJ + 740 kJ = -440 kJ.

Answer: -440 kJ.

Potter14.09.15, 15:11How do you solve this?

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