# Answer to Question #35788 in Inorganic Chemistry for Rachel

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

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.

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## Comments

Assignment Expert07.03.19, 14:16Dear Jaime,

Questions in this section are answered for free. We can't fulfill them all and there

is no guarantee of answering certain question but we are doing our best. And if answer

is published it means it was attentively checked by experts. You can try it yourself by

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experts will surely assist you.

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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