# Answer to Question #35788 in Inorganic Chemistry for Rachel

Question #35788
Given the following thermochemical equations:

X2 + 3Y2 ----&gt; 2XY3 ^H1= -370 kJ

X2 + 2Z2 ----&gt; 2XZ2 ^H2= -170 kJ

2Y2 + 2Y2Z -----&gt; 2Y2Z ^H3= -280 kJ

Calculate the change in enthalpy for the following reaction:

4XY3 + 7Z2 -----&gt; 6Y2Z + 4XZ2

change H= ________ kJ
1
2015-09-14T09:27:42-0400
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.

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Assignment Expert
07.03.19, 14:16

Dear 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 publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Jaime
04.03.19, 04:11

Why is delta Z2 = 0??

Assignment Expert
14.09.15, 16:27

According 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.

Potter
14.09.15, 15:11

How do you solve this?