owlet
  • owlet
Enthalpy I need step-by-step guidance please and thank you in advance. Question Below:
Chemistry
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owlet
  • owlet
Enthalpy I need step-by-step guidance please and thank you in advance. Question Below:
Chemistry
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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owlet
  • owlet
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owlet
  • owlet
Jhannybean
  • Jhannybean
\[\Delta H = \sum {n\cdot \text{products} -\sum m \cdot \text{reactants}}\] m and n are coefficients of the reactants and products,respectively.

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Jhannybean
  • Jhannybean
So you're given \(\Delta H = -89 \frac{kJ}{mol}\)
owlet
  • owlet
yeah i think I manage it now, just to make sure Hf for I2 is zero right?
owlet
  • owlet
**can
Jhannybean
  • Jhannybean
Yes, all diatomics and elements
owlet
  • owlet
thanks!
Jhannybean
  • Jhannybean
\[-89 \frac{kJ}{mol} = -840\frac{kJ}{mol} +2\text{IF} -\left( -942 \frac{kJ}{mol}\right)\]
owlet
  • owlet
that's what i did and i got -94.5 kJ
Jhannybean
  • Jhannybean
- 89 = 2x - 840 + 942 -89 = 2x +102 -89 - 102 = 2x -191 = 2x x = -191/2 = -95.5

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