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owlet

  • one year ago

Enthalpy I need step-by-step guidance please and thank you in advance. Question Below:

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  1. owlet
    • one year ago
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  2. owlet
    • one year ago
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    @Jhannybean

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

  4. Jhannybean
    • one year ago
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    So you're given \(\Delta H = -89 \frac{kJ}{mol}\)

  5. owlet
    • one year ago
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    yeah i think I manage it now, just to make sure Hf for I2 is zero right?

  6. owlet
    • one year ago
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    **can

  7. Jhannybean
    • one year ago
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    Yes, all diatomics and elements

  8. owlet
    • one year ago
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    thanks!

  9. Jhannybean
    • one year ago
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    \[-89 \frac{kJ}{mol} = -840\frac{kJ}{mol} +2\text{IF} -\left( -942 \frac{kJ}{mol}\right)\]

  10. owlet
    • one year ago
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    that's what i did and i got -94.5 kJ

  11. Jhannybean
    • one year ago
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    - 89 = 2x - 840 + 942 -89 = 2x +102 -89 - 102 = 2x -191 = 2x x = -191/2 = -95.5

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