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Author Question: For the reaction SbCl5(g) SbCl3(g) + Cl2(g), Gf (SbCl5) = 334.34 kJ/mol Gf (SbCl3) = 301.25 ... (Read 32 times)

dongbo

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Question 1

For the reaction SbCl5(g) SbCl3(g) + Cl2(g),
         ΔG°f (SbCl5) = –334.34 kJ/mol
         ΔG°f (SbCl3) = –301.25 kJ/mol
         ΔH°f (SbCl5) = –394.34 kJ/mol
         ΔH°f (SbCl3) = –313.80 kJ/mol
Calculate the value of the equilibrium constant (Kp) for this reaction at 298 K.

Question 2

For the reaction SbCl5(g) SbCl3(g) + Cl2(g),
         ΔG°f (SbCl5) = –334.34 kJ/mol
         ΔG°f (SbCl3) = –301.25 kJ/mol
         ΔH°f (SbCl5) = –394.34 kJ/mol
         ΔH°f (SbCl3) = –313.80 kJ/mol
Calculate the value of the equilibrium constant (Kp) at 800 K and 1 atm pressure.


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Marked as best answer by dongbo on Nov 5, 2023

hillard sane

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dongbo

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Reply 2 on: Nov 5, 2023
Gracias!


amcvicar

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Reply 3 on: Yesterday
:D TYSM

 

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