please help Use the following information for questions 10-13 From the following

Question

please help

Use the following information for questions 10-13 From the following standard enthalpies of formation and entropies: KCIO3 (s)aHof-391.2 kJ/mole So 0.143 kJ/mole K KCI (s) Alf_-435.9 kJ/mole So-0.0827 kJ/mole K 02 (g) Hof= 0 So-0.205 kJ/mole K 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) 10. Calculate H° for the reaction a) 89.4 kJ b)-89.4 kJ c) 89.4 J d)-89.4 J e) none of the above 11) Calculate ASo for the reaction a) -0.4944 kJ/K b) 0.4944 J/K 0)0.4944 kJ/K d)-0.4944 J/Ke)none of the above 12) Calculate AG for the reaction a)-236 kJ b)236J c)236kJ d)-236J e) none of the above 13) Is the reaction spontaneous a) yes b)no in geid = 1 8 X 10-5, calculate Go for the dissociation of this substance in 1 Ilmolrone of the a

Explanation / Answer

10)

Given:
Hof(KClO3(s)) = -391.2 KJ/mol
Hof(KCl(s)) = -435.9 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol

Balanced chemical equation is:
2 KClO3(s) ---> 2 KCl(s) + 3 O2(g)

Ho rxn = 2*Hof(KCl(s)) + 3*Hof(O2(g)) - 2*Hof( KClO3(s))
Ho rxn = 2*(-435.9) + 3*(0.0) - 2*(-391.2)
Ho rxn = -89.4 KJ

Answer: b

11)

Given:
Sof(KClO3(s)) = 143 J/mol.K
Sof(KCl(s)) = 82.7 J/mol.K
Sof(O2(g)) = 205 J/mol.K

Balanced chemical equation is:
2 KClO3(s) ---> 2 KCl(s) + 3 O2(g)

So rxn = 2*Sof(KCl(s)) + 3*Sof(O2(g)) - 2*Sof( KClO3(s))
So rxn = 2*(82.7) + 3*(205.0) - 2*(143.0)
So rxn = 494.4 J/K
= = 0.4944 KJ/K
Answer: a

12)

use:
Go = Ho - T*So
Go = -89.4 - 298.0 * 0.4944
Go = -236 KJ
Answer: a

13)
Since Go is negative, this is spontaneous
Answer: a

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