Calculate the G°rxn at 25 0 C using the following information. 4 HNO3(g) + 5 N2H

Question

Calculate the G°rxn at 25 0C using the following information.

                  4 HNO3(g) + 5 N2H4(l) 7 N2(g) + 12 H2O(l)   

HNO3(g)

N2H4(l)

N2(g)

H2O(l)

H°f (kJ/mol)

-133.9

50.6

-285.8

S°(J/molK)

266.9

121.2

191.6

70.0

            (A) +4.90 × 103 kJ     (B) +3.90 × 103 kJ     (C) -2.04 × 103 kJ     (D) -3.15 × 103 kJ

            (E) -3.298 x 103 kJ

Every time I do it I get an extremely high number, Id like to know what im doing wrong.

HNO3(g)

N2H4(l)

N2(g)

H2O(l)

H°f (kJ/mol)

-133.9

50.6

-285.8

S°(J/molK)

266.9

121.2

191.6

70.0

Explanation / Answer

first we find Horxn

Horxn = Hoproduct - Horeactant

Horxn = [ 7 (0) + 12 (-285.8) ] - [ 5(50.6) + 4(-133.9) ] KJ/mol

Horxn = -3429.6 - 253 + 535.6

Horxn = -3147 KJ/mol

then we find S°rxn

S°rxn = S°product - S°reactant

S°rxn = [ 7(191.6) + 12(70.0) ] - [ 4 (266.9) + 5 (121.2) ]

S°rxn = [1341.2 + 840 ] - [1067.6 + 606 ]

S°rxn = 507.6 J/mol.K

now, we find G°rxn

G°rxn = Horxn - T Sorxn{ T = temperature in kelvin = 273.15 + 25 = 298 K}

Gorxn = -3147000 J/mol - ( 298.15 x 507.6 J/mol K)

Gorxn = -3298340 J/mol = -3298 KJ/mol = -3.298x103 KJ

reaction is higly spontaneous at this temperature

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