maxamster
Answered

What is the ∆G for the following reaction under standard conditions (T = 298 K) for the formation of NH4NO3(s)?
2NH3(g) + 2O2(g) NH4NO3(s) + H2O(l)

Given:
NH4NO3(s): ∆Hf = -365.56 kJ ∆Sf = 151.08 J/K.
NH3(g): ∆Hf = -46.11 kJ ∆Sf = 192.45 J/K.
H2O(l): ∆Hf = -285.830 kJ ∆Sf = 69.91 J/K.
O2(g): ∆Hf = 0.00 kJ ∆Sf = 205 J/K.

A.) 186.6 kJ
B.) 6.9 kJ
C.) -10.4 kJ
D.) -126.3 kJ
E.) -382 kJ

Answer :

JavierCollante
Use the formula for ΔHrxn= (sum product) - (sum reactants)
and formula for ΔSrxn= (sum product) - (sum reactants)
and ΔG= ΔH-TΔS

there are different ways to solve this question, but I think the simplest is to find the ΔG of each one of the substance and solve for ΔG as (sum product) - (sum reactants). also because the answers are in Kj, you need to convert the ΔS to Kj by simply dividing the value by 1000. 

ΔG= ΔH-TΔS

ΔGNH4NO3= (-365.56) - (298 X 151.08x10-3)= -410.58
ΔG NH3= (-46.11) - (298 X 192.45x10-3)= -103.46
ΔG H2O= (-285.830) - (298 X 69.91x10-3)= -306.66
ΔG O2= (0) - (298 X 205x10-3)= -61.09

ΔGrxn= product- reactants

ΔGrxn= [∆Gf(NH4NO3) + ∆Gf(H2O)] - [2 ∆Gf(NH3) + 2 ∆Gf(O2)] 
ΔGrxn= [ -410.58 + -306.66] - [ (2 x -103.46) + (2 x -61.09)]= -388.14 kj

the answers is closest to E

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