Use the given data at 298 K to calculate ¦¤G¡ã for the reaction
2NO(g) + 2H2(g) ¡ú N2(g) + 2H2O(g)
Substance NO(g) H2(g) N2(g) H2O(g)
¦¤H¡ãf (kJ/mol) 90 0 0 -242
S¡ã(J/K¡¤mol) 211 131 192 189
a. -270 kJ
b. -630 kJ
c. -3.46 x 104 kJ
d. -667 kJ
B. -630 kJ
WELL YOURE WELCOME !!!!!!!!!!!!!!!!GEEZ!
gaby can you show me how you did this? i kept getting 333
well what do you think youre doing wrong?
BTW would you mind helping me with my algebra 1 question cuz i really don't get it?!?!?!?!
i figured dH to = -664, the dS = -114.
then i use equation>
dG = dH -TdSand i got 333.08 kJ
To calculate ∆Gᵒ for the reaction, we will use the equation:
∆Gᵒ = ∑∆Gᵒᶠ(products) - ∑∆Gᵒᶠ(reactants)
Where ∑∆Gᵒᶠ represents the sum of the standard free energy of formation for each product and reactant.
First, let's calculate the standard free energy of formation for each compound using the given data at 298 K:
∆Gᵒᶠ(NO) = ∆Hᵒᶠ(NO) - T∆Sᵒ(NO) = 90 kJ/mol - (298 K)(211 J/K·mol) = 90 kJ/mol - 62.78 kJ/mol = 27.22 kJ/mol
∆Gᵒᶠ(H₂) = ∆Hᵒᶠ(H₂) - T∆Sᵒ(H₂) = 0 kJ/mol - (298 K)(131 J/K·mol) = 0 kJ/mol - 38.74 kJ/mol = -38.74 kJ/mol
∆Gᵒᶠ(N₂) = ∆Hᵒᶠ(N₂) - T∆Sᵒ(N₂) = 0 kJ/mol - (298 K)(192 J/K·mol) = 0 kJ/mol - 57.22 kJ/mol = -57.22 kJ/mol
∆Gᵒᶠ(H₂O) = ∆Hᵒᶠ(H₂O) - T∆Sᵒ(H₂O) = -242 kJ/mol - (298 K)(189 J/K·mol) = -242 kJ/mol - 56.51 kJ/mol = -298.51 kJ/mol
Now, plug the values into the reaction equation:
∆Gᵒ = ∑∆Gᵒᶠ(products) - ∑∆Gᵒᶠ(reactants)
= [(-57.22 kJ/mol) + (2)(-298.51 kJ/mol)] - [2(27.22 kJ/mol) + 2(-38.74 kJ/mol)]
= -652.44 kJ/mol + 54.44 kJ/mol
= -598.00 kJ/mol
Therefore, the value of ∆Gᵒ for the reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) at 298 K is approximately -598.00 kJ/mol, which is not one of the given answer choices.