Question

Given that ∆G for the reaction below is -957.9kj, what is ∆Gf of

H2O?

4NH3(g)+4NO(g)+6H2O(g)

∆Gf, NH3=-16.66kj/mol
∆Gf, NO=86.71kj/mol

Answer 1

- 228.6 kJ/mol.

Step-by-step explanation:

I got bandz

Answer 2

∆Grxn = sum of ∆Gf (products) - sum of ∆Gf(reactants)

The reaction is

4NH3(g)+ 5O2 ---> 4NO(g)+6H2O(g)

∆Grxn = [ 4 X ∆Gf(NO) + 6 X ∆Gf(H2O)] - [ 4∆Gf(NH3) + 5 ∆Gf(O2)]

∆Grxn = [4 X (86.7) + 6 ∆Gf(H2O)] -[ 4 (-16.66) + 5(0)]

∆Grxn = [346.8 + 6∆Gf(H2O)] - [-16.66 + 0]

∆Grxn = -957.9 = [346.8 + 6∆Gf(H2O)] - [-66.64 + 0]

-957.9 = [346.8 + 6∆Gf(H2O)] - [-66.64 + 0]

∆Gf(H2O) = -90.74

This is based on information provided by you.

However the answer should be nearly -241.8