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

∆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

- 228.6 kJ/mol.

Explanation:

I got bandz

the increase of protons   hope this