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Given that delta.g for the reaction below is –957.9 kj, what is delta.gf of h2o? 4nh3(g) 5o2(g) right arrow. 4no(g) 6h2o(g) delta.gf,nh3 = -16.66 kj/mol delta.gf,no = 86.71 kj/mol –228.6 kj/mol –206.4 kj/mol 46.7 kj/mol 90.7 kj/mol

Respuesta :

The valur of energy of formation ΔGf of H₂O is -228.6 kJ/mol.

How we calculate ΔG ?

ΔG of any reaction will be calculated as:

ΔG = Sum of ΔG of products - Sum of ΔG of reactants

Given chemical reaction is:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

For this reaction, ΔG will be calculated as:

ΔG = [ΔGf 4(NO) + ΔGf 6(H₂O)] - [ΔGf 4(NH₃) + ΔGf 5(O₂)]

Given that:

ΔG = -957.9 kJ

ΔGf of NH₃ = -16.66 kJ/mol

ΔGf of NO = 86.71 kJ/mol

ΔGf of O₂ = 0 (because it is present in elemental form)

ΔGf of H₂O = let x (to find)

On putting all these values on the above equation, we get

-957.9 = [4(86.71) + 6(x)] - [4(-16.66) + 5(0)]

-957.9 = 346.84 + 6x + 66.64

6x = -957.9 - 346.84 - 66.64

6x = -1,371.38

x = -228.56 = -228.6 kJ/mol

Hence, correct option is (1) i.e. -228.6 kJ/mol.

To know more about ΔG, visit the below link:

https://brainly.com/question/2507189

Answer:

A) -228.6 kJ/mol

Explanation:

EDGE 2022

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