Question

Given that `DeltaG` for the reaction `4NH_(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)` is -957.9 kJ, what is `DeltaG)f` of `H_2O`?

`DeltaG_(f_1NH_3)` = -16.66 kJ/mol
`DeltaG_(f_1NO)` = 86.71 kJ/mol

Answer 1

`-228.56333kJ` is the `G_f^@` of water

Explanation:

Here's my explanation

`Delta G` of reaction can be calculated in the same way as `Delta H` that is
`DeltaG "of reaction"` = `Delta G` of formation of reactants - `Delta G` of formation ofproducts

In short we call `Delta H` of formation `Delta H_f^@`
Same is the case for `DeltaG` we call `DeltaG` of formation
`DeltaG_f^@`

Always remember `H_f^@` or `G_f^@` of elements like oxygen, carbon, hydrogen,lithium, sodium and all other element is 0

So if we know the `Delta G_f^@` of the reactants and products of a equation we can find out the `DeltaG` f the reaction

`4NH_3(g) + 5O_2(g) →4NO(g) +6H _2O(g)`

In this reaction we know that what is the `DeltaG` of the reaction but not the `G_f^@` of `H_2O`. Recall that elements have a value of 0 of `G_f^@`. Thus oxygen has 0 as its `DeltaG_f^@`. So the only unknown `DeltaG_f^@` is of `H_2O`. So this can be an algebraic equation. Set up the equation and take the `DeltaG_f^@` as `x`

`"-957.9 kJ" = (4 xx "86.71kJ/mol" + 6x )- (4 xx -16.66 + 0)`

`"-957.9 kJ" = 346.84 + 6x - (-66.64) + 0`

`-957.9 kJ = 346.84 + 6x + 66.64`

`-957.9kJ = 413.48 + 6x`

`6x = -957.9kJ - 413.48`

`6x = -1371.38`

`x = 1371.38/6 = -228.56333kJ`

This is the `DeltaG_f^@` of water