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?

$\Delta {G}_{{f}_{1} N {H}_{3}}$ = -16.66 kJ/mol $\Delta {G}_{{f}_{1} N O}$ = 86.71 kJ/mol

Mar 28, 2017

$- 228.56333 k J$ is the ${G}_{f}^{\circ}$ of water

Explanation:

Here's my explanation

$\Delta G$ of reaction can be calculated in the same way as $\Delta H$ that is
$\Delta G \text{of reaction}$ = $\Delta G$ of formation of reactants - $\Delta G$ of formation ofproducts

In short we call $\Delta H$ of formation $\Delta {H}_{f}^{\circ}$
Same is the case for $\Delta G$ we call $\Delta G$ of formation
$\Delta {G}_{f}^{\circ}$

Always remember ${H}_{f}^{\circ}$ or ${G}_{f}^{\circ}$ of elements like oxygen, carbon, hydrogen,lithium, sodium and all other element is 0

So if we know the $\Delta {G}_{f}^{\circ}$ of the reactants and products of a equation we can find out the $\Delta G$ f the reaction

4NH_3(g) + 5O_2(g) →4NO(g) +6H _2O(g)

In this reaction we know that what is the $\Delta G$ of the reaction but not the ${G}_{f}^{\circ}$ of ${H}_{2} O$. Recall that elements have a value of 0 of ${G}_{f}^{\circ}$. Thus oxygen has 0 as its $\Delta {G}_{f}^{\circ}$. So the only unknown $\Delta {G}_{f}^{\circ}$ is of ${H}_{2} O$. So this can be an algebraic equation. Set up the equation and take the $\Delta {G}_{f}^{\circ}$ as $x$

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

$\text{-957.9 kJ} = 346.84 + 6 x - \left(- 66.64\right) + 0$

$- 957.9 k J = 346.84 + 6 x + 66.64$

$- 957.9 k J = 413.48 + 6 x$

$6 x = - 957.9 k J - 413.48$

$6 x = - 1371.38$

$x = \frac{1371.38}{6} = - 228.56333 k J$

This is the $\Delta {G}_{f}^{\circ}$ of water