Respuesta :
Explanation:
The given reactions are as follows.
[tex]P_{4}(g) + 5O_{2}(g) \rightarrow P_{4}O_{10}(s)[/tex] [tex]\Delta G^{o}298[/tex] = -2748 kJ/mol ............ (1)
[tex]2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)[/tex] [tex]\Delta G^{o}298[/tex] = -474.2 kJ/mol ............ (2)
[tex]P_{4}O_{10}(s) + 6H_{2}O(l) \rightarrow 4H_{3}PO_{4}(l)[/tex] [tex]\Delta G^{o} 298[/tex] = -349 kJ/mol ............. (3)
It is known that heat of formation for the reactions where one mole of a compound is made from its constituent particles means gibbs free energy.
The equation is as follows.
[tex]\frac{1}{4}P_{4}(g) + 2O_{2}(g) + \frac{3}{2}H2(g) \rightarrow H_{3}PO_{4}(l)[/tex] [tex]\Delta G^{o} 298[/tex] = ?
So, from the equations (1), (2) and (3) reactions we get the following.
[tex][1]x \frac{1}{4} + [2] x\frac{3}{4} + [3]x \frac{1}{4}[/tex]
[tex]\frac{1}{4}P_{4}(g) + \frac{5}{4}O_{2}(g) \rightarrow \frac{1}{4}P_{4}O_{10}(s)[/tex] ......... (4)
[tex]\frac{3}{2}H_{2}(g) + \frac{3}{4}O_{2}(g) \rightarrow \frac{3}{2}H_{2}O(l)[/tex] ............. (5)
[tex]\frac{1}{4}P_{4}O_{10}(s) + \frac{3}{2}H2O(l) \rightarrow H_{3}PO_{4}(l)[/tex] ............ (6)
On adding equation (4), (5) and (6) by cancelling the common species we get the following.
[tex]\frac{1}{4}P_{4}(g) + 2O_{2}(g) + \frac{3}{2}H_{2}(g) \rightarrow H_{3}PO_{4}(l)[/tex]
Now, calculate the value of [tex]\Delta G^{o} 298[/tex] by putting the given values into the above equation as follows.
[tex]\Delta G^{o}_{rxn} 298[/tex] = [tex][-2748] \times \frac{1}{4} + [-474.2] \times \frac{3}{4} + [-349] \times \frac{1}{4}[/tex]
= (-687 -355.65 -87.25) kJ/mol
= -1129.9 kJ/mol
Thus, we can conclude that value of standard free energy of formation for the given reaction is -1129.9 kJ/mol.
