Respuesta :
Answer: (a) The value of [tex]K_{p}[/tex] is [tex]4 \times 10^{18}[/tex] and value of [tex]\Delta G^{o}[/tex] is -106.4 kJ/mol.
(b) The value of [tex]K_{p}[/tex] is [tex]2 \times 10^{9}[/tex] and value of [tex]\Delta G^{o}[/tex] is -53.2 kJ/mol.
Explanation:
(a) As the given reaction equation is as follows.
[tex]H_{2}(g) + Br_{2}(l) \rightleftharpoons 2HBr(g)[/tex]
Firs, we will calculate the value of [tex]\Delta G^{o}[/tex] as follows.
[tex]\Delta G^{o} = 2\Delta G^{o}_{F(HBr)} - \Delta G^{o}_{F(Br_{2})}[/tex]
= [tex]2 \times -53.2 kJ/mol - (1) \times (0) - (1) \times (0)[/tex]
= -106.4 kJ/mol
[tex]ln K_{p} = \frac{-\Delta G^{o}}{RT}[/tex]
= [tex]\frac{106.4 \times 10^{3} J/mol}{8.314 J/mol K \times 298 K}[/tex]
[tex]ln K_{p}[/tex] = 42.9
[tex]K_{p} = 4 \times 10^{18}[/tex]
Therefore, the value of [tex]K_{p}[/tex] is [tex]4 \times 10^{18}[/tex] and value of [tex]\Delta G^{o}[/tex] is -106.4 kJ/mol.
(b) As the given reaction equation is as follows.
[tex]\frac{1}{2}H_{2}(g) + \frac{1}{2}Br_{2}(l) \rightleftharpoons HBr(g)[/tex]
So, [tex]\Delta G^{o} = \Delta G^{o}_{F(HBr)} - \frac{1}{2}\Delta G^{o}_{F(Br_{2})}[/tex]
= [tex](1)(-53.2 kJ/mol) - (\frac{1}{2})(0) - (\frac{1}{2})(0)[/tex]
= -53.2 kJ/mol
As, [tex]ln K_{p} = \frac{-\Delta G^{o}}{RT}[/tex]
= [tex]\frac{53.2 \times 10^{3} J/mol}{8.314 J/mol K \times 298 K}[/tex]
= 21.5
[tex]K_{p} = 2 \times 10^{9}[/tex]
Therefore, the value of [tex]K_{p}[/tex] is [tex]2 \times 10^{9}[/tex] and value of [tex]\Delta G^{o}[/tex] is -53.2 kJ/mol.
