Respuesta :
Answer:
a) [tex]\triangle G^{0} = 7.31 kJ/mol[/tex]
b) [tex]K_{-1} = 0.0594 m^{-1} s^{-1}[/tex]
Explanation:
Equation of reaction:
[tex]2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)[/tex]
Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
[tex]K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }[/tex]
[tex]K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}[/tex]
Standard free energy:
[tex]\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol[/tex]
b) value of k−1 at 27 °C, i.e. 300K
[tex]K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}[/tex]
[tex]K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}[/tex]
[tex]K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}[/tex]
