Respuesta :
Answer:
Explanation:
- from ΔG = Summation of ΔGproducts - Summation of ΔGreactants
- Given that the standard free energies of formation of CO(g) = −137.3
- and Ni(CO)4(g)= −587.4 kJ/mol.
- R = gas constant = 8.314
- Temperature = 86.0°C +273 = 359k
- ΔG reaction = −587.4 kJ/mol - (4 x −137.3kJ/mol) = -38.2kJ/mol
- From the vant hoff Isotherm equation ; ΔG = -RTlnk
- lnk = -38.2kJ/mol / 359 x 8.314
- lnk = -0.0127, k = 0.987
The equilibrium constant of the Mond process reaction Ni(s) + 4CO(g) ⇄ Ni(CO)₄(g) at 86 °C is 3.62x10⁵, given that the standard free energies of formation of CO(g) and Ni(CO)₄(g) are -137.3 and -587.4 kJ/mol, respectively.
The reaction of the Mond process is the following:
Ni(s) + 4CO(g) ⇄ Ni(CO)₄(g) (1)
We can calculate the equilibrium constant of the above reaction with the equation:
[tex] \Delta G^{0} = -RTln(K) [/tex] (2)
Where:
R: is the gas constant = 8.314 J/(K*mol)
T: is the temperature = 86.0 °C = 359 K
ΔG⁰: is the change in standard Gibbs free energy
K: is the equilibrium constant =?
The change in free energy is given by:
[tex]\Delta G^{0} = \Sigma n_{p}\Delta G^{0}_{f}_{p} - \Sigma n_{r}\Delta G^{0}_{f}_{r}[/tex]
Where p is for products, r for reactants and n is the stoichiometric factors of the compounds
From reaction (1) we have that n for Ni(CO)₄ and CO is 1 and 4, respectively, so
[tex]\Delta G^{0} = \Delta G^{0}_{f}_{Ni(CO)_{4}} - 4\Delta G^{0}_{f}_{CO}[/tex]
The standard free energy of Ni is zero because it is in its standard state.
[tex] \Delta G^{0} = -587.4 kJ/mol - (-4*137.3 kJ/mol) = -38.2 kJ/mol [/tex]
By solving equation (2) for K and entering the values of the parameters, we have:
[tex] K = e^{-\frac{\Delta G^{0}}{RT}} = e^{-\frac{-38.2 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*359 K}} = 3.62 \cdot 10^{5} [/tex]
Therefore, the equilibrium constant of the reaction is 3.62x10⁵.
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I hope it helps you!
