
Respuesta :
Answer:
656055.062
Explanation:
Data:
- gas constant, R = 8.3144 J/(mol*K)
- temperature, T = 70 °C or 70+273 = 343 K
The standard free energy of reaction is:
ÎG = G of formation of Ni(CO)4(g) - 4 * G of formation of CO(g) - G of formation of Ni(s)
ÎG = â587.4 - 4*(-137.3) - 0 = -38.2 Â kJ/mol or -38200 J/mol
The relationship between the standard free energy of reaction (ÎG) and the  the equilibrium constant (Kp) is:
ÎG = -R*T*ln(Kp)
-38200/(-8.3144*343) = ln(Kp)
e^13.394 = Kp
656055.062 = Kp
Answer:
The equilibrium constant is  6.6 * 10^5
Explanation:
Step 1: Data given
The standard free energies of formation of CO(g) = -137.3 kJ/mol
The standard free energies of formation of Ni(CO)4(g) = â587.4 kJ/mol,
Temperature = 70.0 °C
Step 2: The balanced equation
Ni(s) + 4CO(g) â Ni(CO)4(g)
For 1 mol Ni we need 4 moles CO to produce 1 mol Ni(CO)4
Step 3: Calculate Gibbs free energy
Gibbs free energy= Â standard free energy of formation of products - standard free energy of formation of reactants
Gibbs free energy = Â standard free energy of formation(Ni(CO)4) - standard free energy of formation(Ni(s)) +4Gf(CO) )
Gibbs free energy = Â -587.4 -(4*-137.3)
Gibbs free energy = -587. 4 +549.2
Gibbs free energy = -38.2 kJ/mol
Step 4: Calculate the equilibrium cosntant K
Î G â Â = -RT ln K
-38.2 kJ/mol = -8.314 * 343 * ln K
ln K = 13.4
K = e^13.4
K = 6.6 * 10^5
The equilibrium constant is  6.6 * 10^5