Respuesta :
Explanation:
For the determined reaction we know the equilibrium constant at two different temperatures.
T₁ = 300 K K₁ = 3.9 * 10^5
T₂ = 500 K K₂ = 1.2 * 10^-1
Before we find the change in enthalpy and entropy we have to find the free energy change using this formula.
ΔG = - R * T * ln K
Where ^Δ is the Free Gibbs Energy change, R is the ideal gas constant, T is the temperature and K is the equilibrium constant.
R = 8.3145 J/(K*mol)
Then we can find the GΔ for both conditions:
ΔG₁ = - R * T₁ * ln K₁
ΔG₁ = - 8.3145 J/(K*mol) * 300 K * ln (3.9 * 10^5)
ΔG₁ = -32112 J/mol = -32.1 kJ/mol
ΔG₁ = -32.1 kJ/mol
ΔG₂ = - R * T₂ * ln K₂
ΔG₂ = - 8.3145 J/(K*mol) * 500 K * ln (1.2 * 10^-1)
ΔG₂ = 8814 J/mol = 8.81 kJ/mol
ΔG₂ = 8.81 kJ/mol
Now we also know that:
ΔG = ΔH - T * ΔS
But ΔH and ΔS won't be affected by the temperature. So with our two conditions we can set up a system of equations and solve them for ΔH and ΔS.
ΔG₁ = ΔH - T₁ * ΔS ---> ΔH = ΔG₁ + T₁ * ΔS
ΔG₂ = ΔH - T₂ * ΔS
ΔG₂ = ΔG₁ + T₁ * ΔS - T₂ * ΔS
ΔG₂ - ΔG₁ = ΔS * (T₁ - T₂)
ΔS = (ΔG₂ - ΔG₁)/(T₁ - T₂)
ΔS = (8.81 kJ/mol + 32.1 kJ/mol)/(300 K - 500 K)
ΔS = - 0.205 kJ/(mol*K)
ΔH = ΔG₁ + T₁ * ΔS
ΔH = - 32.1 kJ/mol + 300 K * (- 0.205 kJ/(mol*K))
ΔH = -93.6 kJ/mol
Answer:
ΔS = -0.205 kJ/(mol*K)
ΔH = -93.6 kJ/mol
Answer:
