Question

The equilibrium constant is 3.9×105 at 300 K and 1.2×10−1 at 500 K.Calculate ΔrH∘ for this reaction.Calculate ΔrS∘ for this reaction.

Answer 1

To calculate ΔrG° for the reaction, we can use the equation ΔrG° = -RT ln(K) where K is the equilibrium constant and R is the gas constant. By setting up two equations using the given equilibrium constants at different temperatures, we can solve for ΔrG°.

Step-by-step explanation:

The equilibrium constant, K, is related to the standard free energy change, ΔrG°, by the equation:

ΔrG° = -RT ln(K)

Where R is the gas constant (8.314 J/mol K) and T is the temperature in Kelvin.

Thus, we can write the equation as:

K = e^(-ΔrG°/RT)

Given that the equilibrium constant at 300 K is 3.9×10^5 and at 500 K is 1.2×10^(-1), we can set up two equations:

3.9×10^5 = e^(-ΔrG°/8.314(300))

1.2×10^(-1) = e^(-ΔrG°/8.314(500))

We can solve these equations simultaneously to find ΔrG°.

Answer 2

Step-by-step 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: