Consider the isomerization equlibria of an alkene (C4H8) that co-exists as three isomers.

(These 3 are in an equlibrium triangle)
Reaction 1: cis-2-butene ⇔ trans-2-butene
Reaction 2: cis-2-butene ⇔ 2-methylpropene
Reaction 3: trans-2-butene ⇔ 2-methylpropene

Known values at 300k are known:
cis-2-butene: standard G of formation = 66kJ/mol
Standard enthalpy of formation = -7.0kJ/mol

trans-2-butene: standard G of formation = 63kJ/mol
Standard enthalpy of formation = -11.2kJ/mol

2-methylpropene: standard G of formation =58.0kJ/mol
Standard enthalpy of formation = not know

The temperature of the above system is raised to 400K and equilibrium is re-established. The mole percentage of trans-2-butene at the new equilibrium is determined to be 18%. Calculate the enthalpy of formation and the entropy of formation for reaction 2 and 3.

Thanks

To calculate the enthalpy of formation for Reaction 2 (cis-2-butene ⇔ 2-methylpropene) and Reaction 3 (trans-2-butene ⇔ 2-methylpropene), we need to use the following equation:

ΔG = ΔH - TΔS

where ΔH is the enthalpy of formation, ΔS is the entropy of formation, and T is the absolute temperature (in Kelvin).

Given that we know the equilibrium mole percentage of trans-2-butene at the new equilibrium is 18%, we can use this information to determine how the concentrations of the three isomers have changed.

Let's assume that initially, the mole percentages of the three isomers are x, y, and z, respectively. The sum of these mole percentages is 1 since they represent the entire system.

At the new equilibrium, the mole percentages will be:
cis-2-butene: x%
trans-2-butene: 18%
2-methylpropene: z%

We can calculate the mole percentage of cis-2-butene by subtracting the percentages of trans-2-butene and 2-methylpropene from 100 (since the total sum must be 100):
cis-2-butene: (100 - 18 - z)%

Now, we can set up the equilibrium expressions for Reaction 2 and Reaction 3. Let's define K2 and K3 as the equilibrium constants for Reaction 2 and Reaction 3, respectively.

For Reaction 2:
K2 = (2-methylpropene) / (cis-2-butene)

For Reaction 3:
K3 = (2-methylpropene) / (trans-2-butene)

At equilibrium, the concentrations of the reactants and products can be expressed using the K values as follows:

For Reaction 2:
(cis-2-butene) / (2-methylpropene) = K2

For Reaction 3:
(trans-2-butene) / (2-methylpropene) = K3

Based on these equilibrium expressions, we can determine the concentrations of the isomers at the new equilibrium.

Now, we need to apply the Gibbs-Helmholtz equation, which relates ΔG, ΔH, and ΔS:

ΔG = ΔH - TΔS

We can rearrange this equation to solve for ΔH:

ΔH = ΔG + TΔS

Since we are given the values of ΔG for each isomer, we can calculate the change in Gibbs free energy, ΔG, for each reaction by subtracting the ΔG values of the reactants from the ΔG values of the products.

Finally, substitute the values of ΔG, T (temperature in Kelvin), and ΔH (enthalpy of formation) into the equation to solve for ΔS (entropy of formation):

ΔS = (ΔH - ΔG) / T

Using the given values and following these steps, you can calculate the enthalpy of formation and entropy of formation for each reaction (2 and 3).