For gaseous n-pentane and isopentane, both at 298K, ΔGf°=−194.4kJ/mol and−200.8kJ/mol, respectively. The vapor pressures of the liquids are given by:

Question

For gaseous n-pentane and isopentane, both at 298K, ΔGf°=−194.4kJ/mol and−200.8kJ/mol, respectively. The vapor pressures of the liquids are given by:

n-pentane:log10P(atm)=3.9714−1065/T−41

isopentane:log10P(atm)=3.9089−1020/T−40

Calculate (a) KP for the isomerization in the gas phase at 298 K, and (b) Kx for the isomerization in the liquid phase, assuming ideal solutions.

Answer 1

To calculate (a) KP for the isomerization in the gas phase at 298 K and (b) Kx for the isomerization in the liquid phase, we need to use the vapor pressures of n-pentane and isopentane, along with the equations provided. Let's go step by step:

(a) KP for the isomerization in the gas phase at 298 K:
KP is defined as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to the power of their stoichiometric coefficients. In this case, the reaction is the isomerization of n-pentane to isopentane:

n-pentane ⇌ isopentane

The equilibrium constant, KP, can be calculated using the vapor pressures of the two isomers:

KP = (P(isopentane) / P(n-pentane))

To obtain P(n-pentane) and P(isopentane), we need to substitute the temperature (298 K) into the given equations for the vapor pressures:

P(n-pentane) = 10^(3.9714 - 1065/T - 41)
P(isopentane) = 10^(3.9089 - 1020/T - 40)

Now, substitute these expressions into the KP equation:

KP = (10^(3.9089 - 1020/T - 40)) / (10^(3.9714 - 1065/T - 41))

Simplify the equation by applying the laws of logarithms:

KP = 10^(3.9089 - 1020/T - 40 - (3.9714 - 1065/T - 41))
KP = 10^(0.0625 + 45/T)

Therefore, KP for the isomerization in the gas phase at 298 K is 10^(0.0625 + 45/298).

(b) Kx for the isomerization in the liquid phase, assuming ideal solutions:
Kx is defined as the ratio of the molar concentration of the products to the molar concentration of the reactants, each raised to the power of their stoichiometric coefficients. Since we have ideal solutions, Kx is simply the equilibrium constant:

Kx = KP

Therefore, the value of Kx for the isomerization in the liquid phase is the same as KP, which we calculated in part (a).

Note: Make sure to substitute the temperature correctly into the equations provided and use the correct values of KP and Kx obtained from the calculations. Also, the above calculations assume that the given equations for the vapor pressures and the ideal solution assumption are valid for the system in question.