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Step-by-Step Solution
Step 1: Identify the Process
We are considering the vaporization process of substance S:
S(l) → S(g)
The thermodynamic quantity of interest is the Gibbs free energy change for the above process at 500 K.
Step 2: Calculate ΔG° for Vaporization
Given:
ΔfG°(liquid) = +100.7 kcal/mol
ΔfG°(gas) = +103 kcal/mol
For the vaporization process, we use:
ΔG° = ΔfG°(gas) − ΔfG°(liquid)
Substitute the values:
ΔG° = 103 − 100.7 = 2.3 kcal/mol
Convert this to cal/mol:
2.3 kcal/mol = 2.3 × 10³ cal/mol
Step 3: Use the Relation ΔG° = −RT ln K
The equation linking Gibbs free energy change to the equilibrium constant K is:
$ \Delta G^\circ = - R T \ln K $
Sometimes, it is convenient to use the log base 10 form:
$ \Delta G^\circ = - 2.303 \, R \, T \, \log K $
Step 4: Substitute the Known Values
Given in the question:
R = 2 cal K⁻¹ mol⁻¹
T = 500 K
ΔG° = 2.3 × 10³ cal/mol
Hence:
$ 2.3 \times 10^3 = - (2.303 \times 2 \times 500) \log K $
Step 5: Solve for log K and Then K
First, compute the factor:
2.303 × 2 × 500 = 2.303 × 1000 = 2303
So the equation becomes:
$ 2.3 \times 10^3 = -2303 \, \log K $
Rearrange to find log K:
$ \log K = - \frac{2.3 \times 10^3}{2303} $
Numerically, this is approximately −1. Therefore,
$ \log K = -1 $
$ K = 10^{-1} = 0.1 $
Step 6: Interpret K as Vapour Pressure
For vaporization, the equilibrium constant K is the vapour pressure in atm under standard conditions. Therefore,
Vapour pressure of liquid S at 500 K = 0.1 atm
