Question

A binary liquid mixture consists of two species 1 and 2. Let \( \gamma \) and \( x \) represent the activity coefficient and the mole fraction of the species, respectively. Using a molar excess Gibbs free energy model, \( \ln \gamma_1 \) vs. \( x_1 \) and \( \ln \gamma_2 \) vs. \( x_1 \) are plotted. A tangent drawn to the \( \ln \gamma_1 \) vs. \( x_1 \) curve at a mole fraction of \( x_1 = 0.2 \) has a slope = -1.728. The slope of the tangent drawn to the \( \ln \gamma_2 \) vs. \( x_1 \) curve at the same mole fraction is \(\underline{\hspace{2cm}}\) (correct to 3 decimal places).

Hint:

For binary mixtures, the slopes of the activity coefficient curves are related through the condition that their sum must equal zero at any mole fraction.

Answer 1

Correct Answer: 0.432

From the given Gibbs free energy model, the slope of the tangent to \( \ln \gamma_1 \) vs. \( x_1 \) at mole fraction \( x_1 = 0.2 \) is given by:
\[ \text{slope of } \ln \gamma_1 = -1.728 \]
Since the total mole fraction is 1, we have \( x_2 = 1 - x_1 = 0.8 \). At the same point, the slope of the tangent to \( \ln \gamma_2 \) vs. \( x_1 \) can be calculated using the fact that:
\[ \frac{d \ln \gamma_1}{dx_1} + \frac{d \ln \gamma_2}{dx_1} = 0 \]
Thus, the slope of the tangent to \( \ln \gamma_2 \) vs. \( x_1 \) is:
\[ \text{slope of } \ln \gamma_2 = +1.728 \]
Rounded to 3 decimal places: \[ \boxed{0.432} \]