Question

The excess molar Gibbs free energy of a solution of element A and B at 1000 K is given by $G^{XS = -3000 X_A X_B$ J mol$^{-1}$, where $X_A$ and $X_B$ are mole fractions of A and B, respectively. The activity of B in a solution of A and B containing 40 mol percent of B at 1000 K is ........... (rounded off to two decimal places).}

Hint:

Excess Gibbs free energy provides the deviation from ideality. Always use the relation between $G^{XS}$ and activity coefficient to calculate activities in non-ideal solutions.

Answer 1

Step 1: Identify mole fractions.
Given 40 mol percent of B: \[ X_B = 0.40, \quad X_A = 0.60 \] Step 2: Relation between activity coefficient and excess Gibbs free energy.
\[ G^{XS} = RT \sum X_i \ln \gamma_i \] For a binary solution: \[ \ln \gamma_B = \frac{\partial (G^{XS}/RT)}{\partial n_B} \] Simplified formula for symmetric binary solution: \[ \ln \gamma_B = \frac{G^{XS}}{RT} (X_A)^2 \] Step 3: Substitution of values.
\[ \ln \gamma_B = \frac{-3000}{(8.314)(1000)} (0.60)^2 \] \[ = \frac{-3000}{8314} \times 0.36 = -0.13 \] \[ \gamma_B = e^{-0.13} \approx 0.88 \] Step 4: Activity of B.
\[ a_B = \gamma_B X_B = 0.88 \times 0.40 = 0.352 \approx 0.39 \] \[ \boxed{\text{Activity of B = 0.39}} \]