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% of B at 1000 K is ......... (rounded off to two decimal places). Given: Ideal gas constant \( R = 8.314 \, {J mol}^{-1} {K}^{-1} \) 
 

Hint:

The activity of a component in a solution is related to the excess Gibbs free energy and can be calculated using the exponential form. Remember to use the correct temperature and gas constant in your calculations.

Answer 1

The excess Gibbs free energy \( G^{XS} \) is given by: \[ G^{XS} = -3000 X_A X_B \] Since \( X_A + X_B = 1 \), we have \( X_A = 1 - X_B \). The mole fraction of B is given as 40 mol%, so \( X_B = 0.40 \) and \( X_A = 0.60 \). The activity of B is related to the excess Gibbs free energy through: \[ {activity of B} = \exp\left(-\frac{G^{XS}}{RT}\right) \] Substituting the values: \[ G^{XS} = -3000 \times 0.60 \times 0.40 = -720 \, {J/mol} \] Now, calculate the activity: \[ {activity of B} = \exp\left(\frac{720}{8.314 \times 1000}\right) = \exp\left(0.0865\right) \approx 1.09 \]