Answer to Question #213412 in General Chemistry for khola2709

Question

Find an expression for activity coefficient from the  Two-suffix- Margules equation for excess Gibbs energy.

The expression for Two-suffix-Magules Equation is given by :

$G^E=Ax_1x_2$

Answer 1

Given expression for two-suffix-magules equation is given by-

$\dfrac{G^{E}}{RT}=AX_1X_2$

,$where \ RT=constant$ $and\ X_1\ andX_2\ are \ the fractions\ of \ each \ component.$

Now portion molar excess gibbs free energy is given by the equation -

$G_i^{E}=RTln\gamma_{i}$

The above equation can also be written as-

$\dfrac{G_{i}}{RT}=ln\gamma_{i}$

now by definition of extensive partial molar property -

$ln\gamma_{i}=(\dfrac{\partial({\dfrac{nG^{E}}{(RT)}})}{\partial{n}_{i}})_T,_P,n_{j}$ (partial molar excess gibbs free energy)

$ln\gamma_{1}=(\dfrac{\partial({\dfrac{nG^{E}}{(RT)}})}{\partial{n}_{1}})_T,_P,n_{1}$

Now, by expression for - two - suffix- magules -

$\dfrac{G^{E}}{RT}=AX_1X_2$ $X_{1}=\dfrac{n_{1}}{n_{1}+n_{2}}$ $=\dfrac{n_1}{n}$

$X_{2}=\dfrac{n_{2}}{n_{1}+n_{2}}$ $=\dfrac{n_{2}}{n}$

Now putting $X_1,X_2\$ in the above equation we get

$\dfrac{G^{E}}{RT}=AX_1X_2$

$\dfrac{G^{E}}{RT}=A(\dfrac{n_{1}}{n_{1}+n_{2}})(\dfrac{n_{2}}{n_{1}+n_{2}})$ $=A\dfrac{n_{1}n_{2}}{(n_1+n_2)^{2}}$

now multiplying by n on both side of the , equation we get -

$n\dfrac{G^{E}}{RT}=(n_1+n_2)A(\dfrac{n_{1}}{n_{1}+n_{2}})(\dfrac{n_{2}}{n_{1}+n_{2}})$ $=\dfrac{An_1n_2}{n_1+n_2}$

$ln\gamma_{1}=(\dfrac{\partial({\dfrac{nG^{E}}{(RT)}})}{\partial{n}_{i}})_T,_P,n_{2}$

$ln\gamma_{i}=(\dfrac{\partial({\dfrac{An_1n_2}{n_1+n_2}})}{\partial n_1})_T,_P,_{N_J}$

$ln\gamma_{i}=An_2(\dfrac{\partial({\dfrac{n_1}{n_1+n_2}})}{\partial n_1})_{n_2}$

$ln\gamma_{i}=A(\dfrac{n_2^{2}}{n_1+n_2})$

$ln\gamma_{i}=A(\dfrac{n_2^{2}}{n})$

$ln\gamma_{i}=A(X_2^{2})$

$ln\gamma_{i}=A(X_1^{2})$

so this is our equation .