Thermodynamics, cannot reach the answer (Reichl's book related)

[fluidistic]

Homework Statement

Consider a binary mixture of particles of types 1 and 2 whose Gibbs free energy is given by $G=n_1 \mu _1 ^0 (P,T)+n_2 \mu _2 ^0 (P,T) +RT n_1 \ln x_1 +RTn_2 \ln x_2 +\lambda n x_1x_2$.
Where $n=n_1+n_2$. And $x_1$ and $x_2$ are the mole fractions of particles 1 and 2 respectively.
The book on page 159 states that the conditions $\mu _1 ^I=\mu _1 ^{II}$ and $\mu _2 ^I=\mu _2 ^{II}$ where the upperscript denotes the phase. I have no problem to understand that those conditions must be fulfilled for equilibrium when there's a coexistance of phases.
But now comes the part where I struggle. From the condition $\mu _1 ^I=\mu _1 ^{II}$, I'm supposed to find that $RT \ln x_1 ^I + \lambda (1-x_1 ^I)^2=RT \ln x_1 ^{II}+\lambda (1-x_1 ^{II})^2$.
But I don't reach this.

Homework Equations

$\left ( \frac{\partial g}{\partial n_1} \right ) _{T,P,n_2} =\mu _1$.
$x_2=1-x_1$.

The Attempt at a Solution

$\mu _1 =\left ( \frac{\partial g}{\partial n_1} \right ) _{T,P,n_2}=\mu _1 ^0 (P,T)+RT \ln x_1 + \lambda x_1 (1-x_1)$.
So if I use the condition $\mu _1 ^I = \mu _1^{II}$, I'd get $RT\ln x_1 ^I + \lambda x_1 ^I (1-x_1 ^I )=RT\ln x_1 ^{II} + \lambda x_1 ^{II} (1-x_1 ^{II} )$. Which differs from the equation I'm supposed to find.
I know I simply have to derivate g=G/n with respect to $n_1$ while keeping P, T and $n_2$ constant, but I'm simply failing at that apparently.
I've rechecked the algebra many times, I really don't see where I go wrong. My friend told me he reached the good result, so I know I've went wrong somewhere.

Thank you for any help.

 

[TSny]

x₁ and x₂ are not constants. They depend on n₁ and n₂.

 

[fluidistic]

Thank you TSny, I overlooked this.