Question 1 of 5: Activity Coefficients from Partial Molar Excess Gibbs Energy Data (Cr–Ni)

Question 1: Activity Coefficients from Partial Molar Excess Gibbs Energy Data (Cr–Ni) (25%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. T = 1600 °C = 1873 K; the tabulated Δ̄G_Cr^xs(X_Cr) above, which is the only composition data provided (Ni’s partial excess free energy is not given directly).

Find. γ_Cr° (dilute), γ_Ni, Δ̄G_Ni^xs, ΔG^xs_mix and ΔG_mix, all at X_Ni = 0.7 (X_Cr = 0.3).

Approach. Cr’s activity coefficient follows directly from Δ̄G_Cr^xs = RT ln γ_Cr; Ni’s partial excess free energy is not tabulated and must be recovered from Cr’s data via the Gibbs–Duhem equation, integrated numerically (trapezoidal rule) because only five discrete points are given.

1. Activity coefficient of Cr in the dilute region. Under Henry’s law the partial excess free energy of the solute approaches a composition-independent constant as X_Cr→0; the lowest tabulated point (X_Cr = 0.1) is the best available estimate of that constant. $$\ln\gamma_{Cr}^{\circ}=\frac{\Delta\bar{G}_{Cr}^{xs}}{RT}=\frac{-12714}{(8.314)(1873)}=-0.8165$$ $$\boxed{\gamma_{Cr}^{\circ}=e^{-0.8165}=0.442}$$
   Check

   Treating X_Cr = 0.1 as “the dilute region” is the standard exam-level approximation given only discrete data starting at 0.1; a true Henry’s-law constant would extrapolate the curve to X_Cr→0.
2. Gibbs–Duhem relation for the second component. At constant T, P: $$X_{Cr}\,d\big(\Delta\bar{G}_{Cr}^{xs}\big)+X_{Ni}\,d\big(\Delta\bar{G}_{Ni}^{xs}\big)=0 \ \Rightarrow\ \Delta\bar{G}_{Ni}^{xs}(X_{Cr})=-\int_0^{X_{Cr}}\frac{X_{Cr}}{1-X_{Cr}}\,d\big(\Delta\bar{G}_{Cr}^{xs}\big)$$ using Δ̄G_Ni^xs = 0 at X_Cr = 0 (pure Ni) as the reference.
3. Numerical integration to X_Cr = 0.3. Split the integral at the two tabulated points that bracket it. Over [0, 0.1] the integrand’s increment d(Δ̄G_Cr^xs) is taken as ~0 (Henry’s-law plateau, same assumption as Step 1); over [0.1, 0.3] the trapezoidal rule uses the tabulated values directly, with g(X_Cr) = X_Cr/(1−X_Cr): $$g(0.1)=\frac{0.1}{0.9}=0.1111,\quad g(0.3)=\frac{0.3}{0.7}=0.4286,\quad \bar g=0.2698$$ $$\Delta\big(\Delta\bar{G}_{Cr}^{xs}\big)_{0.1\to0.3}=(-5442)-(-12714)=7272\text{ J}$$ $$\Delta\bar{G}_{Ni}^{xs}(X_{Ni}=0.7)=-\bar g\times 7272=-0.2698\times7272$$ $$\boxed{\Delta\bar{G}_{Ni}^{xs}(X_{Ni}=0.7)\approx-1962\text{ J/mol}}$$
   Check

   This is a two-panel trapezoidal (graphical) integration of sparse data — the standard hand method for this class of problem — and carries a few-percent numerical uncertainty; a finer grid or a smooth fitted curve would shift the result slightly.
4. Activity coefficient of Ni at X_Ni = 0.7. $$\ln\gamma_{Ni}=\frac{\Delta\bar{G}_{Ni}^{xs}}{RT}=\frac{-1962}{(8.314)(1873)}=-0.1260$$ $$\boxed{\gamma_{Ni}=e^{-0.1260}=0.882}$$
5. Integral molar excess Gibbs free energy of mixing at X_Ni = 0.7. $$\Delta G^{xs}_{mix}=X_{Cr}\,\Delta\bar{G}_{Cr}^{xs}+X_{Ni}\,\Delta\bar{G}_{Ni}^{xs} =(0.3)(-5442)+(0.7)(-1962)$$ $$\boxed{\Delta G^{xs}_{mix}=-3006\text{ J/mol}}$$
6. Integral molar Gibbs free energy of mixing at X_Ni = 0.7. Add the ideal entropy-of-mixing contribution to the excess term. $$\Delta G_{mix}=\Delta G^{xs}_{mix}+RT\big(X_{Cr}\ln X_{Cr}+X_{Ni}\ln X_{Ni}\big)$$ $$=-3006+(8.314)(1873)\big[(0.3)\ln(0.3)+(0.7)\ln(0.7)\big]=-3006+(15572)(-0.6109)$$ $$\boxed{\Delta G_{mix}=-12{,}519\text{ J/mol}}$$