The change in chemical potential is -36 kJ mol^-1. This numerical fill-in-the-blank question from thermodynamics assesses the temperature dependence of chemical potential at constant pressure with constant molar entropy.

Thermodynamic Basis

Chemical potential μ equals the molar Gibbs free energy g_m, so (∂μ/∂T)_P = -s_m (∂T/∂μ)_P = -s_m from the Gibbs-Duhem equation, where s_m is molar entropy. Integrating at constant pressure and constant s_m gives Δμ = μ(T₂) – μ(T₁) = -s_m(T₂ – T₁). Here, T₁ = 2000 K, T₂ = 2600 K, and s_m = 60 J K^-1 mol^-1 = 0.060 kJ K^-1 mol^-1, so ΔT = 600 K and Δμ = -0.060 × 600 = -36 kJ mol^-1.

Detailed Solution Steps

Identify conditions: Constant pressure, s_m constant over range.

Apply relation: dμ = -s_m dT (at constant P).

Integrate: ∫_μ1^μ2 dμ = -s_m ∫₂₀₀₀²⁶⁰⁰ dT.

Compute: Δμ = -60 × 10^-3 × 600 = -36 kJ mol^-1.

Why No Options?

This is a numerical response question requiring exact computation, common in exams like CSIR NET Life Sciences for thermodynamics applications in biochemical systems. No multiple choices; fill -36 (or -36.0 if decimals expected).

Key Concepts

Chemical potential μ drives phase stability and reactions. At constant pressure:

dμ = -S dT + V dP, simplifies to dμ = -s_m dT.

Constant s_m assumption holds for narrow ranges or ideal cases.

Step-by-Step Calculation

Note ΔT = 2600 – 2000 = 600 K.

s_m = 60 J K^-1 mol^-1 = 0.060 kJ K^-1 mol^-1.

Δμ = -s_m ΔT = -0.060 × 600 = -36 kJ mol^-1.

This negative value indicates lower μ at higher T, favoring reactions or phase changes.

CSIR NET Relevance

Such problems test integration of partial derivatives in biochemical contexts like protein stability or enzyme kinetics under high-temperature conditions. Practice similar derivations for scoring in Physical Chemistry section.