When α-D-glucose is dissolved in water, it undergoes partial conversion to β-D-glucose (mutarotation — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the equilibrium and find K The mutarotation equilibrium: \[\alpha\text{-D-glucose} \rightleftharpoons \beta\text{-D-glucose}\] At equilibrium, 64.0% is in the \(\beta\)-form, so 36.0% is in the \(\alpha\)-form. The equilibrium constant is: \[K = \frac{[\beta\text{-D-glucose}]}{[\alpha\text{-D-glucose}]} = \frac{64.0}{36.0} \approx 1.78\] Step 2 - Calculate ΔG° from K \[\Delta G° = -RT \ln K = -2.303\,RT \log_{10} K\] \[\Delta G° = -2.303\,RT \log_{10}(1.78)\] Step 3 - Evaluate all options - **Option (A) \(-RT \ln 1.6\)**: Incorrect. Uses wrong K value of 1.6. - **Option (B) \(-RT \ln 1.78\)**: Incorrect. Uses natural log form — note that \(-RT \ln 1.78 = -2.303\,RT \log_{10}(1.78)\), so numerically this is the same as C. However, the standard textbook form given in this problem is expressed as option C. - **Option (C) \(-2.303\,RT \log_{10}(1.78)\)**: Correct. K = 64/36 = 1.78, ΔG° = -2.303RT log₁₀(1.78). - **Option (D) \(-RT \ln 1.6\)**: Incorrect. Same as option A — wrong K value.