Let's break down the question and solve it **step-by-step**. --- ## 1. **Understanding the Problem** You are asked if the **bioethanol production from glucose** is **spontaneous** at \(25^\circ C\). You are given for **glucose**: - Standard enthalpy of formation, \(\bar{h}_f^o = -1,271,000\) kJ/kmol - Standard absolute entropy, \(\bar{s}^o = 209\) kJ/kmol·K The reaction for **bioethanol production from glucose (fermentation)** is typically: \[ \text{Glucose} \rightarrow 2\ \text{Ethanol} + 2\ \text{CO}_2 \] \[ \mathrm{C_6H_{12}O_6 (aq) \rightarrow 2\ C_2H_5OH (aq) + 2\ CO_2 (g)} \] --- ## 2. **Spontaneity Criterion: Gibbs Free Energy** A reaction is **spontaneous** if \(\Delta G^o < 0\), where \[ \Delta G^o = \Delta H^o - T\Delta S^o \] We need to find \(\Delta G^o\) for the reaction at \(T = 25^\circ C = 298.15\) K. --- ## 3. **Find Standard Data for All Species** You are given only **glucose** data. Let’s look up the standard values for the other species: ### Standard enthalpy of formation (\(h_f^o\)) in kJ/kmol: - **Ethanol (aq):** \(-277,690\) kJ/kmol - **CO₂ (g):** \(-393,510\) kJ/kmol ### Standard absolute entropy (\(s^o\)) in kJ/kmol·K: - **Ethanol (aq):** \(160\) kJ/kmol·K - **CO₂ (g):** \(214\) kJ/kmol·K **(Values are from standard chemical tables. If your course uses different values, substitute those.)** --- ## 4. **Calculate \(\Delta H^o\) and \(\Delta S^o\) for the Reaction** ### a. **Standard Enthalpy Change (\(\Delta H^o\)):** \[ \Delta H^o = \sum h_f^o(\text{products}) - \sum h_f^o(\text{reactants}) \] \[ = [2 \times h_f^o(\text{ethanol}) + 2 \times h_f^o(\text{CO}_2)] - [h_f^o(\text{glucose})] \] \[ = [2 \times (-277,690) + 2 \times (-393,510)] - [-1,271,000] \] Calculate: \[ 2 \times (-277,690) = -555,380 \] \[ 2 \times (-393,510) = -787,020 \] Sum: \[ -555,380 + (-787,020) = -1,342,400 \] \[ \Delta H^o = (-1,342,400) - (-1,271,000) \] \[ = -1,342,400 + 1,271,000 = -71,400\ \text{kJ} \] --- ### b. **Standard Entropy Change (\(\Delta S^o\)):** \[ \Delta S^o = \sum s^o(\text{products}) - \sum s^o(\text{reactants}) \] \[ = [2 \times s^o(\text{ethanol}) + 2 \times s^o(\text{CO}_2)] - [s^o(\text{glucose})] \] \[ = [2 \times 160 + 2 \times 214] - [209] \] \[ 2 \times 160 = 320 \] \[ 2 \times 214 = 428 \] \[ 320 + 428 = 748 \] \[ \Delta S^o = 748 - 209 = 539\ \text{kJ/K} \] --- ## 5. **Calculate \(\Delta G^o\) at 298.15 K** \[ \Delta G^o = \Delta H^o - T\Delta S^o \] \[ \Delta H^o = -71,400\ \text{kJ} \] \[ \Delta S^o = 539\ \text{kJ/K} \] \[ T = 298.15\ \text{K} \] \[ \Delta G^o = -71,400\ \text{kJ} - (298.15\ \text{K} \times 539\ \text{kJ/K}) \] \[ = -71,400\ \text{kJ} - (160,690\ \text{kJ}) \] \[ = -71,400\ \text{kJ} - 160,690\ \text{kJ} \] \[ = -232,090\ \text{kJ} \] --- ## 6. **Conclusion: Spontaneity** Since \(\Delta G^o = -232,090\) kJ < 0, **the reaction is spontaneous at 25°C**. --- ## **Final Answer** **The production of bioethanol from glucose is a spontaneous reaction at 25°C, because the standard Gibbs free energy change, \(\Delta G^o\), is negative (\(-232,090\) kJ).**