Let's break down the question and the graph, then answer step by step: --- ### **Given:** - **Graph:** ΔS° (entropy change) vs. T (temperature) for the reaction: \( M + O_2 \rightarrow MO_2 \) - **Given:** \( M \) and \( MO_2 \) are solid at 273 K. - **Task:** 1. **Plot ΔG° vs T schematically,** 2. **Explain the slopes and "brakes" (breaks/changes) in the graph.** --- ## **Step 1: Understanding the Graph** The graph shows ΔS° (y-axis) as a function of T (x-axis), and it has steps: - At certain temperatures, ΔS° changes abruptly (vertical "breaks"). - Between these breaks, ΔS° is constant (horizontal lines). --- ## **Step 2: Relating ΔS° to ΔG°** Recall for a reaction: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] - If ΔH° is constant with temperature (reasonable for a schematic plot), the slope of ΔG° vs T is **–ΔS°**: \[ \frac{d(\Delta G^\circ)}{dT} = -\Delta S^\circ \] --- ## **Step 3: Sketching ΔG° vs T** - **Where ΔS° is constant:** ΔG° vs T will be a straight line with slope = –ΔS° (negative slope, magnitude depends on value of ΔS°). - **Where ΔS° changes (step):** Slope of ΔG° vs T changes abruptly at the same temperature as the "brake" in ΔS°. ### **Schematic Plot:** - Start with a straight line (slope = –ΔS°₁). - At the first temperature where ΔS° changes, the slope of ΔG° vs T suddenly changes to –ΔS°₂. - Repeat for each step in ΔS°. --- ## **Step 4: Explaining Slopes and Breaks** - **Slopes:** The slope of the ΔG° vs T line at any region is **–ΔS°** for that region. - **Breaks:** At temperatures where ΔS° changes (vertical steps in the entropy graph), the ΔG° vs T plot has a "kink" or change in slope. This usually corresponds to a phase transition (e.g., melting, boiling, sublimation) of a reactant or product. ### **Physical Interpretation:** - At these temperatures, (likely) one of the substances involved melts, boils, or undergoes a phase change, causing a sudden change in entropy. - For example, if O₂ melts or sublimes, its entropy increases sharply, changing the overall ΔS° for the reaction. --- ## **Step 5: Final Schematic ΔG° vs T Plot** - At 273 K, start with a certain slope (–ΔS°₁). - At the first "brake" (next temperature where ΔS° jumps), slope becomes (–ΔS°₂). - At the next "brake", slope becomes (–ΔS°₃), and so on. - The plot is a series of connected straight lines with slopes changing at specific temperatures. --- ### **[Schematic Sketch]** ``` ΔG° | | / | / | / |__/ | / |__/ | +----------------- T 273 T1 T2 ``` - Each "/" is a straight segment with a different negative slope (steeper or shallower depending on ΔS°). - The points where the slope changes (**kinks**) correspond to the "brakes" in the ΔS° vs T graph. --- ## **Summary Table** | Region (T) | ΔS° | Slope of ΔG° vs T (–ΔS°) | Explanation | |------------|-----|--------------------------|------------------------------| | 273–T₁ | ΔS°₁ | –ΔS°₁ | Both solid phases | | T₁–T₂ | ΔS°₂ | –ΔS°₂ | Phase change (e.g., O₂ melts)| | T₂–T₃ | ΔS°₃ | –ΔS°₃ | Another phase change, etc. | --- ## **Final Answer** ### **Schematic Plot of ΔG° vs T:** - A piecewise linear graph with kinks at the temperatures where ΔS° changes in the given entropy graph. ### **Explanation of Slopes and Breaks:** - **Slopes:** Determined by minus the value of ΔS° in each region. - **Breaks (kinks):** Occur at temperatures where ΔS° changes abruptly—these correspond to phase transitions of substances in the reaction. If you need a hand-drawn or digital sketch, let me know!

Let's approach this systematically with detailed explanations, calculations, and the final schematic plot. --- ### **Given Data & Concepts:** - Reaction: \[ M + O_2 \rightarrow MO_2 \] - All species are solids at 273 K. - The graph provided shows ΔS° (entropy change) vs. T, with abrupt jumps ("breaks") at certain temperatures. - **Goal:** 1. Sketch the variation of ΔG° with T schematically. 2. Explain the slopes and the "breaks" in the graph. --- ### **Step 1: Understanding the thermodynamic relation** The fundamental relation linking Gibbs free energy change (ΔG°), enthalpy change (ΔH°), and entropy change (ΔS°): \[ \boxed{ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ } \] - **At constant ΔH°,** the variation of ΔG° with T is linear, with the slope: \[ \frac{d (\Delta G^\circ)}{d T} = - \Delta S^\circ \] --- ### **Step 2: Analyzing the ΔS° vs T graph** - The graph shows ΔS° as a **stepwise function** of T: - **Horizontal segments:** ΔS° constant over a temperature range. - **Vertical "breaks":** ΔS° jumps abruptly at specific temperatures, indicating phase transitions (e.g., melting, sublimation). - **Interpretation:** - The sudden increase in ΔS° at a certain T is due to a phase change (e.g., O₂ subliming or melting), adding extra entropy to the reaction. --- ### **Step 3: Relating ΔS° steps to ΔG°** Given: \[ \Delta G^\circ(T) = \Delta H^\circ - T \Delta S^\circ(T) \] - **In each temperature range where ΔS° is constant:** \[ \boxed{ \Delta G^\circ(T) = \text{constant} - T \times \Delta S^\circ } \] - **Graphically:** ΔG° vs T is a straight line with slope = \(-\Delta S^\circ\). - **At a temperature where ΔS° jumps:** - The slope of ΔG° vs T changes abruptly (a "kink" or "break" in the plot). --- ### **Step 4: Sketching the ΔG° vs T curve** - **Between phase transition temperatures:** - The plot is a straight line with slope = \(-\Delta S^\circ\) (constant in each segment). - **At phase transition points:** - The slope changes suddenly due to the change in ΔS°. - **Overall shape:** - Piecewise linear with different slopes, connected at the transition temperatures. --- ### **Step 5: Calculations (Quantitative example)** Suppose the ΔS° values and transition temperatures are: | Temperature | ΔS° (J/K) | Description | |--------------|------------|-------------------------------------| | 273 K | ΔS°₁ | Initial entropy change | | T₁ (say, 300 K) | ΔS°₂ | First phase transition (e.g., sublimation) | | T₂ (say, 350 K) | ΔS°₃ | Second phase transition (e.g., melting) | **Note:** Since exact numerical data isn't provided in the image, we'll proceed with symbolic notation. --- ### **Step 6: Final schematic plot** - **At T = 273 K:** \[ \Delta G^\circ = \text{initial value} \] - **Between 273 K and T₁:** \[ \Delta G^\circ(T) = \Delta G^\circ(273) - \Delta S^\circ_1 (T - 273) \] - **At T₁:** \[ \Delta G^\circ(T_1^+) = \Delta G^\circ(T_1^-) - (\Delta S^\circ_2 - \Delta S^\circ_1) T_1 \] - **Between T₁ and T₂:** \[ \Delta G^\circ(T) = \text{value at } T_1^+ - \Delta S^\circ_2 (T - T_1) \] - **At T₂:** Similarly, the slope shifts again. --- ### **Final Answer:** #### **(a) The schematic plot of ΔG° vs T:** - A **piecewise linear graph** with **different slopes** in each temperature segment. - **Each segment's slope = –ΔS° in that region.** - **At temperatures T₁, T₂, ... where ΔS° jumps:** - The plot exhibits a **kink**—a change in slope—reflecting the phase transition. #### **(b) Explanation of slopes and "breaks":** - **Slopes:** \[ \frac{d (\Delta G^\circ)}{d T} = - \Delta S^\circ \] - Larger ΔS° gives a steeper negative slope. - Smaller ΔS° yields a gentler slope. - **Breaks ("kinks"):** - Correspond to abrupt increases in ΔS°—phase transitions such as melting, sublimation, or vaporization. - These cause a sudden change in the slope of ΔG° vs T, indicating a change in the thermodynamic behavior of the system. --- ### **Summary:** | Aspect | Explanation | |---------|--------------| | **Shape of ΔG° vs T** | Piecewise linear with slopes = –ΔS° in each segment. | | **At phase transitions** | Slope changes abruptly ("kink") due to sudden ΔS° change. | | **Physical significance** | These transitions represent phase changes of reactants/products affecting the entropy. --- **Note:** Without the specific numerical values, this is a qualitative schematic and explanation. If you provide exact ΔS° values and transition temperatures, I can help with precise numerical plotting.