### Related Formula Q = n_textreacted cdot Delta H_textneutralization$Q = n_{\text{reacted}} \cdot \Delta H_{\text{neutralization}}$ Delta T = fracQm cdot c$\Delta T = \frac{Q}{m \cdot c}$ ### Core Logic The temperature rise depends directly on the total heat released (Q$Q$) normalized by the total heat capacity of the resulting mixed volume (m cdot c$m \cdot c$). Let's evaluate the millimoles of mathrmH^+$\mathrm{H}^+$ and mathrmOH^-$\mathrm{OH}^-$ that react in each mixture: 1. **Option 1:** 30text mL of 1mathrmM mathrmHCl + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. Both are strong electrolytes, releasing full neutralization energy (sim -57.3text kJ/mol$\sim -57.3\text{ kJ/mol}$). Total volume = 60text mL$60\text{ mL}$. 2. **Option 2:** 30text mL of 1mathrmM mathrmCH_3COOH + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. However, since acetic acid is a weak acid, part of the heat is consumed in its ionization. Thus, less total heat is evolved compared to Option 1. 3. **Option 3:** 50text mL of 1mathrmM mathrmHCl + 20text mL of 1mathrmM mathrmNaOH$50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textLimiting reagent = mathrmNaOH = 20text mmol$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. Only 20text mmol$20\text{ mmol}$ reacts. Total volume = 70text mL$70\text{ mL}$. 4. **Option 4:** 45text mL of 1mathrmM mathrmCH_3COOH + 25text mL of 1mathrmM mathrmNaOH$45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textLimiting reagent = 25text mmol$\text{Limiting reagent} = 25\text{ mmol}$ weak neutralization profile. Comparing Option 1 and Option 3, Option 1 releases significantly more heat (30text mmol$30\text{ mmol}$ vs 20text mmol$20\text{ mmol}$) into a smaller volume (60text mL$60\text{ mL}$ vs 70text mL$70\text{ mL}$), yielding the largest increase in temperature Delta T$\Delta T$. ### Pattern Recognition To maximize Delta T$\Delta T$, look for the option that maximizes the amount of reacting strong acid and strong base equivalents while keeping the total solution volume as small as possible. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium