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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 23 January · Morning Shift · Chemistry · Thermodynamics (C)

Ice at -5^ C is heated to become vapor with temperature of 110^ C...

Ice at −5∘C-5^{\circ} \mathrm{C} is heated to become vapor with temperature of 110∘C110^{\circ} \mathrm{C} at atmospheric pressure. The entropy change associated with this process can be obtained fromIce at -5^ C is heated to become vapor with temperature of 110^ C at atmospheric pressure. The entropy changeIce at -5^ C is heated to become vapor with temperature of 110^ C at atmospheric pressure. The entropy changeIce at -5^ C is heated to become vapor with temperature of 110^ C at atmospheric pressure. The entropy changeIce at -5^ C is heated to become vapor with temperature of 110^ C at atmospheric pressure. The entropy change

Options

  1. AAt 298K , a 1 litre solution containing 10mmol of Cr _ 2 O _ 7 ^ 2 - and 100mmol of Cr ^ 3 + shows a pH of 3 . 0 .
  2. BAt 298K , a 1 litre solution containing 10mmol of Cr _ 2 O _ 7 ^ 2 - and 100mmol of Cr ^ 3 + shows a pH of 3 . 0 .
  3. CAt 298K , a 1 litre solution containing 10mmol of Cr _ 2 O _ 7 ^ 2 - and 100mmol of Cr ^ 3 + shows a pH of 3 . 0 .
  4. DAt 298K , a 1 litre solution containing 10mmol of Cr _ 2 O _ 7 ^ 2 - and 100mmol of Cr ^ 3 + shows a pH of 3 . 0 .

Correct answer

B. At 298K , a 1 litre solution containing 10mmol of Cr _ 2 O _ 7 ^ 2 - and 100mmol of Cr ^ 3 + shows a pH of 3 . 0 .

Step-by-step solution

\begin{aligned} & \underset{-5^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{ s})} \rightarrow \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{ s})} ; \Delta \mathrm{S}_1=\int_{268 \mathrm{ K}^{273 \mathrm{ K}} \frac{\mathrm{C}_{\mathrm{p}}, \mathrm{ m} \mathrm{dT}}{\mathrm{ T}} \\ & \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{ s})} \rightleftharpoons \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})} ; \Delta \mathrm{S}_2=\frac{\Delta \mathrm{H}_{\mathrm{m}} \text{, fus }}{273} \\ & \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{I})} \rightarrow \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{I})} ; \Delta \mathrm{S}_3=\int_{273}^{373} \frac{\mathrm{C}_{\mathrm{p}}, \mathrm{ m} \mathrm{dt}}{\mathrm{ T}} \\ & \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})} \rightleftharpoons \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{ g})} ; \Delta \mathrm{S}_4=\frac{\Delta \mathrm{H}_{\mathrm{m}} \text{, vap }}{373} \\ & \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{ g})} \rightarrow \underset{110^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{ g})} ; \Delta \mathrm{S}_5=\int_{373}^{383} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{ m}} \mathrm{dT}}{\mathrm{ T}} \\ & \Delta \mathrm{ S}_{\text{total }}=\Delta \mathrm{S}_1+\Delta \mathrm{S}_2+\Delta \mathrm{S}_3+\Delta \mathrm{S}_4+\Delta \mathrm{S}_5 \end{aligned}
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