$1 \mathrm{~mole} \mathrm{~H_2} \mathrm{O}(l)$ at $-10^{\circ} \mathrm{C}$ is converted to 1 mole $\mathrm{H}_2 \mathrm{O}(s)$ at $-10^{\circ} \mathrm{C}$. Entropy change is the sum of all steps but not of

$\Delta S_1=C_v \ln \frac{273}{263}\left(C_v=\right.$ heat capacity of water)

$\Delta S_3=-\frac{\Delta H_{\text {fusion }}}{273}$

$\Delta S_4=C_{\dot{v}}^{\prime} \ln \frac{263}{273}\left(C_{{v}}^{\prime}=\right.$ heat capacity of ice $)$

JEE CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1

1 \mathrm{~mole} \mathrm{~H_2} \mathrm{O}(l)

-10^{\circ} \mathrm{C}

is converted to 1 mole

\mathrm{H}_2 \mathrm{O}(s)

-10^{\circ} \mathrm{C}

. Entropy change is the sum of all steps but not of

View answer and explanation

Correct option: B

\Delta S_2=\frac{\Delta H_{\text {fusion }}}{273}\left(\Delta H_f=\right.

heat of fusion of ice)

The correct answer is

\Delta S_2=\frac{\Delta H_{\text {fusion }}}{273}(\Delta H_f=\text { heat of fusion of ice})

because the process is occurring at -10°C (263K), not 0°C (273K).

Practice All Chapter MCQs →

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

0.18 \mathrm{~g}

4 \mathrm{~K}.

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

{CsI}

{CsCl}

{CsCl}

{Cs} I

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

\Delta H

\Delta u

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}