One mole of ice is converted into water at 273 K. The entropies of $\;{{\rm{H}}_2}{\rm{O}}$ (S) and $\;{{\rm{H}}_2}{\rm{O}}$ (l)are 38.20 and 60.01 J ${\rm{mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$ respectively. The enthalpy change for the conversion is:

$5954{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$59.54{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$595.4{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$320.6{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Application +4 | -1

One mole of ice is converted into water at 273 K. The entropies of

\;{\mathrm{H}_2}\mathrm{O}

(S) and

\;{\mathrm{H}_2}\mathrm{O}

(l)are 38.20 and 60.01 J

\mathrm{mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}

respectively. The enthalpy change for the conversion is:

View answer and explanation

Correct option: B

5954\mathrm{Jmo}{\mathrm{l}^{ - 1}}

Option B (

5954\mathrm{Jmo}{\mathrm{l}^{ - 1}}

) is correct because it is the enthalpy change calculated using the formula ∆H=T∆S at equilibrium.

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}