JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Analyzing +4 | -1

The hydrogen bond between

\mathrm{F}^{-}

and

\mathrm{CHCOOH}

is very strong and its strength may be analysed by setting up a Born-Haber cycle with the following data. Lattice energy of

\mathrm{KF.CH}_3 \mathrm{COOH}=734;

enthalpy of vaporization of

\mathrm{CH}_3 \mathrm{COOH}=20;

enthalpy of solution of

\mathrm{KF}= 35;

solvation energy of

\mathrm{K}^{+}(\mathrm{g}) =-325;

solvation energy of

\mathrm{F}^{-}=-389;

enthalpy of formation of

\mathrm{KF} \cdot \mathrm{CH}_3 \mathrm{COOH}(\mathrm{s})

from

\mathrm{KF}(\mathrm{s})

and

\mathrm{CH}_3 \mathrm{COOH}(l) =-25.

Find the energy of the hydrogen bond between

\mathrm{F}^{-}

and

\mathrm{CH}_3 \mathrm{COOH}

in the gas phase (in

\mathrm{kJ} / \mathrm{mol}

View answer and explanation

Correct option: A

-60 \mathrm{~kJ} / \mathrm{mol}

Option A (-60 kJ/mol) is correct because it is the result of calculating the energy change using the given enthalpy values in the Born-Haber cycle.

More Questions from THERMODYNAMICS

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}