JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Application +4 | -1

Calculate the lattice energy forthe reaction,

\mathrm{Li}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g}) \longrightarrow \mathrm{LiCl}(\mathrm{s})

From the following data:

\Delta_{\mathrm{s}} \mathrm{H}^{\circ}(\mathrm{Li})=160.67 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \quad \frac{1}{2} \Delta_{\mathrm{a}} \mathrm{H}^{\circ}\left(\mathrm{Cl}_2\right)=122.17 \mathrm{~kJ} \mathrm{~mol}^{-1}

\Delta_{\mathrm{i}} \mathrm{H}^{\circ}(\mathrm{Li})=520.07 \mathrm{~kJ} \mathrm{~mol}^{-1}

;

\quad

\Delta_{\mathrm{eg}} \mathrm{H}^{\circ}(\mathrm{Cl})=-365.26 \mathrm{~kJ} \mathrm{~mol}^{-1}

\text { and } \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{LiCl})=-401.66 \mathrm{~kJ} \mathrm{~mol}^{-1}

View answer and explanation

Correct option: A

-839.31 \mathrm{~kJ~mol}^{-1}

Option A (

-839.31 \mathrm{~kJ~mol}^{-1}

) is correct because it is the calculated lattice energy using the given enthalpy values and applying Hess's law.

More Questions from THERMODYNAMICS

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\ \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}