JEE CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1

Calculate the enthalpy of formation of

{I}_2 {O}_5(\mathrm{s})

from the following data.

(i)

{I}_2 {O}_5(\mathrm{s})+{H}_2{O}({l}) \rightarrow 2 {HIO}_3\text{(aq)}; \Delta H=+4.0 \mathrm{~kJ}

(ii)

{KI} \text{(aq)} +3 {HClO}\text{(aq)} \rightarrow {HIO}_3 \text{(aq)} +2 {HCl}\text{(aq)}+{KCl}(\mathrm{aq}); \Delta H=-322.0 \mathrm{~kJ}

(iii)

{NaOH}(\mathrm{aq})+{HClO}(\mathrm{aq}) \rightarrow {NaOCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-44.0 \mathrm{~kJ}

(iv)

{NaOH}(\mathrm{aq})+{HCl}(\mathrm{aq}) \rightarrow {NaCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-57.0 \mathrm{~kJ}

(v)

2 {NaOH}(\mathrm{aq})+{Cl}_2(\mathrm{g}) \rightarrow{NaOCl}(\mathrm{aq})+ {NaCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-100.0 \mathrm{~kJ}

(vi)

2 {KI}(\mathrm{aq})+{Cl}_2(\mathrm{g}) \rightarrow 2 {KCl}(\mathrm{aq})+{I}_2(\mathrm{s}); \Delta H=-224.0 \mathrm{~kJ}

(vii)

{H}_2(\mathrm{g})+1 / 2 \, {O}_2(\mathrm{g}) \rightarrow {H}_2 {O}({l}); \Delta H=-285.0 \mathrm{~kJ}

(viii)

1 / 2 \,{H}_2(\mathrm{g})+1 / 2 \, {Cl}_2(\mathrm{g}) \rightarrow {HCl}(\mathrm{g}); \Delta H= -92.0 \mathrm{~kJ}

(ix)

{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow {HCl}(\mathrm{aq}); \Delta H=-75.0 \mathrm{~kJ}

View answer and explanation

Correct option: A

-169.0 \mathrm{~kJ}

Option A (-169.0 kJ) is correct because the enthalpy of formation of I2O5(s) is calculated by combining the given reactions using Hess's Law.

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}